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 $n(n+1)\bmod 10 \neq 8$.

Here's how I answered this question:
Through induction, I first showed that S(4)=1 (not 2, 4, 7, 9).
Then I went to show S(k)--> S(k+1)


*

*Show that since (k+1)(k+2)/2 cannot end in 2, 4, 7, 9, then (k+1)(k+2) cannot end in 4 or 8.

*Show that anytime you multiply 2 numbers, the last digit of each number is the only necessary information needed to find the last digit of the product.  (2x4=8 12x14=48, etc)

*Therefore the last digit of (k+1) and (k+2) cannot be 1 and 4 or 8, or 2 and 2 or 4. (1x4=4, 1x8=8, 2x2=4, 2x4=8)  

*And since (k+1) and (k+2) has a difference of 1, then the previously mentioned combinations can never be met.

