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CAT Exam: Everything of Number System for CAT Exam CAT MBA 

CAT Exam: Everything of Number System for CAT Exam

This section is about the most fundamental topics in mathematics: Number System. What is maths all about? Numbers, off course. This unit of the website is all about numbers. Starting from absolute basics, this section provides you all the material you need to study the topic: concepts, articles, blogs and tests. Make sure you go through each and every piece in the order provided.

How to use this section? We have simple piece of advice for you: Start with articles (provided in the index) and read them one at a time. Once you are through with an article, solve the topic tests corresponding to that topic. Once you are done with all the articles and the tests, head over to the level-wise tests. Geez, do all this work and you will be a master of the topic.

 

1. Types of Numbers: Part-1

In this first article of our topic ‘Numbers’, we deal with ‘Numbers’ themselves. What are numbers? Where do we see and meet them? Are they of different types? Can we learn these distinct types of numbers?

Well, numbers are all around us, floating all over the place. They demand our attention and careful evaluation. A number here or there can make a huge difference. In fact a misplaced zero or decimal is the difference between one being poor or rich.

Such being their importance, it is critical we begin with the most basic concepts in Mathematics: types of numbers. But before we do that, let’s define what is a number line:

Number line:A number line is line where all the numbers are allocated their positions. The origin of the number line starts from zero and it continues to infinity, on either side.

Few Basic Types of Numbers

Positive Numbers : Numbers which are to the right of zero are said to be positive numbers.

For example 1, 3 ,1.2 , 2.6 , 7 etc.

Negative Numbers : Numbers which are to the left of zero are said to be negative numbers.

For example -1, -5 , -7.2 , -2.5 , -9 etc.

Counting Numbers:Counting numbers are those numbers which are well managed on the number line with the difference of 1. The smallest counting number on the number line is 1.

Natural Numbers: Numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11…and so on are called natural numbers. They are also called positive integers. Also we can say that the other name for counting numbers is natural numbers.The lowest natural number is 1.

Whole Numbers:Whole numbers are numbers without fractions, they take integral values. Whole numbers are those numbers which start from 0 or we can say if 0 is included in set of counting numbers (natural numbers) then we get set of whole numbers.Remember whole numbers would always take non- negative integral values

Types of Numbers Part-2

Integers: It is combination of both positive and negative numbers lying on the number line including zero. Remember zero is an integer. The integers are the natural numbers, their negatives, and the number zero like …,-3,-2,-1,0,1,2,3,… and they go on forever in both directions of the number line.

Rational numbers: Rational numbers are those, which can be written in the form of a ratio of x/ y, where the denominator y is non-zero.

Irrational numbers: Numbers, which are non-terminating and non- recurring (non-repeating) decimals are said to be irrational or we can say irrational numbers are those, which are not rational, that is those numbers that cannot be written in the form of a ratio x/y. For example  = 1.414213….. , √3 =1.732050…..

Fractions: Fractions are those numbers, which are in the form of p/q where q is non-zero. For example 4/5, 5/6, 6/7 etc.

Real Numbers: Any number which can be plotted on the number line is a real number .The number can be positive or negative in nature. For example it may be like as 3, 4, 5, 6, -6, -5, -4, -3, -2…..

Prime Numbers and Composite numbers
In the list of natural numbers, there are two types of natural numbers.
One type is called the PRIME numbers and the other is called the COMPOSITE numbers.
(1 is not included in this list i.e.1 is neither prime nor composite)

Prime numbers: The numbers, which have exactly two factors, 1 and the number itself, are called prime numbers.

For example 2, 3, 5, 7, 11, 13, 31, 43,………………or we can say that the numbers, which are not divisible, by any number except one and itself, are called prime numbers. There are 25 prime numbers between 1 and 100.

 
Composite number: The composite numbers are those numbers which can be written as the product of prime numbers in a unique way.
Or we can say that the number which is the product of two or more than two distinct or same prime numbers is said to be a composite number.
Example 4, 6, 8,…
Note that composite number always has more than two factors and prime number cannot be a composite number.

Rounding numbers:When the number is approximated to the nearest possible integral value to maintain the accuracy of the data then the number is called rounded.

Properties Of Numbers

Properties of Numbers
The following is a handy list of tips that you can remember about numbers (think about each one of these):

  • The number line goes on till infinity in both directions, which is indicated by the arrows.
  • The numbers on the number line are indicated by their respective signs, which shows that the line includes both positive and negative numbers.
  • The integer zero is neutral and neither positive or negative.
  • The number in the halfway of 1 and 2 is 1.5 and in the half way of -1 and -2 is -1.5.
  • Between two rational numbers we can always find a rational number.
  • Numbers, which are terminating and non-recurring are rational numbers. Similarly numbers, which are non-terminating and recurring are rational numbers.
  • Set of natural numbers is contained in set of integers which is contained in set ofrational numbers which is further contained in set of real numbers, which is further contained in set of complex numbers.
  • Addition as well as product of two real numbers is a real number.
  • Two real numbers can be added or multiplied in either order i.e. Addition and multiplication of real numbers is commutative.
  • Two real numbers cannot be subtracted or divided in either order i.e. Subtraction and Division of real numbers is not commutative.
  • In set of real numbers we don’t define square root of negative numbers.

Rather we define set of complex numbers for this purpose. Any complex number z ={ a + ib , where a and b belongs to set of real numbers and i=  }

Properties of Zero

  • a x 0 =0 always for any real number a.
  • a +0 = a always for any real number a.
  • a – 0 = a always and 0 – a = -a for any real number a.
  • 0/a =0 when “a” is a non-zero real number.
  • a/0 is not defined i.e. we don’t define division by zero.
  • a0=1for any non-zero real number a.
  • 00 is not defined.

Prime numbers

Prime numbers 
Numbers, which don’t have any factor other than 1 or itself, are said to be prime numbers.
For example 2, 3, 5, 7, 19, 29, 31………………
Or we can say that the numbers which are not divisible by any other number except 1 and itself, are called prime numbers.
There are 25 prime numbers between 1 and 100. In general we say there are infinite prime numbers.
 
Prime Numbers and Composite numbers 
In the list of natural numbers there are two types of natural numbers
One type is called PRIME numbers and other one is called COMPOSITE numbers.

Properties of Prime Numbers

  • If p is any prime number and p divides a product of two integers say m and n i.e. p|mn (read as p divides mn), then p divides ‘m’ or p divides ‘n’ or both.
  • Number of prime numbers is uncountable. i.e. there are infinite prime numbers.
  • Every Prime number has exactly two factors or divisors.
    For example: 13 is prime number as the divisors of 13 are 1 and 13.
  • There is only one even prime i.e. 2 and all other primes are odd.
  • G.C.D of prime numbers is always 1. The numbers whose G.C.D is 1 are said to be co-prime thus we say two primes are always co-prime.
  • Two numbers are co-prime if their HCF is 1 .For example G.C.D. (21, 25)=1 and hence 21 and 25
    are co- prime.
  • The Twin Primes are pair of primes of the form (p, p+2).
  • It is conjectured that there are an infinite number of twin primes.
  • The term odd prime refers to any prime number greater than 2. For example: 3, 5,7, …………..
  • First Prime Number is 2.
  • 1 is neither prime nor composite number.
  • All prime numbers greater than 3 can be expressed in the form of 6n-1 or 6n+1 i.e.
    all prime numbers,which are greater than 3 leave remainder 1 or 5 when divided by 6.

List of prime numbers with number ranges:

list-of-prime-numbers

Prime Numbers and Composite Numbers

How to check whether a number is prime or not?

To check whether a given number N is prime or not, first find the square root of that number N and then approximate that to immediately lower integer (say n) and write down all the prime numbers less than that integer (n). Then check the divisibility of the given number N by all the prime numbers we have written in previous step, if it is not divisible by any of the prime numbers then given number N is prime.

Let us write algorithm for the same
Step 1: Find square root of N, call it as K (Just find approximate values)
Step 2: Write down all the prime numbers less than K.
Step 3: Check divisibility of N with these prime numbers, which we have got in Step 2.
Step 4: If N is not divisible by any of the prime numbers then N is prime.

Example:
Let us check whether 211 is prime or not?
Solution:
Step 1: We find square root of 211 i.e. K=√211  = 14.52
Step 2: We write all primes less than 14.52 i.e. 2, 3, 5, 7, 11 and 13.
Step 3:Since 211 is not divisible by any of these prime numbers, hence 211 is a prime number.
 

Example:
Let us check whether 313 is prime or not?
Solution:
Step 1: We find square root of 313 i.e. K=√311  = 17.69
Step 2: We write all primes less than 17.69 i.e. 2, 3, 5, 7, 11, 13 and 17
Step 3:Since 313 is not divisible by any of these prime numbers, hence 313 is a prime number.

Composite numbers
Number which is the product of two or more than two distinct or same prime numbers is said to be composite number Or we say that if a number has more than two factors then it is said to be a composite number.
For example 4, 6, 8, 15………..are all composite numbers
We can write 4 = 2 × 2,
6 = 2 × 3.
Some points to remember

  • 1 is neither prime nor composite.
  • 2 is the only prime number which is even. Rest all prime numbers are odd.

Properties of Addition of Natural Numbers

Properties of Addition of Natural Numbers:

Sum of 1st ‘n’ Natural numbers =  1 + 2 + 3 + …. + n = n(n+1)2n(n+1)2

Sum of squares of 1st ‘n’ natural number  =  12 + 22 + 32 + ….. + n2  = n(n+1)(2n+1)6n(n+1)(2n+1)6

Sum of cubes of 1st ‘n’ Natural number  =  13 + 23 + 33 + ….. + n3  = (n(n+1)2)2(n(n+1)2)2

Sum of 1st ‘n’ Odd natural number  =  1+ 3 + 5 + ….. + (2n – 1)  = n2

Sum of 1st ‘n’ even natural numbers  = 2 + 4 + 6+ …….2n = n (n+1)

Sum of squares of 1st ‘n’ Odd natural number = 12 + 32 + 52 + ….. + (2n – 1) 2n(2n1)(2n+1)3n(2n−1)(2n+1)3

Sum of squares of 1st ‘n’ Even natural number =  22 + 42 + 62 + ….. + (2n)22n(n+1)(2n+1)32n(n+1)(2n+1)3

Sum of cubes of 1st  ‘n’ Odd natural number =  13 + 33 + 53 + ….. + (2n -1)3 = n2 (2n2 -1)

Sum of cubes of 1st  ‘n’ Even natural number  =  23 + 43 + 63 + ….. + (2n)3 = 2 [n (n + 1)]2

Some examples for the above properties:

Find the sum of:

(A) First 30 natural numbers

(B) Squares of first 30 natural numbers

(C) Cubes of first 30 natural numbers

(D) First 30 odd natural numbers

(E) Squares of first odd 30 natural numbers

(F) Squares of first even 30 natural numbers

(G) Cubes of first odd 30 natural numbers

(H) Cubes of first even 30 natural numbers

Number System

Tricks with Numbers

Number of times a particular digit (except 0) appears in a certain range:

The concept, the title of the post talks about, is best understood with the help of examples. This is our way of practical learning and using problem types to understand concepts.
Example 1: Calculate how many times digit 3 appears in first hundred natural numbers?
Solution: Well, there are two series of three’s here:
1. Those set of numbers, which have 3 at unit place. 3, 13,……..,83 ,93, Here we have Ten 3s at the unit’s place
2. Those set of numbers, which have 3 at ten`s place. 30,31…39.Here we have ten 3s at the ten’s place. So a total of 20 threes make an appearance here.

