[vc_row][vc_column][vc_column_text]Each of the 12 white squares is to be filled with a single digit. The four 2 digit numbers and the four 3 digit numbers thus created are either perfect squares or perfect cubes. These squares and cubes are all distinct and one of the down answers is both a square and a cube.[latexpage]
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We use $1d$ for ``1 down", etc.
The list of available squares and cubes is
\[ 16, 25, 27, 36, 49, 64, 81, 100, 121, 125, 144, 169, 196, 216, 225, 256, 289, \]\[
324, 343, 361, 400, 441, 484, 512, 529, 576, 625, 676, 729, 784, 841, 900, 961.
\]
The two available numbers that are both square and cube are 64 and 729, and one of these is one of the down answers. Three possibilities lead to a contradiction, as we see next.
If $6d=64$ then there are two cases according to $6a$.
If $6a=625$ then $2d=512$ and $5d=256$ which implies that $7d=64=6d$ which won't do.
If $6a=676$ then there is no way to choose $2d$.
If $2d=729$ then $6a=196$ and $5d=169$ and there is no way to choose $7a$.
If $5d=729$ then, once again, there is no way to choose $7a$.
The only remaining possibility is $3d=64$.
It follows that $4a\in\{144,324,484,784\}$. But $4a\neq324$ because $2d$ can't have 3 as a middle digit.
If $4a=484$ then $5d=841$ and $2d=144=6d$, so this case fails.
If $4a=784$ then $5d=814$ and $6a$ ends in 64, are there are no 3 digit numbers in our list that fit.
Therefore $4a=144$ and $2d=512$ (since $2d=216$ leaves no possible $1a$). It follows that $1a=25$. It quickly follows that $6a=324$ and $5d=441$ and we get to the completed grid as below.
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Problem Page Coordinator: Stephen Lee CMath MIMA – Mathematics in Education and Industry
Acknowledgement: The IMA are indebted to Peter Chamberlin FIMA (University of Reading) for creating and supplying Mathematics Today with this puzzle.
First published in Mathematics Today (December 2015)
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