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CGPSC Prelims CSAT 2014 Solved Paper 2

Question : 15 of 100

Marks: +1, -0

(4764)^{1795}

5

. The remainder is

Solution:

Concept:

To find the remainder when a large power is divided by

5

, reduce the base modulo

5

and use the repeating cycle of powers.

Explanation:

First, reduce the base modulo

5

4764 \equiv 4 \pmod{5}

So,

(4764)^{1795} \equiv 4^{1795} \pmod{5}

Now check the powers of

4

modulo

5

4^1 \equiv 4 \pmod{5}

4^2 \equiv 16 \equiv 1 \pmod{5}

Hence, the cycle is

4, 1

and repeats every

2

powers.

Since

1795

is odd,

4^{1795} \equiv 4 \pmod{5}

Therefore, the remainder is

4

Answer:

Option D:

4

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