What is the remainder when 5^68 is divided by 7?

A) 1
B) 2
C) 3
D) 4
E) 6

\(\frac{5^{68}}{7} = \frac{( 7 - 2 )^{68}}{7}\)

\(\frac{2^1}{7}\)= Remainder 2
\(\frac{2^2}{7}\)= Remainder 4
\(\frac{2^3}{7}\)= Remainder 1

\(\frac{2^4}{7}\)= Remainder 2
\(\frac{2^5}{7}\)= Remainder 4
\(\frac{2^6}{7}\)= Remainder 1

So, We have a pattern here, the cyclicity is of 3.

\(\frac{68}{3}\) = Remainder is 2

Thus, the remainder corresponding to the value will be 4, answer must be (D) 4
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\(\frac{5^{68}}{7}=\frac{(-2)^{68}}{7}=\frac{2^{68}}{7}=\frac{(2^3)^{22}\times 2^2}{7}
=\frac{8^{22} \times 2^2}{7}=\frac{1\times 2^2}{7}=4\)

The answer is D.
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We can do it by expanding the number .. {} remainder value in the below explanation
{5^68 / 7}= {25^34 /7} = {4^34 /7} = {16^17 /7 } ={2^17/7} = {((2^3)^5 * 2^2) /7} = {((8)^5 * 2^2) /7} = 1 * 4 = 4

So Remainder is 4.

The expansion can be written as (126-1)^22 *5^2 divided by 7.

This approach would be very simple and easy to solve this type of questions;

Take 5^3 = 125 and restructure the question like this : (5^3)^22 * 5^2 divided by 7.

The main idea is to expand the numerator with a number close to 7 or multiples of 7 by 1.

If I take 5^3 =125 = 126-1. Here 126 is divisible by 7 or a multiple of 7.

Now it becomes very simple. If we expand this (126-1)^22 using binomial expansion, the remainder term would be (-1)^22, which is 1.

So we left with 5^2 divided by 7 which is 25/7 and the remainder is 4.

Hence always try to expand the numerator such that we convert a big part of the problem into 1 and then find the remainder on the remaining part.

Thanks

Yes Mansoor, you are absolutely correct with your approach
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Hello. Could you please tell me how this approach called? I still don't understand it. I thought we can use cyclicity of a unit digit to find out the remainder. 5 has 1 cyclicity and 5 to any power will end with 5. Therefore, 5/7 where remainder should be 5. Please help me to understand. Thank you.