Alphametic puzzle, involving square and prime numbers [closed]

Question

I have been trying puzzle solving recently and came across this problem.

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.

Answer 1

Is it

$FOUR=3407$, $TEN=529$?

Reason:

Since $SEVEN$ is a prime, $N$ must be in $\{1,3,7,9\}$. Since $TEN$ is a square, $TEN \in \{169, 289, 361, 529, 729, 861\}$.
Since $FOUR$ is a prime, $R$ must be in $\{1,3,7,9\}$. Since $E+R\equiv N\pmod {10}$, we can rule out $TEN=361$ and $TEN=861$ (as $R=5$ in these cases).
Since $SEVEN=THREE+FOUR$, $S=T+1$.
If $TEN=729$ then $SEVEN=82V29$ and $R=N-E=7=T$. Rule out this.
If $TEN=289$ then $S=3$ and $R=N-E=1$. $SEVEN=THREE+FOUR$ translates to $38V89=2H188+FOU1$. Looking at the thousand digit, we have $H+F=17$ or $18$. But $9$ is already used by $N$. Rule out this.
If $TEN=169$ then $S=2$ and $R=N-E=3$. $SEVEN=THREE+FOUR$ translates to $26V69=1H366+FOU3$.

• Looking at the thousand digit, we have $H+F=15$ or $16$. Since $9$ is already used by $N$, it must be that one of $H,F$ is $8$ and the other one is $7$. Since $RUOF$ is a prime, $H=8$ and $F=7$.
• Since $6=6+U$, $U=0$.
• The rest digits are $4,5$, one of which is $O$ and the other is $V$. However $V=3+O$ which is impossible.

So $TEN=169$ is ruled out. We only have $TEN=529$ now. $S=2$ and $R=N-E=7$.

• $SEVEN=THREE+FOUR$ translates to $62V29=5H722+FOU7$.
• Since $2=2+U$, $U=0$. We have $V,H,F,O$ being digits $1,3,4,8$.
• Since $RUOF$ is a prime, $F=1$ or $F=3$. But if $F=1$ then the ten thousand digit wouldn't carry. So $F=3$. We also have $H=8$.
• Then $V\equiv 7+O\pmod{10}$, so $V=1$ and $O=4$. We have $FOUR=3407$.