Example 2: Calculate how many times digit 3 appears in natural numbers, which are less than one thousand?
Solution: In previous example we calculated number of 3 s present till 100 are 20.
Now we look at three digit numbers and calculate how many times digit 3 appears in numbers between 100 and 1000 Well, there are three types of series here:
1. Those set of numbers, which have 3 at unit place.
103, 113, 123, ……..,193
203, 213 ,………………., 293
………………………………….
903,913,………………….993

Here we have 90 numbers having 3 at the unit’s place.
2. Those set of numbers, which have 3 at ten`s place.
130, 131, 132, ……..,139
230, 231, 232, ……..,239
………………………………….
930,931,………………….939
Here we have 90 numbers having 3 at the ten’s place

3. Those set of numbers, which have 3 at hundred`s place.
301, 301, 302, ……..,399
Here we have hundred 3s at the hundred’s place.
Thus total number of times digit 3 appears in natural numbers, which are less than one thousand are 20+90+90+100=300

Key Observations:
From 1-9, the digit 3 is used 1 time
From 1 – 99, the digit 3 is used 20 times.
From 1 – 999, the digit 3 is used 300 times
From 1 – 9999, the digit 3 is used 4000 times and so on

Note: Above is true for any digit except 0.

Unit Digit of Perfect Square

Properties regarding Unit Digit of Perfect Square

Squares: There is a definite relationship between the unit digits when square of a number is considered, we will see that one by one.

– If unit digit of a perfect square is 1 then ten’s digit has to be Even
E.g. 81, 121, 441, 961 all are Perfect Square having unit digit 1 and tens digit is even.

– If unit digit of a number is 2 then it can’t be a perfect square.

– If unit digit of a number is 3 then it can’t be a perfect square.

– If unit digit of a perfect square is 4 then ten’s digit has to be Even.
E.g. 64, 144, 484 all are Perfect Square having unit digit 4 and tens digit is even.

– If unit digit of a perfect square is 5 then ten’s digit has to be 2
E.g. 25, 225, 625,1225 all are Perfect Square having unit digit 5 and tens digit is 2.

– If unit digit of a perfect square is 6 then ten’s digit has to be Odd. e.g. 256, 576, 676, 1296 etc.

– If unit digit of a number is 7 & 8 then it can’t be a perfect square

– If unit digit of a perfect square is 9 then ten’s digit has to be Even.
E.g. 49, 169, 529 all are Perfect Square having unit digit 9 and tens digit is even.

– If unit digit of a perfect square is 0 then ten’s digit has to be 0.
E.g. 100,400,900 all are Perfect Square having unit digit 0 and tens digit is 0.

– If a number ends with 2, 3, 7 or 8 then it can’t be perfect square.

-Square of any natural number has last two digits same as that of last two digits of squares of first twenty–five natural numbers. For example if we calculate square of
88 we get 7744. You can notice that last two digits are 44, which are same as last two digits of the square of 12 i.e. 144.

Magic of Numbers

Magical 1
3 x 37 = 111
33 x 3367 = 111,111
333 x 333667 = 111,111,111
3333 x 33336667= 111,111,111,111
33333 x 3333366667= 111,111,111,111,111
333333 x 333333666667 = 111,111,111,111,111,111
Consider the L.H.S of the above numbers. The number of 3s in both the numbers are same and 6 is one less than the number of 3 and there is only one 7. Then in the final product, N would appear three times the number of threes in the first number on the LHS.

6 when multiplied with 7 give 4 and 2
6 x 7 = 42
66 x67 = 4422
666 x 667 = 444222
6666 x 6667 = 44442222
For multiplication such as the above:
The number of digits in both the numbers is same, the 2nd number has one 6 less than the number of  6’s the 1st number has .
The number of 4’s and 2’s is equal to the number of 6’s in the 1st number

Cyclic numbers
142857 is called as the cyclic number, since its digits are rotated around when multiplied by any number from 1 to 6
142857 x 1 = 142857
142857 x 5= 714285
142857 x 4= 571428
142857 x 6 = 857142
142857 x 2= 285714
142867 x 3 = 428571

The same method helps us calculating fractions involving 7 in denominator

Magic of numbers pic

We know 1/7 = 0.142857
Thus 2/7 will be a value close to 0.28 and directly we will write exact answer as numbers repeat in cyclic order as shown in diagram. Hence exact answer will be 0.285714

Similarly 3/7 will be a value close to 0.42 and directly we will write exact answer as numbers repeat in cyclic orderas shown in diagram. Hence exact answer will be 0.428571

Similarly 4/7 will be a value close to 0.56 and directly we will write exact answer as numbers repeat in cyclic orderas shown in diagram. Hence exact answer will be 0.571428

Similarly 5/7 will be a value close to 0.7 and directly we will write exact answer as numbers repeat in cyclic orderas shown in diagram. Hence exact answer will be 0.714285

Similarly 6/7 will be a value close to 0.84 and directly we will write exact answer as numbers repeat in cyclic orderas shown in diagram. Hence exact answer will be 0.857142

Properties of Consecutive Integers

Properties of Consecutive Integers

We have studied about integers, integers are combination of both positive and negative numbers lying on the number line including zero. There are few important properties regarding integers that you should know:

  • Zero is also an Integer.
  • Integers which follow one another are called consecutive integers.
  • If we multiply a number by an integer and resultant value remains same then that integer is 1.

Consecutive Integers: The integers which follow one another are called consecutive integers.
For example 3, 4, 5, 6 are consecutive integers. An individual random number can never be
a consecutive integer.

Note: If we have consecutive integers in a set M from a to b i.e. M ={a,a+1,a+2,……………………..,b}.
Then number of elements in the set M is b–a+1.

Example: Set M consists of natural numbers from 75 to 199. Find the numbers of elements in set M ?

Solution: Question would have been very easy if we are asked number of natural numbers from 1 to 199. The answer to this is obviously 199. So we can also conclude it same way. From 75 to 199 we are not counting first 74 natural numbers. Hence the answer should be 199 -74= 125
Some more points about Consecutive Numbers
TOOLTIP 1
If there is odd number of digits in the set of consecutive numbers like set of three consecutive numbers (4,5,6,) or 5 digits are there, say (3,4,5,6,7), then in this case the sum of all integers is always divisible by the number of digits present in the set.
For example 2+3+4 = 9 is divisible by number of digits i.e. 3.
For example 1+2+3+4+5 = 15 is divisible by 5.

TOOLTIP 2
On the other hand ,If there are even number of digits in  set of consecutive numbers like set of four consecutive numbers (4,5,6,7) or 6 digits are there, say (3,4,5,6,7,8), then in this case the sum of all integers is never divisible by the number of digits present in the set .
For example 2+3+4+5 = 14 is not divisible by number of digits i.e. 4
For example 1+2+3+4+5+6 = 21 is not divisible by 6.

BODMAS

Order of Operation
While solving questions involving mathematical operations involving addition, subtraction, multiplication and division we follow a particular order while performing a calculation. To perform calculations we follow BODMAS.

BODMAS stands for
B = Bracket
O = of
D = Division
M= Multiplication
A = Addition
S= Subtraction
Let us see examples based on this
Example 1:

What is the value of N= {(4 +5)of4}/3 + 2 ?
Solution:
We will solve bracket first
{(9) x 4}/3 + 2
= 12 + 2
= 14

Example 2:
What is the value of N= {(12÷4× 2 +5) of4}/21 ?Solution:We will solve terms inside bracket first={(12÷4× 2 +5) of4}/21={(3 ×2 +5) of4}/21 (We divided 12 and 4 first)={(6 +5) of4}/21 (We multiplied 3 and 2)={(11) of4}/21 (We added 6 and 5)={44}/21 (We wrote 11 of 4)= 221(We divided 44 and 2)=21(We subtracted 1 from 22)What is the value of N= {(12÷4× 2 +5) of4}/2−1 ?Solution:We will solve terms inside bracket first={(12÷4× 2 +5) of4}/2−1={(3 ×2 +5) of4}/2−1 (We divided 12 and 4 first)={(6 +5) of4}/2−1 (We multiplied 3 and 2)={(11) of4}/2−1 (We added 6 and 5)={44}/2−1 (We wrote 11 of 4)= 22−1(We divided 44 and 2)=21(We subtracted 1 from 22)

Example 3:
What is the value of N = {(18 ÷ 3 × 2 + 1) of 8}/2 + 5?
Solution:
We will solve the terms inside the bracket first
{(18 ÷ 3 × 2 + 1) of 8}/2 + 5
= {6 × 2 + 1) of 8}/2 + 5   (We divided 18 and 3)
= {(12 + 1) of 8}/2 + 5      (We multiplied 6 and 2)
= {13 of 8}/ 2 + 5              (We added 12 and 1)
= (104)/2 + 5                      (We wrote 13 of 8)
= 52 + 5                              (We divide 104 and 2)
= 57                                    (We added 52 and 5)

Example4:What is the value of N= {(9 + 8÷ 2× 3) of8}/2 + 3 ?Solution:We will solve terms inside bracket first{(9 +8÷ 2 ×3) of8}/2 + 3={(9 + 4× 3) of8}/2 + 3(We divided 8 and 2 first)= {(9 + 12) of8}/2 + 3(We multiplied 3 and 4)= {(21) of8}/2 + 3(We added 9 and 12)= {168}/2 + 3(We wrote 21 of 8)= 84+ 3(We divided 168 and 2)= 87(We added 3 in 84 )

Fractions

Basic Applications of Fractions
1. Fraction helps us determine the part of any number
¾ part of 56 = ¾ x 56 = 42
4/5 part of 90 = 4/5 x 90 =72

2. You can be asked to represent a number in the form of fraction.
For example, you can be asked to represent 15 as a fraction of 450.
This can be done as follows:
15/450 = 1/30
We have solved the above example and it can be easily seen that 15 is our numerator and 450 is our denominator

3. Always remember that the major quantity from which we have to extract something is the denominator.
For example, when we say 4/5, we are essentially extracting four parts out of five.

4. Extending the above concept, the quantity that is extracted is our numerator
For example: 15/450 = 1/30
15 is the numerator because we have extracted 15 from 450 and the denominator is 450 because 15 is extracted from 450 so we can say that 1/30th part of 450 is 15

 5. Converting percentages into equivalent fractions makes our calculation easy
For example:
If we remember 37.5% = 3/8 and one is asked to find 37.5% of 24 then we can write
37.5% of 24= 3/8 x 24 = 9
 Remember few equivalent fractions for given percentages:

50% = 1/2
25% = 1/4
12.5% = 1/8
6.25% = 1/16
33.33% = 1/3
66.66% = 2/3
75% = 3/4
20% = 1/5
40% = 2/5
60% = 3/5
80% = 4/5
16.66% = 1/6
83.33% = 5/6
14.28% = 1/7
28.57% = 2/7
42.85% = 3/7
57.14%= 4/7
71.42% = 5/7
85.71%=6/7
37.5% = 3/8
62.5% = 5/8
87.5% = 7/8
11.11% = 1/9
22.22% = 2/9
44.44% = 4/9
55.55% = 5/9
77.77% = 7/9
88.88% = 8/9
9.09% = 1/11
18.18% = 2/11
27.27% = 3/11
36.36% = 4/11
45.45% = 5/11
54.54% = 6/11
63.63% = 7/11
72.72% = 8/11

Decimals

DECIMALS
Are all numbers integers? Well, the obvious answer to that question is a no. All numbers are not integers. Consider the case of 0.333333. What is this number? An integer? Well, it is a decimal. But what are decimals? Decimals are nothing else but the values lying between two integers on the number line.
Relating decimals to fraction: When we solve a fraction of the form p/q where q is non-zero, it is not necessarily it would return an integral value.  When we are left with a remainder, we ultimately convert it into decimal form. Some examples of Decimals are 4.5, 9.6, 6.78, and 99.98, these all are decimal numbers.
Tooltip 1: Forming Decimals from Fractions

How we can write 4/5 in decimal form?
Since the number 4 is smaller than the 5, so the decimal value will be less than one.
Multiple and divide both numbers by 10. We have:
40/ (10 x 5)
Which effectively is (dividing 40 by 5 first) 8/10

Thus, the final result is 0.8  So 4/5 = 0.8 and 0.8 is the decimal form of 4/5.

