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A $3 \times 3$ arrangement of boxes is filled with one of the numbers one, seven, or nine. Prove that of the eight possible sums along the rows, the columns, and the diagonals, two sums must be equal.

I tried to use the pigeonhole principle, but there are ten possible ways to generate different sets/sums with cardinality three , and I am not able to eliminate enough possibilities. In addition, I tried to find a pattern by placing many possibilities without any luck.

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  • $\begingroup$ I am sorry for not writing all the ways I tried to solve it, it is hard for me to write it in English. $\endgroup$ – joem Jul 1 '18 at 13:55
  • $\begingroup$ Even if your English is not good (mine is not), write down what you tried. $\endgroup$ – Claude Leibovici Jul 1 '18 at 13:59
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    $\begingroup$ I don't want to see an atlas of the brain gyri of the OP. The people who are eager to close this question would better give a hint. Fact is they have no idea of one. (Nor do I.) $\endgroup$ – Christian Blatter Jul 1 '18 at 14:27
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    $\begingroup$ It might help -- at least in the form of minimizing mental complexity -- to subtract $1$ from each of the entries and then dividing by $2$, giving $\{0,3,4\}$ instead of $\{1,7,9\}$. This can't change whether two of the sums-of-three are equal or not. To me it makes it somewhat easier to see that "equal sum" in this case is the same as "equal multiset of terms". $\endgroup$ – Henning Makholm Jul 1 '18 at 15:23
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You can’t have two rows with three numbers the same. If you have $$111\\777\\ *\_\_$$ or $$111\\ \_*\_\\777$$ whatever number you put in the $*$ will make two duplicate lines.

Assume we use $111$, so don’t use $777$ or $999$. We have to use all the other possibilities. There needs to be another $1$ to make $117$ and $119$, so there are only five cells for $7$ and $9$. Say there are only two $9$s. You can’t have both $991$ and $997$

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  • $\begingroup$ Great and simple answer thanks a lot! $\endgroup$ – joem Jul 2 '18 at 6:51

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