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

Question

Sadie Alexander · Accepted Answer

See proof below

#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# # herefore 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#

Theodore Amidon · Answer

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#

Anna Allen · Answer

undefined To prove that $1^{99} + 2^{99} + 3^{99} + 4^{99} + 5^{99}$ is divisible by 5, we can use the fact that for any integer $n$, $n^5 - n$ is divisible by 5.

Now, consider the expression modulo 5:
$$
\begin{align*}
1^{99} + 2^{99} + 3^{99} + 4^{99} + 5^{99} &\equiv 1^{99} + 2^{99} + 3^{99} + 4^{99} + 0^{99} \pmod{5} \
&\equiv 1^5 + 2^5 + 3^5 + 4^5 + 0 \pmod{5} \
&\equiv 1 + 32 + 243 + 1024 + 0 \pmod{5} \
&\equiv 1 + 2 + 3 + 4 + 0 \pmod{5} \
&\equiv 10 \pmod{5} \
&\equiv 0 \pmod{5}
\end{align*}
$$

Therefore, $1^{99} + 2^{99} + 3^{99} + 4^{99} + 5^{99}$ is divisible by 5.