Tooltip 2: Adding Decimals
While adding decimals, you should always write the decimals in a vertical column with the decimal points aligned vertically.
Add all these 0.567 +78+8.9+5.06+56
= 78.000
+ 56.00
+   5.06
+0.567
 139.627

Tooltip 3: Subtracting Decimal

In addition we can write the numbers in any order. But while subtracting, we should preferably write the numbers in descending order and the vertical column with decimal points should be aligned to the same decimal points.

Let’s take an example: we have to subtract 0.567 from 5.06. If we write it as:

5.06
-0.567
 4.507
This result is wrong because in case of subtraction we need equal digits in both the quantities, so these blank spaces are filled with 0
So this can be done like as
5.060
-0.567
 4.493 
This is the right approach for the question

Tooltip 4: Multiplying Decimals
As a first step, multiply the given integers in the normal form keeping the decimals aside.  The number of decimal places in the product is then equal to the total of the decimal places in the two decimals. It is as simple as that.
Consider the following example:
5.060
x0.567
2.869020
First multiply 5060 with 567 and get the result as 2869020 and then move the decimal point to 6 places from the left i.e. between 2 and 8. Effectively, we move it to six decimals places, our sum total of the decimal places in the two numbers.

Exponents

Exponents
Exponents are a method used to express products of the same number.
How would you write: 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2?
Hard to repeat the above number in calculations, isn’t it?
Well, exponents provide us the solution to the above problem. The above number simply becomes: 29
The expression AN is called a power and it stands for A x A x A x A ……. x A , where there are N factors of A. A is called the base, and N is called the exponent. By definition, A0= 1
In order to solve any question involving exponents you must remember the 5 basic rules of exponents. These 5 rules are universally applicable and help you solve any exponent problem with ease. These rules are:

rules for exponents-1

Also, keep in mind:  a-m = 1/am
Let’s use some of the rules above and try to solve some problems using the tips given above.
Example 1: Which out of the following is equal to the expression 164 x 43?

  1. 222
  2. 220
  3. 48
  4. 412
  5. 410

Solution:
Let us reduce the expression to a common base. You should notice is that the number 16 can be expressed as 42. Thus, 164 x 43  becomes ( 42)4
( 42)4 is equal to 42×4, that is 48.
Thus, the original product becomes 48 x 43= 48+3 = 411= (22)11 = 222
Thus, the correct answer is option A.
Example 2: If a not equal to zero, a(a3)3/a2 is equal to:

  1. a6
  2. a7
  3. a8
  4. a9
  5. a10

Solution:
Let us first simplify the numerator.
a(a3)3= a (a3×3) =a x a9 = a10
Now the equation becomes a10/a2 = a8
Thus, the correct answer is option C.

Divisibility Rules

A divisibility rule is a short cut for discovering whether a given number is divisible by a fixed divisor without performing the division, usually by examining its digits. Most of the divisibility rules are derived from a concept of remainders.
The two basic rules for finding out whether the number is divisible or not by any particular number:
1.If the divisor is prime number, then we can directly check for it.
2.If the divisor is a composite number then do factorization of divisor into factors, which are co-prime and check the divisibility for each prime factor individually

Divisibility Rules: 2 to 7

Number System

Divisibility Rules: 8 to 12

Divisibility Rules: Special Numbers

Divisibility Rules explained with examples

Divisibility Rules explained with examples:
In this article, we use a few example to understand the concept of divisibility.  Go through the following examples:

Example 1:
Check divisibility of 124 by 2, 3, 4 and 5.
Solution:
124 is divisible by 2 since the last digit is 4.
124 is not divisible by 3 since the sum of the digits is 7 (1+2+4 = 7), and 7 is not divisible by 3.
124 is divisible by 4 since 24 is divisible by 4.
124 is not divisible by 5 since the last digit is 4 it is neither 0 nor 5.
124 is divisible by 2 and 4 not by 3 and 5.

Example 2:
Check divisibility of 225 by 2, 3, 4, 5, 6, 9 and 10.
Solution: 
225 is not divisible by 2 since the last digit is not 0, 2, 4, 6 or 8.
225 is divisible by 3 since the sum of the digits is 9 and 9 is divisible by 3.
225 is not divisible by 4 since 25 is not divisible by 4.
225 is divisible by 5 since the last digit is 5.
225 is not divisible by 6 since it is not divisible by 2.
225 is divisible by 9 since the sum of the digits is 9.
225 is not divisible by 10 since the last digit is not 0.
225 is divisible by 3, 5, 9.

Example 3: 
Check divisibility of 400 by 2, 3, 4, 5, 6, 8, 9 and 10.
Solution:
400 is divisible by 2 since the last digit is 0.
400 is not divisible by 3 since the sum of the digits is 4 and 4 is not divisible by 3.
400 is divisible by 4 since 00 is divisible by 4.
400 is divisible by 5 since the last digit is 0.
400 is not divisible by 6 since it is not divisible by 3.
400 is divisible by 8 since the last 3 digits are 400 and 400 is divisible by 8.
400 is not divisible by 9 since the sum of the digits is 4 and 4 is not divisible by 9.
400 is divisible by 10 since the last digit is 0.
400 is divisible by 2,4,5,8 and 10.

Example 4:
Check divisibility of 750 by 2, 3, 4, 5, 6, 8, 9 and 10.
Solution:
750 is divisible by 2 since the last digit is 0.
750 is divisible by 3 since the sum of the digits is 12 and 1 is not divisible by 3.
750 is not divisible by 4 since 50 is not divisible by 4.
750 is divisible by 5 since the last digit is 0.
750 is divisible by 6 since it is divisible by 2 and 3.
750 is not divisible by 8 since the last 3 digits are 750 and 750 is not divisible by 8.
750 is not divisible by 9 since the sum of the digits is 12 and 12 is not divisible by 9.
750 is divisible by 10 since the last digit is 0.
750 is divisible by 2,3,5,6 and 10.

Example 5: 
Check divisibility of 1200 by 2, 3, 4, 5, 6, 8, 9 and 10.
Solution:
1200 is divisible by 2 since the last digit is 0.
1200 is divisible by 3 since the sum of the digits is 3.
1200 is divisible by 4 since 00 is divisible by 4.
1200 is divisible by 5 since the last digit is 0.
1200 is divisible by 6 since it is divisible by both 2 and 3.
1200 is divisible by 8 since the last 3 digits are 200.
1200 is not divisible by 9 since the sum of the digits is 3 and 3 is not divisible by 9.
1200 is divisible by 10 since the last digit is 0.
1200 is divisible by 2,3,4,5,6,8 and 10.

Tricks for Divisibility

Some basics tips and tricks that you can use for divisibility
All whole numbers are divisible by 1.
A number is divisible by 2 if it’s even.
A non-zero number is divisible by 5 if it ends in 0 or 5.
In order to check the divisibility of a number by a composite number, divide the composite divisor into prime factors, which are co-prime and then check for its divisibility with each. For example, to check the divisibility of a number with 12, break down 12 into 3 and 4.

Tricks for Divisibility
an – bn is always divisible by a-b
8555 is divisible by 8-5= 3

Remember it by:
a3b3 is divisible by a-b
a2b2 is also divisible by a-b
an – bn is divisible by a+b when n is even
710 – 510 is divisible by 7-5= 2
Remember it by:
a3 – b3 is not divisible by a+b
a2 – b2 is also divisible by a+b
a4 – b4 is also divisible by a+b

an + bis divisible by a+b when n is odd
711 + 511 is divisible by 7+5= 12
Remember it by:
a3 + b3 is divisible by a+b
a2 + b2is NOT divisible by a+b
a4 + b4 is NOT divisible by a+b

an + bn+cnis divisible by a+b+c when n is odd.
73 + 53 + 23= 343+125+8=476 divisible by 7+5+2=14

Questions you can solve by using above formula:
Example 1: 3223 + 1723is definitely divisible by….

a. 49
b. 15
c. 49 & 15
d. none of these.
ANSWER  A
As an + bn is divisible by a + b when n is odd
So 3223 + 1723 is divisible by 32 + 17 = 49.

Example 2: 3223 – 1723 is definitely divisible by….
a. 49
b. 15
c. 49 & 15
d. none of these.
ANSWER B
As an – bn is always divisible by a-b.
So 3223 – 1723 is divisible by 32 – 17 = 15.

Example 3: 32232 – 17232 is definitely divisible by….
a. 49
b. 15
c. 49 & 15
d. none of these.
ANSWER C

As an – bn is always divisible by a-b and an – bn is divisible by a + b when n is even
So 32232 – 17232 is divisible by both 32 – 17 = 15 and 32 + 17 = 49.

Example 4: 35 + 55+ 75is definitely divisible by….
a. 8
b. 7
c. 15
d. all of these.
ANSWER C
As an + bn + cn is divisible by a + b + c when n is odd.
So 35 + 55 + 75 is divisible by 3 + 5 + 7 = 15

Shortcuts for HCF

 There are two types of question usually asked in the examination based on H.C.F.

TYPE – 1
When various number a, b, c is divided by number N and gives the same remainder r in each case. Then H.C.F. of a, b, c is given by the H.C.F. of (a – b) or (b – a) and (b–c) or (c – b) and (c –a) or (a –c). Let’s understand with help of example

Example 1: Find the largest possible value of N if all three numbers 59, 77,104 when divided by N leaves same remainder?
Solution: As per given
59 = N x a + r………………………..(1)
77 = N x b + r………………………..(2)
104 =N x c + r………………………..(3)
Subtract equation (1) from (2)
77 – 59 =18 = N x (b -a)……………………..(4)
Subtract equation (2) from (3)
104 – 77=27 = N x (c – b)…………………….(5)
Subtract equation (1) from (3)
104 -59 =45 = N x (c – a)……………………..(6)
From this we can conclude that the N has to be the factor of 18, 27,45
Now we are asked for the largest number so we will take the H.C.F. of 18, 27, 45 and which is 9 .So 9 is the largest number which divides 59, 77, 104 and gives same remainder in each case.

TYPE – 2
When the number is divided by various numbers a, b, c and gives different remainders p, q, r.

Example 2: Find the largest number N, which divides, 162,292,325 and gives remainders 1, 5 and 3 respectively?
Solution: As per given information
162 = N x a + 1………………………..(1)
292 = N x b + 5………………………..(2)
325 = N x c + 3………………………..(3)
Since the three numbers are 162, 292 and 325 and the remainders are 1, 5, 3 so we can say that the number is totally divisible with 161, 287, 322. So finding largest number, which divides 162,292,325, is equivalent to finding H.C.F of 161, 287 and 322.Since HCF (161, 287, 322)=7 so the 7 is the number which when divided by the 162,292,325 gives the remainder 1,5 and 3 respectively.

