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 $n(n+1)\bmod 10 \neq 8$.
 A: Note that $1+2+3+4+5$ is divisible by $5$, and for the same reason $5k+1+5k+2+5k+3+5k+4+5k+5$ is divisible by $5$ for any integer $k$.
Let $5m$ be the largest multiple of $5$ which is $\le n$. Then $n=5m$ or $n=5m+1$ or $n=5m+2$ or $n=5m+3$ or $n=5m+4$.
The sum of the integers from $1$ to $5m$ is divisible by $5$. So when we sum from $1$ to $n$, the remainder on division by $5$ is $0$, or $1$, or $1+2$, or the remainder when $1+2+3$ is divided by $5$, or the remainder when $1+2+3+4$ is divided by $5$. So the possibilities are $0$, $1$, $3$, $1$, and $0$. In particular, the remainders cannot be $2$ or $4$. So the last digit cannot be $2$, $7$, $4$, or $9$.
A: Hint
$n^2+n \in \{ 0, 2 \} \pmod 4$
$n^2+n \in \{ 0, 1, 2 \} \pmod 5$
By CRT we have only $6$ possible values modulo 20:
$$n^2+n \in \{ 0, 10, 16, 6, 2, 12 \} \pmod {20}$$
A: 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 $n^2 + n$ $mod5 \notin (3,4)$.
A: 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.

