susheelh wrote:

What is the remainder when \(8^1+ 8^2+ 8^3……+8^{15}\) is divided by 6

A. 0

B. 1

C. 2

D. 4

E. 5

\(8^1+ 8^2+ 8^3……+8^{15}\) is same as

\(2^3+ 2^6+ 2^9……+2^{45}\)

What is the remainder when expression above is divided by 6?

To simplify, what is the remainder of

\(2^2+ 2^5+ 2^8……+2^{44}\)

when divided by 3? (Once we find the remainder, remember to multiply it by 2. Since we're factoring out 2 to make the expression easier to work with, we have to re-insert the factor of 2 at the end.)

Pattern:

1st term: r=1

2nd term: r=2

3rd term: r=1

etc

This means that the sum of every 2 terms will have r=0 (when div by 3). So, if there is an even number of terms, r=0; if there is an odd number of terms, r=1.

Total # of terms: \(\frac{44-2}{3}+1=15\)

So r=1 when div by 3. Re-introduce the 2 we factored out earlier, and r=2 when div by 6.

Answer: C