Some points to remember
1. LCM x HCF of two numbers is equal to the product of numbers.
2. If HCF (H) of two numbers is given, then the numbers can be assume as HCF x A and HCF x B (A and B are integers)
3. HCF of a given set of numbers is always a factor of LCM

LCM of Fractions:
 LCM of two or more fractions is given by as a fraction:
(LCM of numerators of all the fractions) / (HCF of denominator of all the fractions)
For example:

LCM [ ½ , 5/12 ,7/6] = {LCM of (1,5,7)}/{HCF of (2,12,6)} = 32/2
Assignment:
1.What is the LCM of 18 and 36?
A. 18
B. 36
C. 90
D. 72

Answer: Option B
Explanation:
18= 2x3x3=21x32
36=2x2x3x3=22x32
Primes involved are 2 and 3 and maximum power involved is 2 for both primes. Hence
22 x 32 is required LCM.

2. What is the HCF of 18 and 36?
A.18
B.9
C.12
D.36

Answer: Option A
Explanation:
18= 2x3x3=21x32
36=2x2x3x3=22x32
Primes involved are 2 and 3 and minimum power involved is 1 for both primes. Hence 2×32=18 is required HCF.

3. What is the LCM of 18, 180 and 216?
A. 540
B. 360
C. 720
D. 1080

Answer: Option D
Explanation
18=2x3x3=21x32
180=2x2x3x3x5=22x32x51
72=2x2x2x3x3x3=23x33
Primes involved are 2, 3 and 5 and maximum powers involved are 3, 3 and 1 respectively. Hence 23x33x51 = 1080 is required LCM

4. What is the HCF of 12,18 and 30?
A.10
B.5
C.6
D.8

Answer: Option C
Explanation:
12=2x2x3x3=22×31
18=2x3x3=21×32
30 =2x3x5=21x31x51
Primes involved are 2, 3 and 5 minimum powers involved are 1 ,1 and 0 respectively. Hence
21x31x50=6 is required HCF.

5. What is the LCM of 13 and 11 ?
A.11
B.13
C.143
D.1

Answer: Option C
Explanation:
Since both are prime numbers so LCM is nothing but their product = 13×11 = 143

L.C.M. by Long Division Method

A least common multiple of two numbers is the smallest positive number that is a multiple of both.
Multiple of 3 — 3, 6, 9, 12, 15, 18,…………..
Multiple of 4 — 4, 8, 12, 16, 20, 24,………….
So the LCM of 3 and 4 is 12, which is the lowest common multiple of 3 and 4.

An example of LCM
The LCM of 10, 20,25 is 100. It means that 100 is the lowest common multiple of these three numbers, but there is a question in our mind that can the LCM be (-100)? Since (-100) is lower than 100 and divisible by each of 10, 20, 25, or can it be zero or what will be the LCM of (-10) and 20? Will it be (-20) or (-200)?
For all these questions, there is only one answer that the LCM is only defined for positive numbers and LCM is not defined for 0.

PROCESS OF FINDING LCM

  • We will do prime factorization in first step of all the numbers.
  • Then we calculate the number of times each prime occurs in prime factorization and write each number as power of primes.
  • Then in last step we write all the primes involved and raise each of the primes to highest power present.

Example based on above

Example 1:LCM of   10, 20, 25?
Step 1: 10= 2 × 5
20 = 2 × 2 × 5
25= 5 × 5
Step 2: 10= 21 × 51
20 = 22  × 51
25= 52
Step 3:Primes involved are 2 and 5
Now we raise each of the primes to highest power present i.e.22 × 52 =100. So 100 is required LCM.

Example 2: What is the LCM of 35, 45, 55?
Step-1:
35 = 5 × 7
45 = 3 × 3 × 5
55 = 11 × 5
Step-2:
35 = 51 × 71
45 = 32 × 51
55 = 111 × 51
Step-3: Primes involved are 3, 5, 7 and 11
Now we raise each of the primes to the highest power present i.e. 32 × 51 × 71 × 111
LCM of the given numbers = 3465

Another Method to calculate LCM

Another Method to calculate LCM
Division method is another method to find the L.C.M. of numbers.

  1. First we will write the numbers in ascending order.
  2. Now we will divide the numbers with a common prime factor and continue the same till the time it is possible.
  3. We multiply all common prime factors and numbers obtained in last row to get the LCM.

Remember if a number is not divisible by this prime number, then write the number as it is
Let’s take an example to understand it:
 
Example 1: Find the L.C.M. of 24, 18 and 36?
Solution:
Step 1: We arrange given numbers in ascending order i.e. we write as 18, 24 and 36.
Step 2: Here the common prime factors are 2 and 3. So we divide as shown in table below till the time it is possible.
wordpandit
Step 3: We multiply all common prime factors and numbers obtained in last row to get LCM= 2x2x3x3x2=72
 
Example 2: Find the L.C.M. of 60, 16 and 48?
Solution:
Step 1: We arrange given numbers in ascending order i.e. we write as 16, 48 and 60.
Step 2: Here the common prime factors are 2 and 3. So we divide as shown in table below till the time it is possible.
fr3
Step 3: We multiply all common prime factors and numbers obtained in last row to get LCM= 2x2x3x4x5=240

H.C.F. by Long Division Method

H.C.F. by Long Division Method
To find the H.C.F. of the given number we will follow the following steps:

  1. We divide the bigger number by smaller one.
  2. Divide smaller number in step 1 with remainder obtained in step 1.
  3. Divide divisor of second step with remainder obtained in step 2.
  4. We will continue this process till we get remainder zero and divisor obtained in end is the required H.C.F.

Let’s take few examples for this:
Example 1: Find the H.C.F. of 248 and 492?
To find the solution we will follow the following method i.e. we divide bigger number 492 by smaller one i.e. 248
de
So the divisor in the end was 4 so the H.C.F of the given numbers is 4
Example 2: Find the H.C.F. of 420 and 396?
To find the solution we will follow the following method i.e. we divide bigger number 420 by smaller one i.e. 396
65
So the divisor in the end was 12 so the H.C.F of the given numbers is 12
Some points to remember
– If H is the HCF of two numbers A and B, then H is also a factor of AX and BY, where X and Y are integers. In other words, H is also a factor of multiples of these numbers.
– If HCF (A,B) is H, then H is also the HCF of (A) and (A+B)
– If HCF (A,B) is H, then H is also the HCF of (A) and (A-B)
– If HCF (A,B) is H, then H is also the HCF of (A+B) and (A-B)
– If HCF=LCM for two numbers, then the numbers must be equal to each other.
– HCF of two or more fractions is given by:
(HCF of numerators of all the fractions) / (LCM of denominator of all the fractions)

Shortcut for LCM

Type 1:
Remainder is same when divided by distinct divisors
If the Number N leaves the same remainder R when divided by the numbers a, b, c, then the number N = (LCM of (a, b, c) p +R, where p is any whole number.

Type 2:
Difference between the divisors and remainders is same
If  the number N is divided by the numbers a, b, c and N gives three different remainders p, q, r respectively such that the difference between the divisors and remainders is same i.e.(a-p)=(b-q)=(c-r)=Z, then the number N must be in the form of {(LCM of a, b, c) p – Z} in each case, where p is any whole number.

Type 3:
Neither remainder is same nor is difference between the divisors and remainders same.
When the number N is divided by the two numbers m and n and they leave remainders p, q respectively. In this case we identify the smallest number (say A) that satisfies given conditions (the smallest number which gives remainder p, q when divided by m, n and this number is identified by equating the number and using the fact that quotient is whole number). Once we have done that, the number must be in the form of {(LCM of a, b)p + A} (A is the smallest number)and p is any whole number.

Let’s solve few examples based on above shortcuts:
Example 1: Find the largest 4 digit number, which gives 4 as a remainder when divided by 17 and 19?
 Solution:
Let N be such number which satisfies given conditions.
Since N gives the same remainder 4 in each case. Hence N must be of the form
N = (LCM of 17, 19) p + 4where p is any whole number
Since LCM of 17 and 19 is 323 so the number must be in the form
N=323 p + 4
The largest value of “p” possible so that the N remains 4-digit number is 30.
Therefore, 323p = 323 X 30=9690.
Hence largest 4-digit number, which gives 4 as a remainder when divided by 17 and 19
is 9694.

Example 2: Find the largest 4-digit number, which gives the remainder 7 and 13 when divided by 11 and 17?
Solution:
Let N be such number satisfying given conditions
Here remainders are not same but the difference between the remainders and the divisors is same i.e. 11-7 =17-13=4
Hence N must be of the form
N = (LCM of 11, 17) p – 4where p is any whole number
Since LCM of 11 and 17 is 187so the number must be in the form
N=187 p – 4.
The largest value of “p” possible so that the N remains 4-digit number is 53.
Therefore, 187p= 187 x 53=9911.
Hence the largest 4-digit number, which gives remainders 7 and 13 when divided by 11 and 17, is 9907.

Example 3: Find the largest 3-digit number, which gives the remainder 1 and 2 when divided by 3 and 5?
Solution:
Let N be such number satisfying given conditions
i.e. N = 3a + 1 where a is any whole number
&N = 5b + 2 where b is any whole number.
Where a & b are whole numbers.
Equating
3a + 1=5b + 2
b= (3a-1)/5
Since b is whole number. So, 5 should divide numerator implies a=2.
Thus smallest number satisfying both given conditions is 7.
So the number will be in the form of (LCM of 3,5)k + 7.
Now we have to find the largest 3-digit number N in the form of 15k+7
The largest value of “k” possible so that the N remains 3-digit number is 53.
Hence the largest 3-digit number, which gives the remainder 1 and 2 when divided by 3 and 5 is 15 x 66 + 7 = 997

Basics of Factors

Various Types of Factors
Even factors:
Even factors are the factors of number, which are divisible by 2.

Example 1: Find the number of even factors of 58800?
Solution: We first factorize 58800.
58800 = 24 315272
In this case we have to find number of even factors, an even factor is divisible by 2 or we can say smallest power of 2 should be 1 not 0.
Hence a factor must have
2(1 or 2 or 3 or 4)  — 4 factors

3(0 or 1 )       — 1+1=2 factors

5(0 or 1 or 2) — 1+2=3 factors

7(0 or 1 or 2) — 1+2=3 factors

Hence total number of even factors = (4)(2)(3)(3) = 72
Hence, number of even factors of a number
N=2paqbrcsare p(q + 1)(r + 1)(s + 1)N=2paqbrcsare p(q + 1)(r + 1)(s + 1)

Odd factors:
Odd factors are those factors, which are not divisible by 2.

Example 2: Find the number of odd factors of 58800.
Solution: We first factorize 58800.
58800 = 24 315272
Total number of factors of 58800 is 5x2x3x3=90 and in previous example we calculated total number of even factors is 72.
Number of odd factors = Total number of factors – Number of even factors
=90- 72 = 18

Alternate way: Since odd factors should have power of 2 as 0. Hence odd factor must have
2(0 )  ———- 1 factor

3(0 or 1 )     — 1+1=2 factors

5(0 or 1 or 2) — 1+2=3 factors

7(0 or 1 or 2) — 1+2=3 factors

Total number of odd factors = (1 )(2)(3)(3) = 18
Hence, number of odd factors of a number
N=2paqbrcsare p(q + 1)(r + 1)(s + 1)N=2paqbrcsare p(q + 1)(r + 1)(s + 1)

Perfect square factors:
If a number is perfect square then its prime factors must have even powers.

