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For example:

$10 = 1 + 2 + 3 + 4$

$11 = 1 - 2 + 3 \times 4$

$12 = 3 \times 4$

$13 = -(1 - 2) + 3 \times 4$

$14 = 2 + 3 + 4 + 5$

$15 = 1 + 2 + 3 + 4 + 5$

$16 = (2/3)(4/5)(6 + 7 + 8 + 9)$

$17 = 1 + (2/3)(4/5)(6 + 7 + 8 + 9)$

$18 = (1 + 2)3!$

$19 = 4! - 5$

Obviously any triangular number has such an expression, as well as numbers one less than a triangular number. But how to prove it for other numbers? Or is there a counterexample?

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    $\begingroup$ $n=(n+1)+(n+2)-(n+3)$ ? $\endgroup$ May 26, 2014 at 17:32
  • $\begingroup$ So painfully simple...!+1 $\endgroup$
    – DonAntonio
    May 26, 2014 at 17:39
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    $\begingroup$ I presume the OP means consecutive positive integers strictly smaller than $n$. A fact that you can use: Every number that is not a power of $2$ is a sum of consecutive positive integers (all smaller than itself). For instance, $11 = 5 + 6$, $12 = 3 + 4 + 5$, $13 = 6 + 7$, $17 = 8 + 9$, $18 = 5 + 6 + 7$, $19 = 9 + 10$, etc. (For proof, see here or here.) So it's enough to ask the (suitably modified) question for powers of $2$. $\endgroup$ May 26, 2014 at 17:42
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    $\begingroup$ Why do you presume such a thing, @ShreevatsaR ? Some of the examples of the OP don't even use integers...or even positive integers. $\endgroup$
    – DonAntonio
    May 26, 2014 at 17:48
  • $\begingroup$ $(2-1)+(2-1)+(2-1)...$ $n$ times $\endgroup$ May 26, 2014 at 17:56

1 Answer 1

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Every integer $n$ can be written as $$n= (n+1)+(n+2)-(n+3)$$ where $(n+1)$ and $(n+2)$ are of course two consecutive integers.

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  • $\begingroup$ Rather, $(n+1)$, $(n+2)$ and $(n+3)$ are three consecutive integers. +1. $\endgroup$ May 27, 2014 at 2:30

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