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

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Puzzles and Curious Problems by Henry Ernest Dudeney: $175$

Cross-Number Puzzle
Dudeney-Puzzles-and-Curious-Problems-175.png
Across:
1. a square number
4. a square number
5. a square number
8. the digits sum to $35$
11. square root of $39$ across
13. a square number
14. a square number
15. square of $36$ across
17. square of half $11$ across
18. three similar figures
19. product of $4$ across and $33$ across
21. a square number
22. five times $5$ across
23. all digits alike, except the central one
25. square of $2$ down
27. see $20$ down
28. a fourth power
29. sum of $18$ across and $31$ across
31. a triangular number
33. one more than $4$ times $36$ across
34. digits sum to $18$, and the three middle numbers are $3$
36. an odd number
37. all digits even, except one, and their sum is $29$
39. a fourth power
40. a cube number
41. twice a square
Down:
1. reads both ways alike
2. square root of $28$ across
3. sum of $17$ across and $21$ across
4. digits sum to $19$
5. digits sum to $26$
6. sum of $14$ across and $33$ across
7. a cube number
9. a cube number
10. a square number
12. digits sum to $30$
14. all similar figures
16. sum of digits is $2$ down
18. all similar digits except the first, which is $1$
20. sum of $17$ across and $27$ across
21. a multiple of $19$
22. a square number
24. a square number
26. square of $18$ across
28. a fourth power of $4$ across
30. a triangular number
32. digits sum to $20$ and end with $8$
34. six times $21$ across
35. a cube number
37. a square number
38. a cube number


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.

$\blacksquare$


Sources

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