Example 3: Find the number of factors of 58800 which  are perfect square?
Solution:
We know that for a number to be a perfect square, its factor must have the even number of powers.
We first factorize 58800.
58800 = 24 315272
Hence perfect square factors must have
2(0 or 2 or 4)—– 3 factors

3( 0 )     —–   1  factor

5(0 or 2) ——- 2 factors

7(0 or 2) — 2 factors

Hence, the number of factors which are perfect square are 3x1x2x2=12

Remember
-If number of factors is odd then the number is a perfect square and vice versa  is also true i.e. if a number is a perfect square then number of factors is  odd.
This is because if number is a perfect square then p, q, and r are even and hence
(p + 1) (q + 1) and (r+ 1) are odd and so product of these numbers is also an odd number.
-If number of factors is even then number is not a perfect square.

Factors that end with zero

Factors whose unit digit is zero or we can say factors which are divisible by 10.

Example 1: Find the number of factors of 58800 that end with 0.
Solution:  If a number ends with 0 then it must have at least 21 and 51.
We first factorize 58800.
58800 = 24 315272
Hence odd factor must have
2(1 or 2 or 3or 4)—– 4 factors
3(0 or 1 )     —– 1+1=2 factors
5(1 or 2) ——- 2 factors
7(0 or 1 or 2) — 1+2=3 factors
Hence, total number of factors ending with 0 = (4)(2)(2)(3) = 48
Alternate way:
To find number of factors, which are divisible by 10 we divide the given number by 10 and then find the number of factors of the quotient.
Divide 58800=24 315272 by 10 to get 23 3151 72 and its number of factors are 4x2x2x3=48

Factors not ending with zero
Example 2: Find the number of factors of 58800 that are not ending with 0?
Solution: We first factorize 58800.
58800 = 24 315272
Total number of factors of 58800 is 5x2x3x3=90 and in previous example we calculated total number of factors ending with 0 are 48.
Number of odd factors = Total number of factors – Number of factors ending with 0 =90- 48 =42

Factors divisible by 12
Example 3: Find the number of factors of 58800, which are divisible by 12?
Solution:
Since we have to find the number of factors, which are divisible by 12,then it must have at least 22 and 31.
We first factorize 58800.
58800 = 24 315272
Hence factors divisible by 12 must have
2(2 or 3or 4)—– 3 factors
3( 1 )     —–   1  factor
5(0 or1 or 2) ——- 3 factors
7(0 or 1 or 2) — 1+2=3 factors
Hence, total number of factors, which are divisible by 12 = (3)(1)(3)(3) = 27
Alternate way:
To find number of factors, which are divisible by 12 we divide the given number by 12 and then find the number of factors of the quotient.
Divide 58800=24 315272 by 12 to get 22 52 72 and its number of factors are 3x3x3=27

Perfect cube factors

If a number is a perfect cube, then the power of the prime factors should be divisible by 3.
Example 1:Find the number of factors of293655118 that are perfect cube?
Solution: If a number is a perfect cube, then the power of the prime factors should be divisible by 3. Hence perfect cube factors must have
2(0 or 3 or 6or 9)—– 4 factors
3(0 or 3 or 6)  —–  3  factors
5(0 or 3)——- 2 factors
11(0 or 3 or 6 )— 3 factors
Hence, the total number of factors which are perfect cube 4x3x2x3=72
Perfect square and perfect cube
If a number is both perfect square and perfect cube then the powers of prime factors must be divisible by 6.
Example 2: How many factors of 293655118 are both perfect square and perfect cube?
Solution: If a number is both perfect square and perfect cube then the powers of prime factors must be divisible by 6.Hence both perfect square and perfect cube must have
2(0 or 6)—– 2 factors
3(0 or 6) —– 2 factors
5(0)——- 1 factor
11(0 or 6)— 2 factors
Hence total number of such factors are 2x2x1x2=8
Example 3: How many factors of293655118are either perfect squares or perfect cubes but not both?
Solution:
Let A denotes set of numbers, which are perfect squares.
If a number is a perfect square, then the power of the prime factors should be divisible by 2. Hence perfect square factors must have
2(0 or 2 or 4 or 6 or 8)—– 5 factors
3(0 or 2 or 4 or 6)  —– 4 factors
5(0 or 2or 4 )——- 3 factors
11(0 or 2or 4 or6 or 8 )— 5 factors
Hence, the total number of factors which are perfect square i.e. n(A)=5x4x3x5=300
Let B denotes set of numbers, which are perfect cubes
If a number is a perfect cube, then the power of the prime factors should be divisible by 3. Hence perfect cube factors must have
2(0 or 3 or 6or 9)—– 4 factors
3(0 or 3 or 6)  —–  3  factors
5(0 or 3)——- 2 factors
11(0 or 3 or 6 )— 3 factors
Hence, the total number of factors which are perfect cube i.e. n(B)=4x3x2x3=72
If a number is both perfect square and perfect cube then the powers of prime factors must be divisible by 6.Hence both perfect square and perfect cube must have
2(0 or 6)—– 2 factors
3(0 or 6) —– 2 factors
5(0)——- 1 factor
11(0 or 6)— 2 factors
Hence total number of such factors are i.e.n(A∩B)=2x2x1x2=8
We are asked to calculate which are either perfect square or perfect cubes i.e.
n(A U B )= n(A) + n(B) – n(A∩B)
=300+72 – 8
=364
Hence required number of factors is 364.

Product of Two Factors

To find the number as the products of two factors, use the following steps :
Step1: Write Prime factorisation of given number i.e. convert the number in the form ap bq cr
where a ,b,c are prime numbers and the p,q,r are natural numbers as their respective powers.
Step 2:Find Number of factors which can be expressed as( p+1)(q+1)(r+1).
Step 3: Number of ways to express the number as a product of two numbers is exactly half its number of factors i.e.½ *(p+1)(q+1)(r+1).
Let’s have an example on this :
Example 1: In how many ways can you express 54 as a product of two of its factors?
Solution:  We will do the above problem step by step:
Step 1: Prime factorization of 54 i.e. we write 54 = 2133
Step 2: Number of factors of 54 will be (1+1)(3+1) = 2 x 4= 8
Step 3:  Hence number of ways to express 54 as a product of two numbers is exactly half its number of factors i.e. ½ *8 = 4 ways.
In fact we can list these 4 ways as well
Factors of 54 are 1,2,3,6,9, 18,27,54.
Now it is very simple to find the factors from 1 to 9 but it is difficult to find the ones that are greater than 10. So number of ways to express 54 as a product of two of its factors is
1 x 54 = 54
2 x 27 = 54
3 x 18 = 54
6 x 9   = 54
Example 2: In how many ways you can express 120  as a product of two of its factors?
Solution:
Step 1: Prime factorization of 120 i.e. we write 120  = 233151
Step 2: Number of factors of 120 will be (3+1)(1+1)(1+1) = 4 x 2 x2 = 16
Step 3:  Hence number of ways to express 120 as a product of two numbers is exactly half its number of factors i.e. ½ *16 = 8
In fact we can list these 8 ways as well
Factors of 120 are 1,2,3,4,5,6,8,10,12,15,18,24,30,40,60,120.
So number of ways to express 120 as a product of two of its factors is
1 x 120 = 120
2 x 60 = 120
3 x 40 = 120
4 x 30 = 120
5 x 24 = 120
6 x 20 = 120
8 x 15 = 120
10 x 12=120

Sum of Factors

The concept of cyclicity is used to identify the last digit of the number.Let’s take an example to understand this:

Example 1:  Find the unit digit of 354.
Solution:  Now it’s a very big term and not easy to calculate but we canfind the last digit by using the concept of cyclicity. We observe powers of 3
31 = 3
32 = 9
33 = 27
34 = 81
35 = 243
36= 729

So now pay attention to the last digits we can observe that the last digit repeats itself after a cycle of 4 and the cycle is 3, 9, 7, 1 this repetition of numbers after a particular stage is called the cyclicity of numbers. Therefore when we need to find the unit digit of any number like 3we just need to find the number on which the cycle halts. So we divide power n by 4 to check remainder

  • If remainder is 1 then the unit digit will be 3
  • If remainder is 2 then the unit digit will be 9
  • If remainder is 3 then the unit digit will be 7
  • If remainder is 0 then the unit digit will be 1

We divided the power by 4 because cycle repeats itself after 4 values .Now the main question was that how much is the last digit of 354 and we know the cycle repeat itself after 4 so we will divide the 54 with 4,so on dividing 54 by 4 the remainder becomes 2 .Now as we discussed above if the remainder is 2 the last digit would be 9.Hence unit digit of 354 is 9.

Example 2:What will be the unit digit of 34745
Solution:
Lets observe unit digit of 347 x 347 = 9.
The main purpose of the above is that the unit digit of any multiplication depends upon the unit digit of numbers, whatever is the number big or small the unit digit always depends upon the multiplication of the last digit.
So the last digit of 34745 can be found by calculating last digit of 745
We observe unit digit while calculating powers of 7
71  = 7
72 = 49
73 = 343
74 = 2401
75 = 16807
So on dividing 45 with 4, 1 will be the remainder and the last digit would be 7.

Product of Factors

Perfect square as a product of two factors
In case of perfect square number we have odd number of factors i.e. the number of factors are odd hence in that case required number of ways in which we can write perfect square number as a product of its two factors are (n – 1)/2 if we do not include the square root of the number and
(n + 1 )/2 if we include the square root of the number. So number of ways to express a perfect square as product of two different factors (that means excluding its square root) is
½ {(p + 1)(q + I)(r + I) … – 1)}. And if we include the square root then required numbs 1/2 {(p+1)(q+1)(r+1) … +1}
Let’s take one example to understand this.
Example 2: In how many ways you can express 36 as the product of two of its factors?
Solution :
Step 1: Prime factorization of 36 i.e. we write 36 = 22 x 32
Step 2: Number of factors of 36 will be (2+1)(2+1)=9 (i.e. factors are 1,2,3,4,6,9,12,18,36)
Step 3: Since we are asked total number of ways hence we include square root of 36 i.e. 6 as well. Thus number of ways you can express 36 as the product of two of its factors is (9+1)/2=5
 
Product of two co-prime numbers
To express the number as a product of co-prime factors we will use the following steps:
Step 1:Write Prime factorisation of given number i.e. convert the number in the form where p1 ,p2,p3…..pn are prime numbers and a,b,c….. are natural numbers as their respective powers.
Step 2: In the above step we have n prime numbers then the number of ways to express the number as the product of two co prime numbers =2n-1
Because two primes are always co-prime and after we pick 1 prime the other prime can be picked in 2n-1ways.Hence number of ways in which we can write given number as a product of two co prime factors =2n-1
Example 1: In how many ways you can write 315 as product of two of its co-prime factors.
Solution:

Step 1: Prime factorization of 315 i.e. we write 315  = 32 x 51 x 71
Step2: In above step 3 prime numbers are used. Hence number of ways are 23-1 = 4.Infact we can mention these cases as well 9×35, 5 x 63, 7 x 45, 15 x 21.
Remember: The number of numbers, which are less than N= pa qb rc(where p , q, r are prime numbers and the a,b,c are natural numbers as their respective powers)and are co-prime to N is given by N{(1- 1/p)(1-1/q )(1-1/r)}.
 
