If $S_n = 1+ 2 +3 + \cdots + n$, then prove that the last digit of $S_n$ is not 2,4 7,9. If $S_n = 1 + 2 + 3 + \cdots + n,$ then prove that the last digit of $S_n$ cannot be 2, 4, 7, or 9 for any whole number n.
What I have done:
*I have determined that it is supposed to be done with mathematical induction.
*The formula for an finite sum is $\frac{1}{2}n(n+1)$.
*This means that since we know that $n(n+1)$ has a factor of two, it must always end in 4 and 8.
*Knowing this, we can assume that $n(n+1)\bmod 10 \neq 4$ or $n(n+1)\bmod 10 \neq 8$.

First show that for $n^2$ the last digit will always be from the set $M=1,4,5,6,9,0$ (I don't know how to create those brackets with your version of TeX, \left{ doesn't seem to work).
Then consider all cases for the last digit of $n$ (last digit is a $1$, I get a $2$ as last digit for $n(n+1)= n^2 + n$ and so on). If you do all of this in $mod5$, you only have 5 easy cases to check and show that $n^2 + n$ $mod5 \notin (3,4)$.
