Prove that 1^99+2^99+3^99+4^99+5^99199+299+399+499+599 is divisble by 5.?

2 Answers

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

Jul 28, 2018

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