Product of all the factors
To find product of all the factors we follow the following steps:
Step1: Prime factorisation N= apbqcr
Step2: Number of factors   (say X)
Step3: Product of all the factors is given by =  NX/2
Example 2: Find the product of all the factors of 120 ?
Solution:
Step 1: Prime factorisation : = 23X51X31
Step2: Number of factors are ( 3+1)(1+1)(1+1)= 16
Step 3: Product of all the factors =( 23X51X31)16/2 =(120)8
 
Assignment:
Questions:
1: In how many ways can you express 216 as a product of two of its factors?
a) 4
b) 6
c) 8
d) 9
Answer: c
Solution:
Step 1: 216 = 2333
Step 2: Number of factors are (3+1)(3+1) = 4 x 4= 16
Step 3:  So number of ways ½ (4)(4) = 8

2: Find the sum of factors of 216.
a. 950
b) 850
c) 600
d) 1000
Answer: c
Solution:
Step 1: 216 = 2333
Step 2: Sum of factors = (20+21+22+23)(30+31+32+33)= 15 x 40 = 600

3: Find the product of all the factors of 400
a. 805
b) 804
c) 807
d) 806
Answer: a
Solution:
Step 1: Prime factorisation of 400 = 24X52
Step2: Number of factors (4+1)(1+1)= 10
Step 3: Product of all the factors =( 24X51)10/2 =(80)5

4:In how many ways you can write 200 as product of two of its co-prime factors.

a) 1
b) 2
c) 4
d) 8
Answer: b
Step1: 200 = 2352
Hence number of ways are 22-1 = 2.Infact we can mention these cases as well 8 x 25,
1 x 100.

5: In how many ways you can write 10890 as product of two of its co-prime factors.
a) 7
b) 3
c) 31
d) 15
Answer: d
Solution:
We will do the solution with step by step
Step1: first prime factorization i.e.21 x 32 x 51x 112
Step2: 3 prime number are used in this so the number of ways are 24-1=15

Basic Concept of Cyclicity

The concept of cyclicity is used to identify the last digit of the number.Let’s take an example to understand this:

Example 1:  Find the unit digit of 354.
Solution:  Now it’s a very big term and not easy to calculate but we canfind the last digit by using the concept of cyclicity. We observe powers of 3
31 = 3
32 = 9
33 = 27
34 = 81
35 = 243
36= 729

So now pay attention to the last digits we can observe that the last digit repeats itself after a cycle of 4 and the cycle is 3, 9, 7, 1 this repetition of numbers after a particular stage is called the cyclicity of numbers. Therefore when we need to find the unit digit of any number like 3we just need to find the number on which the cycle halts. So we divide power n by 4 to check remainder

  • If remainder is 1 then the unit digit will be 3
  • If remainder is 2 then the unit digit will be 9
  • If remainder is 3 then the unit digit will be 7
  • If remainder is 0 then the unit digit will be 1

We divided the power by 4 because cycle repeats itself after 4 values .Now the main question was that how much is the last digit of 354 and we know the cycle repeat itself after 4 so we will divide the 54 with 4,so on dividing 54 by 4 the remainder becomes 2 .Now as we discussed above if the remainder is 2 the last digit would be 9.Hence unit digit of 354 is 9.

Example 2:What will be the unit digit of 34745
Solution:
Lets observe unit digit of 347 x 347 = 9.
The main purpose of the above is that the unit digit of any multiplication depends upon the unit digit of numbers, whatever is the number big or small the unit digit always depends upon the multiplication of the last digit.
So the last digit of 34745 can be found by calculating last digit of 745
We observe unit digit while calculating powers of 7
71  = 7
72 = 49
73 = 343
74 = 2401
75 = 16807
So on dividing 45 with 4, 1 will be the remainder and the last digit would be 7.

Application of Concept of Cyclicity

How to calculate unit digit if a number contains power of a power

Example 1:What will be the last digit of (12^23)^45

Solution:
To find the last digit of this type of number we will start the question from the base the base is given to be 12. It means we will see the cyclicity of 2 because the last digit depends upon the unit digit of 12 i.e. 2. We observe unit digit while calculating powers of 2
21 = 2
22 = 4
23 = 8
24 = 6
25 = 2

Step 1: Now we know that cyclicity of last digit of 12 i.e. 2 is of 4, hence we divide the power of 12 i.e. 2345 with 4.

Step 2: Now lets calculate the remainder of 2345 when divided by 4 and then we will determine the last digit.

Step 3:  The remainder will be 3 because we can write remainder of 23 by 4 as 3 or -1.Hence we can write 2345 as (-1) 45 but odd power of -1 will be again -1 and thus 2345 when divided by 4 will give us remainder as -1 or 3.

Hence unit digit of 122345122345 will be same as unit digit of 23i.e. 8

Example 2: Find the unit digit of 322595322595  ?  

Solution: To find the last digit of this type of number we will start the question from the base the base is given to be 32. It means we will see the cyclicity of 2 because the last digit depends upon the unit digit of 32 i.e. 2. We observe unit digit while calculating powers of 2
21 = 2
22 = 4
23 = 8
24 = 6
25 = 2
Step 1: Now we know that cyclicity of last digit of 32 i.e. 2 is of 4, hence the divide the power of 12 i.e. 2595 with 4.
Step 2: Now lets calculate the remainder of 2595 when divided by 4 and then we will determine the last digit.
Step 3:  The remainder will be 1 because we can write remainder of 25 by 4 as 1. Hence we can write 2595 as (1) 95 but any power of 1 will be again 1 and thus 2595 when divided by 4 will give us remainder as 1 .
Hence unit digit of 322595322595   will be same as unit digit of 2i.e. 2

Assignment:
Questions:
1: Find the unit digit of 4686
a) 4
b)  6
c) 2
d)  3
Answer: b
Solution: We know
41 = 4
42 = 6
43 = 4
44 = 4
Here, we see that powers of four repeat after a cycle of 2 i.e.
Odd power of 4 gives unit digit as 4 and even power of 4 gives unit digit as 6
So unit digit of 4686 is 6.

2 : What will be the unit digit of 651234
a) 5
b)  0
b)  2
d) 3
Answer: a
Solution We know
51 = 5
52 = 25
53 = 125
54 = 625
So, we observe that unit digit of any power of 5 is 5.Hence unit digit of 651234 is 5.

3:  Find the unit digit of 36512
a) 4
b) 6
c) 2
d) 3
Answer: b
Solution We know
61 = 6
62 = 36
63 = 216
So, we observe that unit digit of any power of 6 is 6.Hence unit digit of 36512 is 6.

4: Find the unit digit of 185693
a) 4
b) 6
c) 2
d) 8
Answer: d
Solution: We observe unit digit while calculating powers of 8
Unit digit in 81 = 8
Unit digit in 82 = 4
Unit digit in 83 = 2
Unit digit in 84 = 6
Unit digit in 85 = 8
So on dividing 5693 with 4, 1 will be the remainder and hence the last digit would be 8.

How to find the Unit Digit of a number

For the concept of identifying the unit digit, we have to first familiarize with the concept of cyclicity. Cyclicity of any number is about the last digit and how they appear in a certain defined manner. Let’s take an example to clear this thing:

The cyclicity chart of 2 is:
21 =2
22 =4
23 =8
24=16
25=32

Have a close look at the above. You would see that as 2 is multiplied every-time with its own self, the last digit changes. On the 4th multiplication, 25 has the same unit digit as 21. This shows us the cyclicity of 2 is 4, that is after every fourth multiplication, the unit digit will be two.

Cyclicity table:
The cyclicity table for numbers is given as below:
cyclicity table for numbers
How did we figure out the above? Multiply and see for yourself. It’s good practice.
Now let us use the concept of cyclicity to calculate the Unit digit of a number.

What is the unit digit of the expression 4993?
Now we have two methods to solve this but we choose the best way to solve it i.e. through cyclicity
We know the cyclicity of 4 is 2
Have a look:
41 =4
4=16
4=64
4=256

From above it is clear that the cyclicity of 4 is 2.  Now with the cyclicity number i.e. with 2 divide the given power i.e. 993 by 2 what will be the remainder the remainder will be 1 so the answer when 4 raised to the power one is 4.So the unit digit in this case is 4.

For checking whether you have learned the topic, think of any number like this, calculate its unit digit and then check it with the help of a calculator.

Note : If the remainder becomes zero in any case then the unit digit will be the last digit of  acyclicity number 

where a is the given number and cyclicity number is shown in above figure.

Lets solve another example:
The digit in the unit place of the number 7295 X 3158 is
A.  7
B.  2
C.  6
D.  4

Solution
The Cyclicity table for 7 is as follows:
7=7
7=49
73 = 343
74 = 2401

Let’s divide 295 by 4 and the remainder is 3.
Thus, the last digit of 7295 is equal to the last digit of 73 i.e. 3.
The Cyclicity table for 3 is as follows:
31 =3
32 =9
3= 27
3= 81
3= 243

Let’s divide 158 by 4, the remainder is 2. Hence the last digit will be 9.
Therefore, unit’s digit of (7925 X 3158) is unit’s digit of product of digit at unit’s place of 7925 and 3158 = 3 * 9 = 27. Hence option 1 is the answer.

Factorials

A factorial is a non-negative number which is equal to the multiplication of numbers that are less than that number and the number itself. It is denoted by (!)
Let’s take an example to understand this
What will be the value of 5!
So in the above definition we discussed that the multiplication of the numbers which all are less than that number and the number itself. Hence number less than 5 are 1,2,3,4 and 5 is number itself so
5! = 5 x 4 x 3 x 2 x 1 = 120
Always remember we define the value of 0! =1

Lets take some more examples for this
Value of 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720
Value of 4! = 4 x 3 x 2 x 1 = 24
Value of 3! = 3 x 2x 1= 6

We can also write n! = n x (n – 1)!
Concept of factorial is very important, there are few basic types in which this concept is used. We will discuss each type with the help of examples.
 
Type 1: Highest power of p (p is a prime number) which divides the q!
Let’s take an example to understand this type
 
What is the highest power of 3 that divides 13! ?
Solution:
13! = 13 x 12 x 11 x 10 x 9 x 8 x7 x 6 x 5 x 4 x 3 x 2 x 1
And we have to find highest power of 3 that can divide the above term
So first we need to know how many times 3 is multiplied in 13! .
So to find the number of 3’s we will divide 13 with 3.
13 divided by 3 gives 4 as quotient and 1 remainder. We will keep remainder aside and move further till the quotient cannot be divided further
Again we divide 4 by 3 we get quotient as 1 and again remainder is 1 .Now we stop at this stage because quotient 1 cannot be divided by 4.
Now add all the quotients
4+1 =5.
So the maximum power of 3 is 5, which can divide the 13! .

We can also generalize the formula for calculating
Maximum power of any prime p in any n! is calculated as

[np]+[np2]+[np3]+[np4]+[np]+[np2]+[np3]+[np4]+………………

Where [ ] represents integral part
Maximum power of 3 in 13!

[133]+[1332]+..[133]+[1332]+………..
= 4+ 1 =5

Other method
Since the number 13! is not very big number in-fact we can write and check maximum power of 3
13 x 12 x 11 x 10 x 9 x 8 x7 x 6 x 5 x 4 x 3 x 2 x 1
13 x 3 x4 x 11 x 10 x 3 x 3 x 8 x 7 x 3 x 2 x 3 x 2 x 1
Here the number of 3’s is 5 so we can cancel it with 35. So maximum power of 3 is 5, which can divide 13!

