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There is a 6-sided die with 6 different positive integer numbers on it. The die is rolled four times and it is guaranteed that the multiplication of these four numbers ($O_1 \times O_2\times O_3 \times O_4$) is divisible by the sum of the numbers ($D_1+D_2+D_3+D_4+D_5+D_6$) on the die.

What is the minimum value of the sum of the numbers on the die?

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The minimal value of the sum $D_1+D_2+D_3+D_4+D_5+D_6$ is

$81$.

Proof

First, note that

the product of the $O_j$ must be divisible by $\sum D_i$ even if $O_j=\min_i(D_i)$ for all $j$.

  • Clearly, then, $\min_i(D_i)$ can't be $1$ or $2$, since $1$ and $16$ can't be multiples of any sum of six distinct natural numbers.
  • So let's try $\min_i(D_i)=3$. Then $\sum D_i$ must be a factor of $81$, and therefore a power of $3$ (and clearly not $3^0=1$), which means every $D_j$ has to be a multiple of $3$, so $\sum D_i=81=3\times27$.

This is achievable with e.g.

$\{D_i\}_i=\{3,6,9,12,15,36\}$. Clearly any product of four of these will be a multiple of $81$, which is their sum.

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  • $\begingroup$ "... therefore all Di are multiples of 6". I don't see how this follows, can you elaborate? $\endgroup$ – ffao Aug 8 '17 at 21:47
  • $\begingroup$ @ffao All fixed. $\endgroup$ – Rand al'Thor Aug 8 '17 at 22:01
  • $\begingroup$ A simpler argument is that $3+4+5+6+7+8>27$, leaving $81$ as the only possibility for the sum. Though you'd still need your argument to construct a working example. $\endgroup$ – Jaap Scherphuis Aug 8 '17 at 22:18

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