# CAT Remainder Theorem Questions MBA Preparation

# Most Important CAT Remainder Theorem Questions

Questions like 21 divided by 25, 101 divided by 16, 300 divided by 25, 222 divided by 6, 222 divided by 332, 124 divided by 160, 30 divided by 325, 99 divided by 9 ate very familiar to those who have appeared for CAT, GMAT or GRE. Here is the article which will help you better understand this concept and clear of your doubts on Remainder Theorem.

### CAT- Remainder Theorem Questions:

CAT focuses on 3 sections and among them Quantitative Aptitude is highly interesting and scoring of all. Quantitative aptitude includes very basic concepts that one have done in their 10th , 12th. You just need to have a conceptual clarity to crack this section with 99 percentile. One of the very interesting topic in this section which has high probability to appear in the exam is **Remainder Theorem** Questions. Here are most important question types you must go through:

# Remainder Theorem Questions :

**Question 1.** After the division of a number successively by 3, 4 and 7, the remainders obtained are 2, 1 and 4 respectively. What will be the remainder if 84 divides the same number?

(a) 80

(b) 75

(c) 41

(d) 53

**Answer is d:**

**Explanation:** As the number gives a remainder of 4 when divided by 7, then the number must be in form of (7 x+4).

The same gives remainder 1 once it divided by 4, so the number must be in the form of {4*(7 x +4)+1}.

Also, the number when divided by 3 will gives remainder 2, thus number must be in the form of [3*{4*(7 x+4)+1}+2]

Now, on simplifying, we get 84 x+53

Hence, we get our final answer 53. So, if the number is divided by 84, the remainder will be 53.

**Question 2.** How many positive integers are there from 0 to 1000 that leave a remainder of 3 on division by 7 and a remainder of 2 on division by 4?

(a) 32

(b) 36

(c) 24

(d) 19

**Answer is b:**

**Explanation:** Firstly, when a number is divided by 4 and leaves a remainder of 2 then the number is definitely a multiple of 2 ,

Now when the same number is divided by 7, it leaves the remainder of 3 so, if we subtract the original even(explained above) no, then we will get an odd number.

If we carefully observe, this two condition only met when you subtract 2 from the parent number you will have a multiple of 4 , and subtract 3 from the parent number you will have a multiple of 7.

So you have a consecutive pair… Like : 38 , subtract 3 from it you get 35 which is divisible by 7 and subtract 2 from it you get 36 which is divisible by 4.[ 35 and 36 is that pair ]

Here you get

7(2 k+1)+1 = 4 x

or, 4 x = 14 k + 8

x = 7/2 k + 2

As x is an integer , hence k has to be an even number.

Now , up to 1000 , let’s see what can be maximum value of k ,

7(2 k+1)<= 1000

Or , k <= 70 [ you put 71 , you get 1001>1000]

So total number of even k possible within this range is

= 36 [ 0 to 70 you have total 36 even numbers]

Hence…total 36 numbers are there within 1000 [like 10,38,66… And so on]

**Question 3.** Find the sum of all 2 digit numbers that leaves a remainder of 3 when divided by 7

(a) 503

(b) 676

(c) 777

(d) 832

**Answer is b:**

**Explanation:** Because remainders of consecutive whole numbers form a repeating consecutive list, we know that all numbers having a certain remainder will form an arithmetic sequence. This sequence starts as follows (remember, we just want two-digit numbers):

10, 17, 24, 31, …

To find out the last number in the sequence, find the remainder of 99 ÷ 7, which is 1. So, working backward, the last number in the sequence whose remainder will be 3 must be 94.

Then you can use the formula for a sum of an arithmetic sequence. How many terms will be there? well if *a *_{1} = 10 = 7(1) + 3, and *a _{n}* = 7(

*n*) + 3 = 94, then solve to get

*n*= 13.

Sum = (*n*/2) (*a *_{1 } + *a _{n}*) = (13/2)(10 + 94) = 676.

**Question 4: **What will be the remainder when (16*27+37) is divided by 17?

(a) 2

(b)1

(c) 3

(d) 0

**Answer is a:**

**Explanation:** [Rem [(16*27+37)/17]

= Rem [16*27/17] + Rem [37/17]

= Rem [(-1)*27/17] + 3

= -1 + 3

= 2

**Question 5:** **What is the remainder when 97! divided by 101?**

(a) 5

(b) 9

(c) 6

(d) 1

**Answer is c:**

**Explanation:** Lets take the remainder from 1 to 97 when divided by 101

1*2*3*…*97/101

[Rem(1/101)+Rem(2/101)+….+Rem(97/101)]Rem[(1*2*3*…*97)/101]

Rem(4753/101)

6 is the answer.

**Question 6: What is the remainder of 21^21^21…21 divided by 25?**

(a) 20

(b) 15

(c) 50

(d) 21

**Answer is d:**

**Explanation:**

remainder (rem) if 21^21^21…21 divided by 25

21^1 rem 21

21^2 rem 16

21^3 rem 11

21^4 rem 6

21^5 rem 1

21^6 rem 21

21^7 rem 16

……

like that 21^21 rem 21

so 21^21^21 rem will be 21

finally remainder for 21^21^21….21 divided by 25 is 21

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