Question

Show that
${1}^{99} + {2}^{99} + {3}^{99} + {4}^{99} + {5}^{99 }$
is divisible by 5

Answer 1

(1+2+3+4+5)=15 to the power 99 which is divisible by 5

Answer 2

in the given expression,
${1}^{99} + {2}^{99} + {3}^{99} + {4}^{99} + {5}^{99 }$
the number (this, down below) is already divisible by 5.
${5}^{99}$
so we need to only show if
${1}^{99} + {2}^{99} + {3}^{99} + {4}^{99}$
is divisible by 5 or not.
here, note that
$1 + 4 = 5 \\ {1}^{2} + {4}^{2} = 1 + 16 = 17 \: not \: divisible \\ {1}^{3} + {4}^{3} = 1 + 64 = 65 \\ .......$
so
${1}^{n} + {4}^{n}$
is only divisible by 5 if n is odd.
Same rule will hold for
${2}^{n} + {3}^{n}$
and since 99 is obviously odd, so the expression
${1}^{99} + {2}^{99} + {3}^{99} + {4}^{99} + {5}^{99 }$
is divisible by 5