Question

The remainder when 5^99 is divided by 13 is?

Answer 1

Answer:

It simply follows a cycle..

5^1 / 13  - Remainder is 5

5^2 / 13  - Remainder is 12

5^3 / 13  - Remainder is 8

5^4 / 13  - Remainder is 1

5^5 / 13  - Remainder is 5

5^6 / 13  - Remainder is 12

.......... and so on. Hence,

5^96 / 13  - Remainder is 1

5^97 / 13  - Remainder is 5

5^98 / 13  - Remainder is 12

5^99 / 13  - Remainder is 8

Answer : 8

Answer 2

Answer:

5

Explanation

${5}^{99} = 5 \times {5}^{98} = 5 \times ( {5}^{2} ) {}^{49} = 5 \times {25}^{49} \\ {5}^{99} =5 (26 - 1) {}^{49} \\ use \: binomial \\ 5(26 - 1) {}^{49} = 5 \bigg \{ \binom{26}{0} {26}^{0} ( - 1) {}^{26 - 0} + \binom{26}{1} {26}^{1} ( - 1) {}^{25} + .... \bigg \}$

All the terms except first has coefficient of 26 , so they are divisible by 13 .

$5(26 - 1) {}^{49} = 5(13k + 1) = 65k + 5$

So remainder is 5 .