Alfred E. Neuman alphametic

Question

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.

An alphametic for the lovers of Mad magazine:

ALFRED $ \, \, \times \, \, $ E = NEUMAN

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

Answer 1

Written vertically:

ALFRED
x    E
------
NEUMAN

Trivially, we can see that $E \gt 1$ and $A \in \{1,2,3,4\}$ since other values would either create a leading zero or too many digits. Also, $N$ is greater than both $A$ and $E$.

If $A=4$, then $E=2$ and $N=8$. But no value for $D$ can result in $D \times E$ yielding 8.

If $A=3$, then $E=2$. But $D \times E$ can have at most a carry over of 1, and $E\times E =E^2=4 \implies A\in\{4,5\}$.

Thus, $A\in\{2,3\}$.

Assume $A=2$

Then $E\in\{3,4\}$. If $E=4$, then $E^2=17$, which means $D\times E$ must carry over 4, which is impossible.

Thus $E=3$, and $E^2=10$ which means $D \times E$ must carry over 2, which means $D=\in\{7,8\}$ and $N\in\{3,6\}$. Only the second pair of these doesn't result in a conflict, so $D=8$ and $N=6$.

This means that $L\times E$ must result in no carry over. This is only possible if $L\in\{0,1\}$ of the remaining valid values. If $L=0$ then we need the carry over of $F\times E$ to be $E$ which is impossible. Thus, $L=1$ and the carry over of $F\times E$ is 2.

Thus, $F=7$. Since $F\times E=7\times 3=23$, and the only remaining values for $U$ are $\{0,4,5\}$, we know that the carry over must be 1 and $R=5$.

But then we have $217538 \times 3=645726 \implies U=7$.

Thus $A=1$

The ones column is $D \times E$ which yields a carry over less than both $D$ and $E$. Lets look at the second column where adding this carry over to $E\times E$ yields $A$

• if $E=2$, then $E^2=4$, requiring carry over of 6.
• if $E=3$, then $E^2=10$, requiring carry over of 1. Thus, $D\in\{4,5\}$ and $N\in\{3,6\}$. Since $3$ is already taken, we know $D=5$ and $N=6$. But this requires a carry over of 3 in the $L\times E$ spot.
• if $E=4$, then $E^2=17$, requiring carry over of 3.
• if $E=5$, then $E^2=27$, requiring carry over of 3.
• if $E=6$, then $E^2=40$, requiring carry over of 1. Thus, $D=2 \implies N=3$.
• if $E=7$, then $E^2=54$, requiring carry over of 6. Thus $D=8 \implies N=2$.
• if $E=8$, then $E^2=71$, requiring carry over of 0. Thus $D=0 \implies N=0$.

So, the only ones that don't lead to immediate contradiction are $E\in\{4,5\}$.

Assume $E=4$

$D=7 \implies N=1$, so $D=8$ and $N=5$ is the only way to get our carry over of 3. We then need a carry over of 1 in the $L\times E$ spot.

So, we need to have $L\in\{2,3\}$. If $L=2$, then $L\times E=8$, so we need a carry over of 5 to get $E$. This is impossible, so $L=3$.

Then, $L\times E=13$ so in order to yield $E$ again, we need another carry over of 1 from the $F \times E$ spot. Thus, $F=2$ and we need a carry over from the $R \times E$ spot. If this carry over is greater than 1, then $U$ will conflict, so it needs to be 1 and $U=0$.

But now we've run out of numbers that will result in a carry over of 1!

Thus, $E=5$

We need a carry over of 3, so $D\in\{6,7\}$.

If $D=6$, then $D\times E=6 \times 5=33$. This means $N=3$ which is too small.

Therefore, $D=7$ and $D\times E=7\times 5= 38 \implies N=8$. This means we also need a carry over of 3 from $L\times E$, so $L=6$.

$L\times E=6\times 5=33$ and we need this to yield $E=5$, so we need a carry over of 2 from the $F \times E$ spot. This means we need $F\in\{3,4\} \implies F\times E\in\{16,22\}$. If $F=3$, then we need $R\times E$ to yield a carry over of 4. But 8 and 7 are already taken, and $R=6$ is insufficient. Thus, $F=4$.

The remaining values are $\{0,2,3\}$. If $R=2$, then $M=4$. If $R=3$, then $M=0$ and $U=4$. Thus, $R=0$, then $M=3$ and $U=2$.

164057
x    5
------
852318