I made a small number crossword for my mathletes club that I thought I'd share. It's not too hard, I think it's just right for a coffee break. Have fun!
(Edit: No answer has a leading zero)
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5
asked Apr 25, 2019 at 21:13
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$\begingroup$ Is the last across clue intended to be enigmatic? And isn't 4A true of every integer? $\endgroup$– noedneCommented Apr 25, 2019 at 21:32
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$\begingroup$ @noedne How is (e.g.) 3 divisible by two distinct perfect cubes? $\endgroup$– GentlePurpleRainCommented Apr 25, 2019 at 21:42
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1$\begingroup$ @GentlePurpleRain $1^3$ and $(-1)^3$ $\endgroup$– noedneCommented Apr 25, 2019 at 21:43
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1$\begingroup$ I'm pretty sure that when discussing divisibility, it is generally assumed that one is dealing only with positive integers. $\endgroup$– GentlePurpleRainCommented Apr 25, 2019 at 21:45
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1$\begingroup$ @noedne yes, I only want positive factors. $\endgroup$– greenturtle3141Commented Apr 25, 2019 at 22:06
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2 Answers
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The numerical answer is
$$\begin{array}{|c|c|c|} \hline 7 & 1 & 8 \\ \hline 1 & 3 & 5 \\ \hline \end{array}$$
My thinking went something like this:
3D means that the second digit of 1D must be a $1$, because any number 20 or larger multiplied by 5 would end up as 3 digits. (It can't be a 0 because leading 0s aren't allowed for 4A.) Since 1 is reversible on a 7-segment display, the other digit of 1D must not be. The only digits that don't form a digit when turned upside down are 3, 4, and 7. Thus 3D must be $65$, $70$, or $85$. For 4A, we now have a first digit of $1$, and a last digit of $5$ or $0$. The middle digit must be odd, because 2D is prime, and no 2-digit prime is even. This makes the number of possibilities for 4A very small. By a little trial and error (multiplying 8, 27, 64), it easy to discover that the only possible solution is $135$, which is a multiple of both $1^3$ and $3^3$. Now we have that 1A is either $3x6$ or $7x8$ (where $x$ is the unknown digit). Since two of the digits must sum to the third, the only possibilities are $336$, $396$, or $718$. The first two options do not make 2D prime (33 and 93 are both composite), so the solution for 1A must be $718$, making 2D $13$.
The final answer (thanks to PiIsNot3, below), is found by
using A1Z26 to translate each number into a letter. The dotted line indicates that those two digits should be treated as a single number.
Thus we have 7-18-1-3-5, which translates to
GRACE
answered Apr 25, 2019 at 21:55
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$\begingroup$ This is the correct numerical answer. $\endgroup$ Commented Apr 25, 2019 at 22:06
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5$\begingroup$ Notice that the line separating 2 and 3 is dotted, meaning that the 1 and 8 should be read together as 18. Doing A1Z26 gives rot13(TENPR), which is the “elegant name.” $\endgroup$– HTMCommented Apr 25, 2019 at 22:08
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$\begingroup$ I have added an explanation and the final answer, as supplied by @PiIsNot3. $\endgroup$ Commented Apr 26, 2019 at 15:09
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Speaking strictly, there is another (numerical) solution:
145
135
That's because
On a 7-segment indicator, an upside-down 1 is not the same that the "original" 1.
answered Apr 26, 2019 at 11:32
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$\begingroup$ There is no such requirement in the puzzle. $\endgroup$– jarnbjoCommented Apr 26, 2019 at 13:16
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$\begingroup$ You are technically right, that clue may be somewhat vague. In spirit, what I wanted to get at was the fact that numbers such as 219, 6901, 55, etc. can be cleanly read as numbers when turned upside-down, especially when entered into a 7-segment display. However, I will argue here that even an "11" would still technically represent a base 10 numeral, even though the placement of the ones don't follow the standard convention. The same cannot be argued for numbers such as 3, 4, or 7, and assigning values to "E", "h", or "L" seems a bit of a stretch, so I think we're in the clear here. $\endgroup$ Commented Apr 26, 2019 at 16:05