Finding Remainders - Practice Test (1) UPSC Prelims Paper 2 CSAT MCQs

(xⁿ + aⁿ) is divisible by (x + a) when n is odd
∴ (25²⁵ + 1²⁵) is divisible by (25 + 1)
⇒ (25²⁵ + 1) is divisible by 26
⇒ On dividing 25²⁵ by 26, we get (26 - 1) = 25 as remainder

​The given expression is in the form x⁶⁷ + x ……….(a polynomial in x)
Now 68 = 67 + 1; means x +1

So according to the remainder theorem when a polynomial is divided by another of the form          x + 1, the remainder is equal to p(­1) where p is the polynomial itself.
So the remainder is ­1⁶⁷ + (­1) = ­1 + (­1) = ­2

But the remainder should not be described negative of a number; in such a situation it is added to the divisor to find the actual.

So the remainder is ­2 + 68 = 66 (option ‘C’)

The given expression is in the form (x¹⁹) + c; where ‘c’ is a constant ……….(a polynomial in x).

Now 8 = 9 ­ 1; means a polynomial in the form of x ­ 1

So according to the remainder theorem when a polynomial is divided by another of the form     x ­ 1, the remainder is equal to p(1) where p is the polynomial itself. So using remainder theorem, the remainder is (1¹⁹) + 6 = 1 + 6 = 7 (option ‘B’)

1. 2¹¹/5
In questions like this we should avoid using the remainder theorem as it can really be difficult when the power of a number (greater than 1) which is derived from the remainder theorem is so high. Better convert the base in powers of such numbers which are easily divisible by the divisor, like:

2¹¹/5 = 2⁴ x 2⁴ x 2³= 16 x 16 x 8

On dividing 16 by 5 we get 1 as the remainder; and if 8 is divided by 5 we get 3
So the multiplication of all the remainders
= 1 x 1 x 3 = 3 which is our answer (option ‘A’)

7⁷/2⁴ = 7² x 7² x 7² x 7/16
= 49 x 49 x 49 x 7/16
Now the remainder on dividing 49 by 16 =1

The multiplication of all the remainders 1 x 1 x 1 x 7 = 7 (option ‘C’)