Henry Ernest Dudeney/Puzzles and Curious Problems/175 - Cross-Number Puzzle/Solution

by : $175$

 * Cross-Number Puzzle

Solution

 * Dudeney-Puzzles-and-Curious-Problems-175-solution.png

Proof
$4$ across is $1$, $4$ or $9$.

As $28$ down is a $4$-digit number, it follows that $4$ across is $9$ and so $28$ down is $6561$.

Hence $28$ across is constrained to be $625$.

As $36$ across is odd, $33$ across must be $53$ and $36$ across must be $13$.

Hence $14$ down, and hence $18$ across.

This gives $14$ across.

Then $7$ down is a cube ending in $61$, which constrains it to being $9261$.

The last digit of $32$ down is given as $8$.

$39$ across is a fourth power beginning with $1$, constraining it to $1296$.

$37$ down follows.

$6$ times $21$ across is a $3$ digit number.

As $21$ across is square, that means $21$ across is $121$.

$34$ down and $18$ down follow.

$31$ across follows, which alerts us to the fact that $31$ down starts with $0$, which we may not have been expecting.

Hence $32$ down follows.

$23$ across follows, as we have sufficient digits for this.

This of course gives $24$ down as $1$, and its square nature is confirmed.

$21$ down follows, and $27$ across then appears.

$34$ across can be completed.

$35$ down follows.

$41$ across is either $6498$ or $6728$.

As $38$ down is a cube, this means $38$ down is $64$ and $41$ across is $6498$.

$26$ down follows from knowing $18$ across.

$29$ across follows from knowing $18$ across and $31$ across.

$30$ down is constrained, and hence follows $37$ across.

$29$ down follows, which gives us $15$ across (which also follows from $36$ across).

$2$ down is evaluated.

$25$ across is evaluated.

$11$ across is evaluated, which leads to $1$ down.

$12$ down follows.

$3$ down is evaluated.

$1$ across follows.

$17$ across is evaluated.

$20$ down is evaluated.

$6$ down is evaluated.

$9$ down is constrained.

The intersecting digits of $13$ across and $10$ down constrain $13$ across.

$5$ across can be deduced.

$8$ down follows.

The remaining $2$ digits of $8$ across sum to $5$, which constrains $10$ down.

$8$ across then follows.

$4$ down follows.

Finally, $22$ across can be evaluated.