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My 11-year-old son has made the following conjecture, and wants to know if it's already known to be true or false:

if n is a whole number, define an operation #(n) = 1+2+3…..+n (like factorial but with addition) The conjecture: if n is odd, then #(n-1) = X*n (where X may be any natural number). That is, #(n-1) should be evenly divisible by n.

Olórin
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Mike Levin
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    When I turned 15 a relative gave me a check for 120 dollars on my birthday. I asked where the amount came from, since it was unusual (not like a more round figure of 100 dollars). I was told that 120 is "15 sumtorial," i.e., 1+2+3+...+15. OK. Then the relative told me that, knowing I like math, the original plan was to give me 15 factorial dollars until realizing how huge such an amount is. – KCd Feb 16 '15 at 17:01

3 Answers3

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$2\#(n)=(1+2+\cdots+n)+(n+n-1+\cdots+1)=(1+n)+(2+n-1)+\cdots+(n+1)=(n+1)+(n+1)+\cdots+(n+1)=n(n+1)$

So $\#(n)=\frac{1}{2}n(n+1)$.

Now $\#(n-1)=\frac{1}{2}(n-1)n$

This is divible by $n$ if $\frac{1}{2}(n-1)$ is an integer.

If $n$ is odd then $n-1$ is even so $\frac{1}{2}(n-1)$ is indeed an integer, therefore the conjecture is true!

Uncountable
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Hint: $1+2+...+n=\sum\limits_{k=1}^nk=\dfrac{n(n+1)}{2}$

Scientifica
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That's a great observation for an 11 year old to make!

I'm not sure this is a proof (perhaps someone can check) but I think I can justify it.

Because $n$ is odd there are an even number of numbers in the sum you're considering.

This means they can always be paired up.

It remains to show that the sum is divisible by $n$ to do this I choose to pair them up outside in.

Notice that each pair totals $n$.

(It always will be if we step inwards by the same amounts from both ends)

Here is a specific example $n=7$ $$1+2+3+4+5+6+7$$

Pair up $$1+6 \\ 2+5 \\ 3+4$$ All the pairs sum to $7$

Karl
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