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Finding Remainders - Practice Test (1) - UPSC MCQ


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

Finding Remainders - Practice Test (1) for UPSC 2024 is part of UPSC Prelims Paper 2 CSAT - Quant, Verbal & Decision Making preparation. The Finding Remainders - Practice Test (1) questions and answers have been prepared according to the UPSC exam syllabus.The Finding Remainders - Practice Test (1) MCQs are made for UPSC 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Finding Remainders - Practice Test (1) below.
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Finding Remainders - Practice Test (1) - Question 1

2525 is divided by 26, the remainder is?

Detailed Solution for Finding Remainders - Practice Test (1) - Question 1

(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

Finding Remainders - Practice Test (1) - Question 2

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

Detailed Solution for Finding Remainders - Practice Test (1) - Question 2

​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’)

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Finding Remainders - Practice Test (1) - Question 3

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

Detailed Solution for Finding Remainders - Practice Test (1) - Question 3

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’)

Finding Remainders - Practice Test (1) - Question 4

Find the remainders in
211/5

Detailed Solution for Finding Remainders - Practice Test (1) - Question 4

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’)

Finding Remainders - Practice Test (1) - Question 5

Find the Remainder

77/24

Detailed Solution for Finding Remainders - Practice Test (1) - Question 5

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’)

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