# CAT Exam: 13 Short Cuts for Number System

**(i)** Every number has the same unit’s digit at its fifth power as it has at its first power, thus the standard method that can be followed is to divide the power given by 4, find the remaining power and check the unit’s digit in that number. This short cut can be applied because you will always get the same unit’s digit as otherwise.

**(ii)** For calculating number of zeroes at the end of factorial of a number, you should divide the number by 5, the quotient obtained is again divided by 5 and so on till the last quotient obtained is smaller than 5. The sum of all the quotients is the number of 5s, which then becomes the number of zeroes in the given number.

**(iii)** The digital root of a number is the sum of the digits, over and over again, till it becomes a single digit number. For example, the digital root of 87984 will be 8 + 7 + 9 + 8 + 4 ⇒ 36 = 3 + 6 ⇒ 9.

**(iv)**When the concept of Euler number is used and the dividend and divisor happen to be co-prime, the remainder questions become very easy.

**(v)** The product of 3 consecutive natural numbers is divisible by 6.

**(vi)** The product of 3 consecutive natural numbers, the first of which is an even number is divisible by 24.

**(vii)** The sum of a two-digit number and a number formed by reversing its digits is divisible by 11. E.g. 28 + 82 = 110, this is divisible by 11. At the same time, the difference between those numbers will be divisible by 9. e.g. 82 – 28 = 54, this is divisible by 9.

**(viii)** ∑n = n(n+1)/2, ∑n is the sum of first n natural numbers.

**(ix)** ∑n^{2} = n(n+1)(n+2)/6, ∑n^{2} is the sum of first n perfect squares.

**(x)** ∑n^{3} = n^{2}(n+1)^{2}/4 = (∑n)^{2}, ∑n^{3} is the sum of first n perfect cubes.

**(xi)** x^{n} + y^{n} = (x + y) (x^{n-1} – x^{n-2}.y + x^{n-3}.y^{2} – … +y^{n-1}) when n is odd. Therefore, when n is odd, x^{n} + y^{n} is divisible by x + y. e.g. 3^{3} + 2^{3} = 35 and is divisible by 5, which is (3 + 2).

**(xii)** x^{n} – y^{n} = (x + y) (x^{n-1} – x^{n-2}.y + … y^{n-1}) when n is even. Therefore, when n is even, x^{n} – y^{n} is divisible by x + y. e.g. 7^{2} – 3^{2} = 40, is divisible by 10, which is (7+3).

**(xiii)** x^{n} – y^{n} = (x – y) (x^{n-1} + x^{n-2}.y + …. + y^{n-1}) for both odd and even n. Therefore, xn – yn is divisible by x – y. For example: 9^{4} – 2^{4} = 6545, is divisible by 7, which is (9 – 2).