I thought this class would be fun but it turned out to be not the thing for me. It's too late to drop the course so I'm stuck doing homework that I don't understand.
Any help with any of these would be appreciated.
Hello qtpipi(a) Suppose that $\displaystyle n$ is represented by the sequence of digits $\displaystyle a_na_{n-1}...a_1$, and that $\displaystyle s = a_n + a_{n-1}+...+a_1$
Then $\displaystyle n = 10^{n-1}a_n + 10^{n-2}a_{n-1} + ... + a_1$
$\displaystyle \Rightarrow n-s = (10^{n-1}-1)a_n +(10^{n-2}-1)a_{n-1} + ... +(1-1)a_1$
Now $\displaystyle 10 = 1 \mod 9 \Rightarrow 10^i = 1 \mod 9,\forall i \in \mathbb{N}$
$\displaystyle \Rightarrow 10^i-1 = 0 \mod 9$
$\displaystyle \Rightarrow n-s = 0 \mod 9$
$\displaystyle \Rightarrow n$ and $\displaystyle s$ leave the same remainder when divided by $\displaystyle 9$
$\displaystyle \Rightarrow 3|n \iff 3|s$ (and of course, $\displaystyle 9|n \iff 9|s$ as well).
(b) In any three consecutive integers, $\displaystyle (x-1), x, (x+1)$, one is always a multiple of $\displaystyle 3$. And $\displaystyle (x-1)$ and $\displaystyle (x+1)$ are primes, and $\displaystyle (x-1) \ne 3 \Rightarrow$ neither $\displaystyle (x-1)$ nor $\displaystyle (x+1)$ is a multiple of $\displaystyle 3$
$\displaystyle \Rightarrow 3|x$
And, of course, $\displaystyle x$ is even whenever $\displaystyle (x-1)$ is a prime $\displaystyle > 2$.
Hence $\displaystyle 6|x$
(c) Consider the remainders when (x-1) and (x+1) are divided by 3, and then use the results of (a).
Grandad