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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?

user21820
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4 Answers4

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$210$ minus the missing number (call it $m$) $=20n$.

$210-m=20n$, where $1\le m \le 20$.

Can you take it from here?

J. W. Tanner
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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.

Bram28
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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 $20$.

Toby Mak
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    Related insight: from $\frac{n(n+1)}{2}$, if $n$ is odd, the sum is a multiple of $n$; if it's even, the sum is a multiple of $\frac{n}{2}$, so certainly subtracting $\frac{n}{2}$ will also leave a multiple of $\frac{n}{2}$ – MichaelChirico May 04 '20 at 03:59
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    Your argument proves that removing $10$ works, but it does not prove that removing some other number can't work, so your "must remove" is not justified. But it is easy to extend it to show that no other removal works. This is a special case of the additive form of Wilsons's theorem – Bill Dubuque May 05 '20 at 01:36
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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 $k =10$.

fleablood
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