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I have been trying puzzle solving recently and came across this problem (sorry if this is not entirely appropriate on a maths stack exchange, I have also posted this on the puzzle one).

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Any hints on how to do this?

So far, I have found that N, R and F must be odd (as they are the last digit in a 5 digit prime number), and that N must be specifically either 1,5 or 9, because all square numbers end in either 1,4,9,5 or 6, and N can’t be 4 or 6, because it must be odd. Finally, I think E must be even because N-E=R, and since N and R are odd, E must be even.

However, these observations don’t seem to be getting me anywhere.

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  • $\begingroup$ $U$ must be $0$ if $N < E$ or $9$ if $N > E$, because it is the result of $E - E$ with a possible borrowing. Also, $N$ can't be $5$, because $SEVEN$ would be divisible by $5$. $\endgroup$ – eyeballfrog Aug 23 at 15:30
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    $\begingroup$ Notice $N$ can't be $5$ because $SEVEN$ is prime. $\endgroup$ – Matt Samuel Aug 23 at 15:33
  • $\begingroup$ @eyeballfrog Do you have the inequality reversed? If $N>E$ then $R=N-E$ and $U=E-E=0.$ $\endgroup$ – saulspatz Aug 23 at 15:41
  • $\begingroup$ @saulspatz Yes I do. $\endgroup$ – eyeballfrog Aug 23 at 15:47
  • $\begingroup$ Since $SEVEN-THREE=FOUR$ we must have $S=T+1$ or the difference would have $5$ digits. Then $TEN$ is a three-digit square, with $3$ different digits, ending in an odd number, whose first digit is not $9$. I can't think of anything better than to test each case. For example, the smallest possibility is $TEN=169,$ which leads to $SEVEN = 26V69,$ which is not possible, by examination of a table of primes. $\endgroup$ – saulspatz Aug 23 at 16:03
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From your reasoning, we know that $N$ is $1$ or $9$. If it is $1$, then $U$ is $9$, and if it is $9$, $U$ is $0$, because $U$ is a result of the $E-E$ with possible borrowing from $SEVEN - THREE$. This means no other letter can be $9$.

We know that $S = T+1$ and $H > E$, because there had to be a borrowing to get a four-digit number. Since none of these letters can be $9$, we know that $E$ and $T$ also can't be $8$. The condition that $TEN$ is a square, leaves $169,361,529,729$. However, in the last case, $R = N-E$ would be $7$, not distinct from $T$, and in the second case, $R = 10 + N - E = 5$, making $FOUR$ divisible by $5$. So $N = 9$, $U = 0$, and $TEN = 169,529$. This means $SEVEN$ is either $26V69$ or $62V29$. However, in the first case, that would mean $16 - H = F$, but that's not possible as they must be distinct and neither can be $9$ (there can't be a borrow because $R = 3$ and $1$ and $2$ are already taken as values for $V$). So $SEVEN = 62V29$.

So $SEVEN - THREE = FOUR$ is now $62V29 - 5H722 = FO07$. We know $F + H = 11$ or $F + H = 12$, depending on whether there's a borrow. Since $RUOF$ is prime and the only odd digits left are $1$ and $3$, $F$ must be one of them, but it can't be $1$ and satisfy either equation. So $F$ is $3$. $H$ must then be $8$, because $9$ is taken. Thus, there must have been a borrow in $V - 7 = O$, which is satisfied by $V = 1$, $O = 4$.

Thus, $SEVEN - THREE = FOUR$ is $$ 62129 - 58722 = 3407 $$ and checking the prime table shows $62129$, $3407$, and $7043$ are indeed prime. Meanwhile, $TEN = 529 = 23^2$ is a perfect square, as desired.

As a fun aside, $FOURTEEN$ is prime but $SEVENTEEN$ is not.

EDIT: removed reliance on prime table.

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  • $\begingroup$ So in the puzzle, the information that $RUOF$ is prime is superfluous, correct? $\endgroup$ – saulspatz Aug 23 at 16:28
  • $\begingroup$ @saulspatz it eliminated $F = 8$. $\endgroup$ – eyeballfrog Aug 23 at 16:29
  • $\begingroup$ So it did. Thanks. $\endgroup$ – saulspatz Aug 23 at 16:30
  • $\begingroup$ @saulspatz it turns out the prime table isn't necessary. see edited answer. $\endgroup$ – eyeballfrog Aug 23 at 16:35
  • $\begingroup$ Elegant solution. +1 $\endgroup$ – saulspatz Aug 23 at 16:52

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