Let’s take one more example like this
What is the highest power of 7 that exactly divides 49!
To find the highest power of 7 that exactly divides 49! . We need to know the number of 7’s in the 49!
So when we divide 49 with 7, the quotient will be 7 and there is no remainder, since the quotient can further divided by 7 we will divide 7 with 7 with quotient 1 and remainder 0 .
Now add all quotients and get the answer as 8
So the highest power which will divide the 49! will be 8.

Highest Power of a prime in a Factorial

Type : Highest power of p which divides the q! ,where p is not a prime number
The approach for this type is same as that for calculating maximum power of prime in any factorial buthere first we will break p into product of primes.
Lets take an example to understand this

Example 1
What will be the maximum power of 6 that divides the 9!
In order to find maximum power of 6 we will first write as product of 2 and 3.
Maximum power of 2 in 9!
[92]+[922]+[923]+..= 4 + 2 + 1 + 0=7Maximum power of 3 in 9![93]+[932]+..= 3+ 1+0=4[92]+[922]+[923]+…..= 4 + 2 + 1 + 0=7Maximum power of 3 in 9![93]+[932]+…..= 3+ 1+0=4

And there are seven 2’s and four 3’s are there, so we need to make the pairs of 2 and 3.
Now we know that 6 = 2 x3. So we need equal number of 2 and equal number of 3 in 9!
Hence there are maximum four pairs that can make 6.So the maximum power of 6 that
can divide 9! is 4.

Other method
Since the number 9! is not very big number in-fact we can write and check maximum power of 3
9! = 9x8x7x6x5x4x3x2x1 =3×3x2x2x2x7x2×3x5x2×2x3x2x1
So there are four pairs of 2 x 3, which can be formed
So the maximum power of 6 that can divide the 9! is 4 .

Example 2
What will be the highest power of 12 that can exactly divide 32!

We can write 12 = 2x2x3 i.e. we need pair of 22 x 3
Maximum power of 2 in 32!

[322]+[3222]+[3223]+[3224]+[3225]+..= 16 + 8 + 4 + 2 +1=31Maximum power of 3 in 32![323]+[3232]+[3233]+..= 10+ 3+ 1= 14[322]+[3222]+[3223]+[3224]+[3225]+…..= 16 + 8 + 4 + 2 +1=31Maximum power of 3 in 32![323]+[3232]+[3233]+…..= 10+ 3+ 1= 14

Now in 32! Number of 2 are 31
And the number of three’s = 14
So therefore if we take 14 three’s then we need 28 two’s because for each three we need two 2’s so therefore we need 28 two’s from 31 two’s to make the equal pairs of 22 x3
So the maximum power of 12 that can divide the 32! is 14

Number of zeros at the end of a factorial

Problem Type: Number of zeros at the end of p!
To solve such type of problems we see only pair of 2 x5 in p! .
Because if a number is divisible by 10 then it will have 0 in the end and 10= 2×5 so finding number of zeros is equivalent to finding maximum power of 10 in p! which is same as counting number of pairs of 2 and 5.
Lets take an example for this

Example
Find the number of zeros at the end of 45!
Solution:
Zero mainly comes from the combination of (5x 2) or by the presence of 10, and the number of zeros depends upon the number of times 10 is involved.So we check power of 2 and 5 first
Maximum power of 2 in 45!

[452]+[4522]+[4523]+[4524]+[4525]+..= 22 + 11 + 5 + 2 +1= 41Maximum power of 5 in 45![455]+[4552]+[4553]+..= 9+ 1= 10[452]+[4522]+[4523]+[4524]+[4525]+…..= 22 + 11 + 5 + 2 +1= 41Maximum power of 5 in 45![455]+[4552]+[4553]+…..= 9+ 1= 10

So in 45! Number of 2’s are 41
And the number of 5’s is 10
So maximum pair of 2 and 5 that can be made are 10 so the number of zeros at the end of the 45! is 10.

Example
Find the number of zeros in 500!
Solution: Zero mainly comes from the combination of (5x 2) or by the presence of 10, and the number of zeros depends upon the number of times 10 is involved.So we check power of 2 and 5 first
Maximum power of 2 in 500!

[5002]+[50022]+[50023]+[50024]+[50025]+[50026]+[50027]+[50028]= 250 + 125 + 62 + 31 +15 +7 + 3 + 1= 494Maximum power of 5 in 500![5005]+[50052]+[50053]+[50054]+..= 100 + 20 + 4= 124[5002]+[50022]+[50023]+[50024]+[50025]+[50026]+[50027]+[50028]…= 250 + 125 + 62 + 31 +15 +7 + 3 + 1= 494Maximum power of 5 in 500![5005]+[50052]+[50053]+[50054]+…..= 100 + 20 + 4= 124
Number of 5’s in 500! are  124
And the numbers of 2’s is  494
So the maximum possible pairs of 2 and 5 that can be made are 4 so the number of zeros in 500! are 124 .So the number of zeros in the end of the 500! are 124.

Example
What will be the number of zeros in (24!)3! ?
Now first calculate the number of zeros in 24!
Maximum power of 2 in 24!
[242]+[2422]+[2423]+[2424]+..= 12 + 6 + 3 + 1= 22Maximum power of 5 in 24![245]+[2452]+..= 4 + 0= 4[242]+[2422]+[2423]+[2424]+…..= 12 + 6 + 3 + 1= 22Maximum power of 5 in 24![245]+[2452]+…..= 4 + 0= 4
Number of 5’s in 24! is 4
And the numbers of 2’s are more than 4
So the maximum possible pairs of 2 and 5 that can be made are 4 so the number of zeros in 24! are 4
Now 24! also have power of 3! i.e. 6 . So the whole expression would have 4 x 6 = 24 zeros in the end.

Extra Problems for ‘Number of Zeros’

Extra Problems for ‘Number of Zeros’ Question Type

Example 1: Find the number of zeros in 2145 x 5234  .
Solution:When we see the question it looks like a very difficult question but this type of question involving number of zeros is very simple and can be solved in seconds. Here the question is asking for the number of zeros, now  we have to look for  the pairs of 2x 5, so the maximum pairs we can form are 145 because we have the  maximum power  of 2 is 145 , so the number of zeros  in this case are 145.

Example 2: How many numbers of zeros are there in following expression
211 x 1253 x 711+ 311 x  711x  210 x  51+  1112 x  213 x 5125
Solution: To explain the solution we must know something about the nature of zeros, before doing this lets have a simple example, the number of zeros in the end of
10+100+1000+……………..+ 10000000000000000000 are 1 because we can take 10 common out of this expression and write it as 10(1+10+100+……………..+ 1000000000000000000) And clearly the term present inside the bracket has unit digit 1 which when multiplied with 10 results into 1 zero at its end.Thus number of zeros in any such expression will depend on the least zero holder in the expression.
Now using the same logic we can say least number of zeros will be contributed by second term i.e. 311 x  711x  210 x  51  as it contains only single power of 5 .Hence it contains only 1 zero, so the number of zeros for whole expression is 1 .

Example 3:Find the number of zeros at end of
5 x 10 x 15 x 20 x 25 x 30 x 35…………………………………… x 240 x 245 x 250
Solution: We can take 5 common out of this expression and write it as
550 (1 x 2 x 3 x 4……………………………………  x 49 x 50)
550 (50!)
To find number of zeros we first find
Maximum power of 2 in 50!

[502]+[5022]+[5023]+[5024]+[5025]+..= 25 + 12 + 6 + 3 +1= 47Maximum power of 5 in 50![505]+[5052]+[5053]+..= 10 + 2= 12[502]+[5022]+[5023]+[5024]+[5025]+…..= 25 + 12 + 6 + 3 +1= 47Maximum power of 5 in 50![505]+[5052]+[5053]+…..= 10 + 2= 12

Thus maximum power of 5 present in given expression is 62 and maximum power of 2 present in given expression is 47. Hence number of zeros will be 47.

Assignment:
Questions:
1 : Find the number of zeros in 75!
a) 16
b) 18
c) 20
d) 21
Ans: b
Solution:
Maximum power of 5 in 75!
[755]+[7552]+[7553]+..= 15 + 3= 18Hence number of zeros will be 18.[755]+[7552]+[7553]+…..= 15 + 3= 18Hence number of zeros will be 18.

2: Find the number of zeros in 255!
a) 63
b) 52
c) 62
d) 65
Ans: a
Solution:
Maximum power of 5 in 255!
[2555]+[25552]+[25553]+[25554]+..= 51 + 10 + 2= 63Hence number of zeros will be 63.[2555]+[25552]+[25553]+[25554]+…..= 51 + 10 + 2= 63Hence number of zeros will be 63.

3: Find the number of zeros in 135!

a) 25
b) 30
c) 33
d) 32
Ans: c
Solution:
Maximum power of 5 in 135!
= 27 + 5 + 1
= 33
[1355]+[13552]+[13553]+..= 27 + 5 + 1= 33Hence number of zeros will be 33.[1355]+[13552]+[13553]+…..= 27 + 5 + 1= 33Hence number of zeros will be 33.

4: Find the number of zeros in n! where n is number between 66 to 69
a) 12
b) 14
c) 15
d) 16
Ans: c
We pick any value of n between 66 and 69. Number of zeros will be same for any value we pick between 66 and 69 say 68
Maximum power of 5 in 68!
= 13 + 2
= 15

[685]+[6852]+[6853]+..= 13 + 2= 15[685]+[6852]+[6853]+…..= 13 + 2= 15
Hence number of zeros will be 15.

5: Find the number of zeros in 350!

a) 84
b) 85
c) 86
d) 87
Ans: c
Solution:
Maximum power of 5 in 350!
= 70 + 14 + 2
= 86

[3505]+[35052]+[35053]+[35054]+..= 70 + 14 + 2= 86[3505]+[35052]+[35053]+[35054]+…..= 70 + 14 + 2= 86
So the number of zeros in the end of the 350! are 86.

Remainders (Part-1)

When a particular number is divided by any another number(divisor) we get quotient and remainder.
For example: On dividing 9 with 4 we get 2 as the quotient and 1 as the remainder. Here 9 is dividend and 4 is divisor.

Tool tip:The value of the remainder is always less than the divisor.
Find the remainder when 39 is divided by6 ?
We can write 39 = 6(6) + 3. Here quotient is 6 and 3 is remainder. Note that remainder 3 is less than divisor 6.

Find the remainder when 79 is divided by11 ?
We can write 79 = 11(7) + 2. Here quotient is 7 and 2 is remainder. Note that remainder 2 is less than divisor 7.

If divisor is same then it hardly creates any difference whether we add numbers and then divide by divisor to calculate remainder or we separately calculate remainders and then add.

Let’s understand first of all whether we deal numbers separately while calculating remainders or we deal together it doesn’t create any difference. Lets understand meaning of above with help of example

Suppose we have 53 sticks in a bundle and we are told to divide it in groups of 5. So we all know that while dividing 53 in groups of 5, three sticks will be left.

Let’s consider another bundle of 21 sticks and we are told to divide it in groups of 5. So we all know that while dividing 21 in groups of 5, one stick will be left.
Now if we add the above two bundles we have total of 53 + 21=74 and again divide it in groups of 5
Then we will be left with 4 sticks.

Hence instead of mixing we can treat them by two groups as well and we can make groups of 5 separately and total numbers of remaining sticks are simply 4(3+1)i.e. addition of remaining sticks in the two groups separately.
Since 4 is lesser than 5 we cannot make another group of 5 from remaining sticks

But if instead of 21 sticks we had 22 sticks then the remainder would be 2 and the total remaining sticks would have been 5, and then we can make a new group of 5 with the remaining sticks with no remnant left.

