a congruence problem

[afif mansib ch]

what is the remainder when the following sum is divided by 7?
\[1^7+2^7+3^7+...+100^7\]

[sourav das]

Use Farmet's little theorem

[Abdul Muntakim Rafi]

Yeah... 3

[nayel]

Follow-up: What is the remainder of the above sum upon division by $8$?

[Abdul Muntakim Rafi]

Bhaiya, did u mean what is the remainder when
$1^7+ 2^7+ 3^7 +.............................. + 100^7 $
is divided by 8?

or,

What is the remainder when

$1^8+ 2^8 + 3^8 + ....................................... + 100^8 $

is divided by 8?

[nayel]

I meant the first one. But since you mentioned you can do both!

[afif mansib ch]

yaa i found the ans of my own pr.
\[1^7+2^7+....+100^7\equiv (1+2+...100)(mod7)\equiv 3(mod7)\]
so the ans is 3

[afif mansib ch]

ok, i can't solve nayel via's probs.need help.

[Abdul Muntakim Rafi]

\[1^7+2^7+3^7+......................+100^7\equiv ?(mod 8)\]

Here is my approach... First we divide the sequence in two parts... Even numbers and odd numbers...

Any even number is of the form $2n$ ...
\[(2n)^7=2^7.n^7=2^3.2^4.n^7=8.2^4.n^7\]
Clearly all the 7th power of even numbers are divisible by 8...
So,
\[1^7+2^7+3^7+......................+100^7\equiv ?(mod 8)\] is same as

\[1^7+3^7+5^7+......................+99^7\equiv ?(mod 8)\]

And we can see that
\[(2n+1)^7\equiv 2n+1 (mod 8)\]

That's why
\[1^7+3^7+5^7+......................+99^7\equiv ?(mod 8)\] is same as
\[1+3+5+......................+99\equiv ?(mod 8)\]
=2500
\[2500\equiv 4 (mod 8)\]

I think the way is right... Though there might be some mistake in calculation... (I am likely to make mistakes in calculation)
And I have invented some theorems while solving this problem...
Unfortunately maybe some mathematicians have found out them earlier... Damn those...

[afif mansib ch]

ok. beeeeeeaaautifullllllll sol
bt what's the pr with my one?
i'm gonna post the sol later.