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Is it possible to find 2014 distinct positive integers whose sum is divisible by each of them?
I'm not really sure how to even approach this question.
Source: Washington's Monthly Math Hour, 2014

Arvin Ding
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2 Answers2

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Hint: $2,4,6$ are all divisors of $2+4+6=12$. Similarly $2+4+6+12$ are all divisors of $2+4+6+12=24$

So too are $2,4,6,12,24$ all divisors of $2+4+6+12+24$

$~$

Claim: Let $x_1=2, x_2=4, x_3=6$ and let $x_{n+1} = \sum\limits_{k=1}^n x_k$ for each $n\geq 3$. You have that $x_i\mid \sum\limits_{k=1}^nx_k$ for all $i\leq n$ for all $n\geq 3$.

JMoravitz
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    Why did you start with 2, 4, and 6 rather than 1, 2, and 3? – Arcanist Lupus Mar 11 '19 at 01:12
  • @ArcanistLupus My first instinct was to try to find an example of four numbers which add up to something with nice clustered divisors by inspection. My initial target was $24$ as I knew it had many divisors, $1,2,3,4,6,8,12$, which as it so happened led me to spot $2,4,6,12$. In finding an example with four numbers satisfying the required condition that they all divide evenly into the sum, that gave me hope that it could be done for $2014$ numbers too and indeed by looking at the example I spotted the pattern I describe above. – JMoravitz Mar 11 '19 at 01:15
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    @ArcanistLupus it was just personal preference that I chose $24$ as my target number instead of $12$, no deep mathematical reason beyond that $24$ had an extra few divisors and seemed an easier target at the time, but honestly I hadn't really given $12$ much consideration. – JMoravitz Mar 11 '19 at 01:17
  • {1,2,3...} should be preferred as it is canonically smaller; this is just 2 . {1,2,3...}. But can you prove it is canonically the smallest? My instinct was to try primes, factorials and primorials. Obviously we want higher multiplicity of smaller primes, rather than p_n# which is unnecessarily large. The sum you propose here will only be divisible by 3 and powers of 2; pretty soon $3\cdot 2^$ becomes very large. Perhaps we can do better, throw in some low-multiplicity power of 5, then 7 etc. – smci Mar 11 '19 at 02:07
  • Vaguely related: Highly composite number – smci Mar 11 '19 at 02:08
  • 1,2,3 and so on only work because of an even number of odd values sonthey can sum yo an even value. –  Mar 11 '19 at 15:49
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$$\begin{align} 1+2+3&=6\\ 1+2+3+6&=12\\ 1+2+3+6+12&=24\\ \vdots \end{align}$$

saulspatz
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