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?
 A: $210$ minus the missing number (call it $m$) $=20n$.
$210-m=20n$, where $1\le m \le 20$.
Can you take it from here?
A: Here is where your approach goes wrong:
The sum of all numbers with the exception of the one skipped one is a multiple of $20$.
So, if $k$ is the skipped number, what you have is: $210-k = 20n$
Also, what you need to solve for is $k$, not $n$.  The fact that in your case, $n$ happened to be fairly close to the $k$ that they were looking for is complete happenstance.
A: You don't actually need to calculate the sum to find the missing number.
Using Gauss's trick again, $1 + 19 = 20$, $2 + 18 = 20$, $3 + 17 = 20$ and so on, all the way up to $9 + 11 = 20$. Observe that each pair is a multiple of $20$.
The numbers which have not been paired up are $10$ and $20$. Of these two numbers, $20$ is a multiple of $20$, but $10$ is not a multiple of $20$.
Therefore we must remove $10$ from the sum in order for all the other numbers form multiples of $20$.
A: 
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 $k =10$.
