# Remainder when dividing by 49.

• May 29th 2011, 01:39 PM
rtblue
Remainder when dividing by 49.
Question:

Let $\displaystyle a_n=6^n+8^n$.

Determine the remainder upon dividing $\displaystyle a_{83}$ by 49.

My attempt:

I thought that I might be able to get somewhere using the binomial theorem, but I'm not entirely sure if this is the right approach. This is what I was thinking about:

We're looking for the remainder that is obtained when $\displaystyle 6^{83}+8^{83}$ is divided by 49. We can express this as:

$\displaystyle \displaystyle(6+8)^{83}-(\sum_{n=1}^{82}\binom{83}{n}6^{83-n}8^{n})$ all divided by 49 (I would have included the division by 49 in the latex, but it was messing up the summation symbol. Sorry about that).

I'm not sure where to go from here. Any help would be appreciated!
• May 29th 2011, 03:18 PM
Also sprach Zarathustra
Quote:

Originally Posted by rtblue
Question:

Let $\displaystyle a_n=6^n+8^n$.

Determine the remainder upon dividing $\displaystyle a_{83}$ by 49.

My attempt:

I thought that I might be able to get somewhere using the binomial theorem, but I'm not entirely sure if this is the right approach. This is what I was thinking about:

We're looking for the remainder that is obtained when $\displaystyle 6^{83}+8^{83}$ is divided by 49. We can express this as:

$\displaystyle \displaystyle(6+8)^{83}-(\sum_{n=1}^{82}\binom{83}{n}6^{83-n}8^{n})$ all divided by 49 (I would have included the division by 49 in the latex, but it was messing up the summation symbol. Sorry about that).

I'm not sure where to go from here. Any help would be appreciated!

Hint:

$\displaystyle 6^{83}+8^{83}=(7-1)^{83}+(7+1)^{83}$