Yesterday I posed an Alfred E. Neuman alphametic with multiplication, and today I pose an Alfred E. Neuman alphametic with division for the lovers of Mad magazine:

Every letter and every question mark stands for a digit in base-9 representation. Different letters stand for different digits. Leading digits are always non-zero.


Which digit does each letter represent? (Please present the full analysis how these digits can be determined.)

  • $\begingroup$ Are the two puzzles related, as in they both can be solved separately right? $\endgroup$
    – ABcDexter
    Commented Mar 4, 2016 at 12:14

1 Answer 1


I hate base-9.

Lets re-write it as follows:

x    E

OK, so first off, we know $E\notin\{0,1\}$. Since the result is a 6 digit number, we know that $N\in\{1,2,3,4\}$. Since $E \times N$ shows up twice, once with a $D$ result and once with an $A$ result, we know that the fifth column ($E \times E$) must have a carry over.

Also, we will use the fact that the carry over for multiplying any two numbers must be less than the two numbers.

  • If $N=4$, then $E=2$ and $A=8$. But then $D=8$ too, which is impossible.
  • If $N=3$, then $E=2$ and $D=6 \implies A \in \{7,8\}$. Since $E^2=4$, we need a carry over of at least 4 from the $U \times E$ column, which is impossible.
  • If $N=1$, then $E \times N=D \implies E=D$.

Thus $N=2$

So $E=3$ and $D=6 \implies A \in \{7,8\}$. The only value that works for the second column where $A \times E$ yields $E$ is 7 since $7\times 3=23$.

Looking at the fifth column, $E \times E$ yields L. Since $E^2=10$, we know that $L$ is simply the carry over from the previous column, which cannot be greater than 2. Thus, $L\in\{0,1\}$ since 2 is already taken.

Assume, $L=1$

So $U \times E$ has a carry over of 1, $U\in \{4,5\}$.

The remaining values are $\{0,4,5,8\}$. And since $A=7$ the carry over from the second column into the third is 2.

  • If $M=0$, then $M \times 3 +2 =2 \implies R=2$, which is already taken.
  • If $M=4$, then $M \times 3 +2 =13+2 \implies R=5$, leaving nothing for $U$.
  • If $M=8$, then $M \times 3 +2 =26+2 \implies R=8$ too.
  • If $M=5$, then $M \times 3 +2 =16+2 \implies R=8$. And $U=4$. This leave $F=0$, but doing the math gets $234572 \times 3=714836 \implies F=4$.

Thus $L=0$

$U \times E$ has a carry over of 0, so $U=1$.

The remaining values are $\{4,5,8\}$.

  • $M=8 \implies 231872 \times 3 = 705836$, so $R=8$
  • $M=4 \implies 231472 \times 3 = 704536$, so $F=4$

Thus, $M=5 \implies 231572 \times 3 = 704836$ so $R=8$ and $F=4$.

So we have

x    3
  • 1
    $\begingroup$ I think the problem is on the F≤1+2=3 part. U x E = 3, plus the carry over, F can be 3, 4 or 5. Following that logic, then M can be 4 or 5, as every other possible number produces duplicate numbers. M 4 or 5 has a carry over of 1, leaving F = 4. So M = 5 and then R = 8 $\endgroup$
    – BianB BB
    Commented Mar 4, 2016 at 13:19
  • 1
    $\begingroup$ @BianBBB was just about to say the same :). You are entirely correct so the solution is A=7 L=0 F=4 R=8 E=3 D=6 N=2 U=1 M=5 $\endgroup$
    – Ivo
    Commented Mar 4, 2016 at 13:21
  • 1
    $\begingroup$ @BianBBB Thanks! I fixed the mistake! $\endgroup$
    – Trenin
    Commented Mar 4, 2016 at 14:35

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