1. ## congruence

The sum of the digits of the number 8215 is 8+2+1+5=16 congruent to 7 (mod)9. Observe also that 8215=9(912) +7 congruent to 7 (mod)9. Does this hold for any number?Is the congruence class of any integer (mod 9) the same as the sum of its digits (mod) 9?

2. Yes. To sum it up, every integer is congruent to the sum of their digits modulo 9. With congruences, this ends up being pretty easy to prove.

Let $n$ be some integer with the representation of its digits being: $d_md_{m-1}d_{m-2}\cdots d_1d_0$

Note two things:
(1) $\color{red} 10 \equiv 1 \ (\text{mod }9)$
(2) $n$ can be written as a sum of the powers of 10 (as with any other integer): $n = d_m10^m + d_{m-1}10^{m-1} + d_{m-2}10^{m-2} + \cdots + d_110^1 + d_010^0$

Now we apply congruences:
\begin{aligned}n & \equiv d_m{\color{red}10}^m + d_{m-1}{\color{red}10}^{m-1} + d_{m-2}{\color{red}10}^{m-2} + \cdots + d_1{\color{red}10}^1 + d_0{\color{red}10}^0 & (\text{mod } 9) \\ & \equiv \qquad \cdots \qquad \cdots \qquad \cdots \qquad \cdots \qquad \cdots \qquad \cdots \qquad \cdots & (\text{mod } 9) \end{aligned}

Can you figure it out from here?

3. ## Appreciate the help

I am not one of those people for whom math comes easy. When the prof. is writing these things in the board, I think I have it - only to get befuddled when I try to work on my own.