Reminder of Remainders

Number Theory Level 2

What is the remainder when

$\large 1^{7}+2^{7}+3^{7}+\cdots + 99^{7}+100^{7}$

is divided by 4?

The answer is 0.

1 solution

Adarsh Kumar
Sep 8, 2014

$For\ understanding\ this\ you\ will\ need\ to \ know\ that\ a^{m}+b^{m}\ is\ divisible\ by \ a+b\ for\ odd\ m.$ $\Longrightarrow$ $1^{7}+99^{7}\ is\ divisible\ by\ 100$ $\Longrightarrow$ $1^{7}+99^{7}\ is\ divisible\ by\ 4.$ $Same\ with\ (2,98)\ (3,97)\ (4,96) .......(49,51)$ $Thus\ we\ just\ need\ to\ find\ the\ remainder\ when\ 100^{7} \ is\ divided\ by\ 4.$ $Which\ is\ zero.$ $Thus,the\ answer\ is\ 0.$

Very true.. what can be done otherwise ;) ? (think )

Krishna Ar - 6 years, 9 months ago

$1^7+2^7+3^7+\cdots+99^7+100^7$

$\equiv 33(1^7+2^7+3^7)$

$\equiv 1^7+2^7+(-1)^7\equiv 2^7\equiv \boxed{0}\pmod {4}$

mathh mathh - 6 years, 9 months ago

Exactly. You're super cool!

Krishna Ar - 6 years, 9 months ago

what did you do in first step?

Manish Mayank - 6 years, 6 months ago

@Manish Mayank – He took modulo $4$ . $4^7,~5^7,~6^7,\cdots$ become $1^7,~2^7,~3^7,\cdots$ .

Omkar Kulkarni - 6 years, 3 months ago

Reminder of Remainders

The answer is 0.

1 solution

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