Question

Prove that `1^99+2^99+3^99+4^99+5^99` is divisble by 5.?

Answer 1

See proof below

Explanation:

`1=1\mod 5`

`1^{99}=1^99\mod 5`

`1^{99}=1\mod 5`

Similarly,

`2^{99}=2^{99}\mod 5`

`3^{99}=(-2)^99\mod 5`

`4^{99}=(-1)^99\mod 5`

`5^{99}=0\mod 5`

`\therefore 1^99+2^99+3^99+4^99+5^99`

`=(1+2^99+(-2)^99+(-1)^99+0)\ mod 5`

`=(1+2^99-2^99-1)\ mod 5`

`=0\ mod 5`

hence the given number is divisible by `5`

Answer 2

We Know by divisibility rule that `a^n+b^n` is divisible by `a+b` when `n` is odd. Since inserting `a=-b` the value of `a^n+b^n= (-b)^n+b^n=0`

For similar reason in our problem

`1^99+4^99` is divisible by `1+4=5`

Again `2^99+3^99` is divisible by `2+3=5`

And `5^99` is divisible by `5`

Hence the sum `1^99+2^99+3^99+4^99+5^99` must be divisible by `5`