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

See proof below

#### Explanation:

$1 = 1 \setminus \mod 5$

${1}^{99} = {1}^{99} \setminus \mod 5$

${1}^{99} = 1 \setminus \mod 5$

Similarly,

${2}^{99} = {2}^{99} \setminus \mod 5$

${3}^{99} = {\left(- 2\right)}^{99} \setminus \mod 5$

${4}^{99} = {\left(- 1\right)}^{99} \setminus \mod 5$

${5}^{99} = 0 \setminus \mod 5$

$\setminus \therefore {1}^{99} + {2}^{99} + {3}^{99} + {4}^{99} + {5}^{99}$

$= \left(1 + {2}^{99} + {\left(- 2\right)}^{99} + {\left(- 1\right)}^{99} + 0\right) \setminus \mod 5$

$= \left(1 + {2}^{99} - {2}^{99} - 1\right) \setminus \mod 5$

$= 0 \setminus \mod 5$

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 ${a}^{n} + {b}^{n} = {\left(- b\right)}^{n} + {b}^{n} = 0$

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 $5$