Divisibility by 30 How to show that if $30$ divides $a_1+a_2+ \ldots + a_n$, then $30$ divides $a_{1}^{5}+a_{2}^{5}+ \ldots + a_{n}^{5}$.
Since $30$ divides $a_1+a_2+ \ldots + a_n$, then $a_1+a_2+ \ldots + a_n \equiv 0 \pmod{30}$, thus $(a_1+a_2+ \ldots + a_n )^5 \equiv 0 \pmod{30}$ I don't know.
 A: Induction works. The proof for $n=2$ is identical to the one presented in the general case.
Say $30$ | $a_1 + \dots a_n$ implies $30$ | $a_1^5 + \dots a_n^5$ .
Now $a_1 + \dots + a_{n+1} = a_1 + \dots a_{n-1} + b$  if $b=a_n + a_{n+1}$ therefore $$30 \;|\; a_1^5 + \dots a_{n-1}^5 + b^5 $$ $$30 \; | \; a_1^5 + \dots a_{n-1}^5 + a_n^5 + a_{n+1}^5 + 5 a_n^4 a_{n+1}+10 a_n^3 a_{n+1}^2+10 a_n^2 a_{n+1}^3+5 a_n a_{n+1}^4$$
So to conclude we have to show that $30$ divides $5 a_n^4 a_{n+1}+10 a_n^3 a_{n+1}^2+10 a_n^2 a_{n+1}^3+5 a_n a_{n+1}^4$
Obviously, $5$ divides it. Also this is an even number (sum of even number of odd terms if $a_n$ and $a_{n+1}$ are both odd, if at least one of them is even all are since $a_n$ and $a_{n+1}$ are both present in all terms), so $2$ divides it. 
Now, say $3$ doesn't divide both $a_n$ and $a_{n+1}$ (else we're done).
We have three cases mod $3$:
$a_n=a_{n+1}=1$ : $\quad$
Then $5+10+10+5 = 30 = 0$
$a_n=a_{n+1}=2$ : $\quad$
Then since $2^2 = 1$, we have $10+20+20+10 = 60 = 0$
$a_n=1, a_{n+1}=2$ or its symmetrical : $\quad$
Then $10+10+20+5 = 45 = 0$
So $2,3,5$ divide that, so $30$ does.
