# Finding Remainders - Practice Test (1)

## 5 Questions MCQ Test UPSC Prelims Paper 2 CSAT - Quant, Verbal & Decision Making | Finding Remainders - Practice Test (1)

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This mock test of Finding Remainders - Practice Test (1) for UPSC helps you for every UPSC entrance exam. This contains 5 Multiple Choice Questions for UPSC Finding Remainders - Practice Test (1) (mcq) to study with solutions a complete question bank. The solved questions answers in this Finding Remainders - Practice Test (1) quiz give you a good mix of easy questions and tough questions. UPSC students definitely take this Finding Remainders - Practice Test (1) exercise for a better result in the exam. You can find other Finding Remainders - Practice Test (1) extra questions, long questions & short questions for UPSC on EduRev as well by searching above.
QUESTION: 1

### 2525 is divided by 26, the remainder is?

Solution:

Use remainder theorem. Here 2525 is the polynomial and the divisor is 26.We can write 26 = 25+1 = 25 – ( -1)So the remainder is ( -1)25= -1. But we don’t take the remainder a negative term; so add it to the divisor.So the remainder is 26 +( -1) = 25 (option ‘D’)

QUESTION: 2

### When (6767 + 67) is divided by 68, the remainder is?

Solution:

​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’)

QUESTION: 3

### What is the remainder when [(919) + 6]  is divided by 8

Solution:

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’)

QUESTION: 4

Find the remainders in
211/5

Solution:

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’)

QUESTION: 5

Find the Remainder

77/24

Solution:

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’)