Sum of all integers from $1$ to $20$ except one is a multiple of $20$ 
Sam was adding the integers from $1$ to $20$. In his rush, he skipped one of the numbers and forgot to add it. His final sum was a multiple of $20$. What number did he forget to add?

My idea was to use Gauss's trick to find this relatively simply so I proceeded as follows.
We have $S=1+2+3+ \dots+ 18+19+20$. Using Gauss's trick we get $\frac{n(n+1)}{2} = \frac{20(21)}{2} = 210$. Since we want this to equal some multiple of $20$ we have that $210 = 20n$, but solving for $n$ results in $\frac{21}{2} = 10.5$.
The correct answer for this was $10$, but it seems that I'm missing something?


but it seems that I'm missing something?

Oh, I can't resist. 
You aren't missing anything.  That's the problem.
(....laughs in the corner to himself for hours....)
You were supposed to skip a number but you included them all.
The numbers add to $1+2+3 + 4+.... + 20 = 210$ and you did that correctly.  But Sam did not do it correctly.  Sam left out a number.  So Sam did not get $210$.
So what did Sam get.  If the number he skipped was $k$ then same got $1 + 2 + 3 + .... + (k-1) + (k+1) + .... +20 = 210 - k$.
So we have $210 - k = 20n$.  Now $k$ can be as small as $1$ and is $210-k = 20n$ can be as big as $209$.  And $k$ can be as big as $20$ so $210-k = 20n$ can be as small as $190$.  So $190 \le 210-k = 20n \le 209$.  Then only number in that range that is divisible by $20$ is ... $200$.  So $20n = 210 -k = 200$. And that means $k$ is .... $10$.
And indeed, if $1 + 2 + 3 + 4 + ..... + 9 + 11 + 12 + 13 + .... + 20 = $
$(1 + 2+ .... + 9) + (11+12 + .... + 20) = $
$(\underbrace{1 + \underbrace {2 + .... +8}+9}) + (\underbrace{11 + \underbrace {12 + .... +19}+20})=$
$(4*10+5) + (5*31) =$
$45 + 155 = 200$.
Then $9+8+7+4 +..... + 1 + 20 + 12 + 13 + .... + 20 = N$ and 
$10+10+10+ .... + 10 + 31 + 31+.... + 31 = N+N$
$90 + 310 = 2N$
$400 = 2N$
$N = 200$.
....
More elegant, if you know modular arithmetic is:
$20$ divides $(1+2+3+ ..... + 20) -k$ so
$(1+2+3 + ..... + 20) -k \equiv 0 \pmod {20}$
So $210 -k \equiv 0 \pmod {20}$
$k \equiv 210 \equiv 10 \pmod {20}$.
And as $1\le n \le 20; k \equiv 10 \pmod{20}$ then $k =10$.