Remainders (Part-2)

Type: Difference between divisor and remainder is same
On dividing by p, q and r the number gives different remainders a, b and c where
p-a=q-b=r-c
Let’s take an example to understand this

Example
What is the minimum two-digit number which when divided by 3, 4, 5 leaves 1, 2, 3 as the remainder respectively?
Solution:
Let P be the required two digit number .As given in the question
P = 3 d + 1 =3 (d + 1) -2
P = 4 e + 2= 4( e + 1) -2
P = 5 f + 3= 5 (f+1) – 2

In each of above cases we observe difference between divisor and remainder is same.
Hence we have same remainder -2 when divided by 3,4,5
Thus number N must be in the form L.C.M. (3,4,5) k – 2 =60 k – 2
Since we are looking for two digit number putting k =1 we get 58 as minimum two-digit number which when divided by 3, 4, 5 leaves 1, 2, 3 as the remainder respectively.

Example
What is the minimum three-digit number which when divided by 2, 5, 6   leaves 1, 4, 5 as the remainder respectively?
Solution:
Let P be the required three digit number .As given in the question
P = 2 d + 1 =2 (d + 1) -1
P = 5 e + 4 = 5( e + 1) -1
P = 6 f + 5 = 6 (f+1) – 1
In each of above cases we observe difference between divisor and remainder is same.
Hence we have same remainder -1 when divided by 2,5,6
Thus number N must be in the form L.C.M. (2,5,6) k – 1 =30 k – 1
Since we are looking for two digit number putting k =34 we get 1019 as minimum three-digit number which when divided by 2, 5, 6 leaves 1, 4, 5 as the remainder respectively.

Assignment:
QUESTIONS:
1. What is the remainder when 72 + 51 is divided by 8 ?
a)1
b)2
c)3
d)5
Answer – c
Solution : Since 72 is multiple of 8 hence it leaves remainder zero and 51 leaves remainder 3 when divided by 8.Hence 72 + 51 leaves remainder 3 when divided by 8.

2. What is the remainder when 111 + 68 + 48 is divided by 11 ?
a)5
b)7
c)9
d)None
answer – b
Solution:
111 leaves remainder 1 when divided by 11
68 leaves remainder 2 when divided by 11
48 leaves remainder 4 when divided by 11
Thus 111 + 68 + 48 leaves remainder 1+2+4 =7 when divided by 11
3. What is the remainder when 42×56 is divided by 8 ?
a)0
b)1
c)2
d)3
answer – a
Solution :
56 is a multiple of 8 hence remainder is zero.

4.What is the remainder when 48×65 is divided by 7 ?
a)1
b)2
c)3
d)5
Answer – 5
48 leaves remainder 6 and 65 leaves remainder 2 so 48 65 divided by 7 gives 6×2=12 but 12 is greater than 7 hence we further divide by 7 and we obtain 5 as remainder.

5. What is the remainder of 84x71x41 by 9?
a)2
b)3
c)6
d)8
answer-b
84 leaves remainder 3, 71 leaves remainder 8 and 41 leaves remainder 5 when divided by 9. So 84x71x42 divided by 9 gives 3 8 =120 but 120 is greater than 9 hence we further divide by 9 and we obtain 3 as remainder.

Remainders (Part-3)

Problem Type 1: What will be the remainder when p + q +r +… is divided by d

Example
What will be the remainder when 63 + 67 +81 is divided by 11 ?
Solution: Instead we add up all numbers, lets do it separately
63 when divided by 11 gives 5 as quotient and 8 as remainder.
67 when divided by 11 gives 6 as quotient and 1 as remainder.
81 when divided by 11 gives 7 as quotient and 4 as remainder.
Now the remainders are 8, 1 and 4 respectively.
By adding all the remainders 8+1+4=13, the sum of remainders is greater than the divisor
Therefore again the sum of remainders is divided by the divisor i.e. by 11
On dividing 13 by 11,quotient will be 1 and 2 will be the remainder
Hence the remainder on dividing 63 +67 +81 by 11 will be 2.

Tooltip :
Now the question arises that by dividing 53 sticks in groups of 5 we are left with 3 extra sticks. Then how many more sticks should be added to this bundle so that the left out sticks are again 3?

Answer
If we want left out sticks to be 3 again then we should add minimum 5 more sticks or
10,15,20 ,….., 5n.
But remember the additional sticks should be multiple of 5.
53 when divided by 5 gives 10 as quotient and 3 as remainder and adding multiple of 5 is equivalent to adding remainder zero every time.
Let’stake an example to understand this

Problem Type 2:When the remainder is same on dividing a number ‘a’ by different divisors.
Example
What is the minimum two-digit number which when divided by 3, 4 and 5 leaves 2 as the remainder each time?

Solution:
The question would have been easy if we were asked about minimum number, which is completely divisible by three numbers 3,4 and 5 then we can say it should be least common multiple of 3,4 and 5 i.e. 60. Again 60 is the number which is completely divisible by 3,4,5 but if we want 2 remainder in each case then add 2 in 60 i.e. 62 will be the minimum two digit number which when divided by 3, 4 and 5 leaves 2 as the remainder each time.

Remainders (Part-4)

In this article, for calculating remainders we will use concept of cyclicity.
Let’s take an example of questions based on concept of remainders.

Example
Find the remainder when 47123 is divided by 7?
Solution:
We will do this question step by step
Step 1: We divide 47 with 7 to get remainder as 5.
Step 2: Here in the question 47 is multiplied by itself 123 times. Now understand every time when we will divide 47 with 7 the remainder will be 5.
Step 3: Hence we can say finding remainder of 47123is equivalent to finding remainder of 5123 when divided by 7.
So we start observing powers of 5.
52 gives remainder 4.
53 gives remainder 6.
54 gives remainder 2.
55 gives remainder 3.
56 gives remainder 1.
Hence after 6th power remainders will start repeating. So we have a cycle of 6 and thus we divide 123 by 6

i.e. 123= 6(20) + 3
Hence 5 123

-> 5 6(20) +3

-> 5 6(20) .5 3 -> 1.6  =6   (As 56 gives remainder 1 and 53 gives remainder 6)
Hence the remainder of 47123 when divided by 7 is 6.

Example
Find the remainder when 79644 is divided by 9 ?
Solution:
We will do this question step by step
Step 1: We divide 79 with 9 to get remainder as 7.
Step 2: Here in the question 79 is multiplied by itself 644 times. Now understand every time when we will divide 79 with 9 the remainder will be 7.
Step 3: Hence we can say finding remainder of 79644 is equivalent to finding remainder of 7644 when divided by 9.
So we start observing powers of 7.
72 gives remainder 4.
73 gives remainder 1.
Hence after 3rd power remainders will start repeating. So we have a cycle of 3 and thus we divide 644 by 3 i.e. 644= 3(214) + 2
Hence 7 644  -> 7 3(214) + 2 -> 73(214) .72-> 1.4  = 4   (As 73 gives remainder 1 and 72 gives remainder 4)
Hence the remainder of 79644 when divided by 9 is 4.
 
Assignment:
QUESTIONS

1. Find the remainder when 257 is divided by 3?
a)0
b)1
c)2
d) None of these
Solution:
Answer – c
So we start observing powers of 2.
21 gives remainder 2.
22 gives remainder 1.
23 gives remainder 2.
24 gives remainder 1.
Hence we can observe odd power of 2 gives remainder 2 and even power gives remainder 1.
Hence 257 will give remainder 2 when divided by 3.

2. Find the remainder when 391 is divided by 5 ?
a)1
b)2
c)3
d)4
Solution:
Answer-b
So we start observing powers of 3.
31 gives remainder 3.
32 gives remainder 4.
33 gives remainder 2.
34 gives remainder 1.
Hence after 4th power remainders will start repeating. So we have a cycle of 4 and thus we divide 91 by 4 and 391 will give same remainder as 33 i.e. 2.
Hence the remainder of 391 when divided by 5 is 2.

3. What is the remainder when 1548 is divided by 4?
a)1
b)2
c)3
d)0
Solution :
Answer – b
Finding remainder of 1548 is equivalent to finding remainder of 348 when divided by 4.
So we start observing powers of 3.
31 gives remainder 3.
32 gives remainder 1.
33 gives remainder 3.
34 gives remainder 1.
Hence we can observe odd power of 3 gives remainder 3 and even power gives remainder 1.
Hence 348 will give remainder 1 when divided by 4.
Hence the remainder of 348 when divided by 4 is 1.

4. Find the remainder when 1348is divided by 9?
a)1
b)2
c)3
d)4
Solution :
Answer- a
Finding remainder of 1348 is equivalent to finding remainder of 448 when divided by 9.
So we start observing powers of 4.
41 gives remainder 4.
42 gives remainder 7.
43 gives remainder 1.
Hence after 3rd power remainders will start repeating. So we have a cycle of 3 and we know 48 is divisible by 3 so answer will same as remainder of 43.
Hence the remainder of 1348 when divided by 9 is 1.

5. What is the remainder when 2871 is divided by 7?
a)1
b)2
c)3
d)0
Solution :
answer – d
Since 7 is a factor of 28 hence remainder will be zero.

Remainders (Part-5)

In this article we will discuss calculating remainders that involve hefty mathematical expressions. We will discuss each and every mathematical expression one by one.

Suppose we have two numbers “a” and “b”. If “a” is in form of 7n +4 and “b” is in form of 7m+1.

Addition
Find remainder when a+b is divided by 7?
Given a = 7n +4
and b = 7m + 1
So when we add them we will have a +b = 7n + 7m + 4 + 1 = 7 (n + m) + 5
That means we have a multiple of 7 and a remainder 5, because five is smaller than the 7 so it cannot be divided further and hence this 5 is our remainder.

Difference
What will be the remainder when a – b is divided by 7 ?
Given a = 7n +4
and b = 7m + 1
So when we subtract them we will have a – b= (7n +4) – (7m + 1)
= 7(n – m ) + 4 – 1
=  7(n – m ) + 3.
That means we will have a multiple of 7 with addition of 3 and because three
is smaller than the 7 so it cannot be divided further and hence this 3 is our remainder .
 
Multiplication
What will be the remainder when a x b is divided by 7 ?
Given: a = 7n +4
and b = 7m + 1
So when we multiply them we will have a x b  = (7n +4) x (7m + 1)
= 7n(7m + 1)  + 4(7m + 1)
= 49nm + 7n  + 28nm + 4. Here we can see that each and every number is a multiple of 7 except 4. Hence this 4 is our remainder.
Let’s see application of above in problems

Example
What will be the remainder when
(576 + 453 -34 – 566 +6734 x 234 x 345 x 4564) is divided by 8 ?
Solution:
So to find the remainder we will not solve all the expression we will divide it one by one separately and apply same operations on remainders as we have in the numbers.
First divide 576 by 8 and remainder is 0
Then divide 453 by 8 and remainder is 5
Then divide 34 by 8 and remainder is 2
Then divide 566 by 8 and remainder is 6
Then divide 6734 by 8 and remainder is 6
Then divide 234 by 6 and remainder is 2
Then divide 345 by 8 and remainder is 1
Then divide 4564 by 8 and remainder is 4
Now apply same operations between remainders and solve
0 + 5 -2 -6 +6 x 2 x 1 x 4
= 45
Now 45 is greater than the 8 so we will again divide the result with 8.
So divide 45 by 8 gives remainder 5 so the remainder of the whole expression is 5.

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