Alphametic between Kennedy and Nixon

Question

Every letter stands for a digit in base-10 representation, different letters stand for different digits, and leading digits are always non-zero.

  LYNDON * B = JOHNSON

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

Answer 1

The answer is

$570140 * 6 = 3420840$

This is the same answer that Marius got. However, the fact that his answer was not accepted leads me to believe an exhaustive analysis is required to prove that it is the only solution.

From $N * B = N$ we deduce we must have one of the following cases:

- $N = 0$
- $B = 1$ (impossible because the result would be $LYNDON$)
- $N = 5, B \in \{3, 7, 9\}$
- $B = 6, N \in \{2, 4, 8\}$

- Assume $B = 6$:
$N * B$ produces a carry digit, $c_{1} \in \{1, 2, 4\}$, and we have $O * B + c_{1} = O \pmod{10}$. But we note that $c_{1} = O * (B - 1) \pmod{10}$ is a multiple of $5$ and thus can never take any of the values in $\{1, 2, 4\}$. Contradiction.

- Assume $N = 5$:
Similar to the logic above, we require $c_{1} = O - ((O * B) \pmod{10}) \pmod{10}$, but since $B$ is odd, $c_{1}$ must be even. That's because for any value of $O$, we either get $c_{1} = even - (even * odd) = even$, or $c_{1} = odd - (odd * odd) = even$. The only case for which $c_{1}$ is even from $N * B$ is for $B = 9$. However, we notice that the 4th multiplication is also $N * B = N$, which means that $D * B = S$ must result to no carry digit. But since we have $B = 9$, the only way to achieve that would be for $O = S = 1$. Contradiction.

Therefore, $N = 0$. Using the same logic as for $N * B = N$, from $O * B = O$ we deduce that:

- $O = 5, B \in \{3, 7, 9\}$
- $B = 6, O \in \{2, 4, 8\}$

- Assume $O = 5$:
We don't want $D * B = S$ to overflow, which is only possible for $B = 3$. From that we get two valid pairs for $(S, D)$, namely $\{(1, 4), (2, 7)\}$. If we tried the first pair, we'd be left with the values $\{2, 6, 7, 8, 9\}$. $Y * B$ can't produce a carry larger than 2, so $L * B \le 29$. The only solution would then be $L = 6, J = 2$. However, from the remaining numbers left no combination of two values would make $Y * B = H \pmod{10}$ valid, which would lead to a contradiction. Similarly, if we took the second pair for $(D, S)$, we'd be left with $\{1, 4, 6, 8, 9\}$. With the same logic as above, this would force $J = 1$, but we'd be left with no number for $L$ to make $L * 3 + c_{5} = 15$ valid, with $c_{5} \le 2$ being the carry digit of $Y * B$. Contradiction.

Therefore, $B = 6$. Since we still want $D * B = S$ not to overflow, this instantly forces $D = 1$. And because we can't have $S * B + c_{2} \ge 10$, this restricts $c_{2} \le 3$, which means $O \in \{2, 4\}$.

- Assume $O = 2$:
This means $S = 1$ and we are now only concerned with $LY * 6 = JOH$, with $\{3, 4, 5, 8, 9\}$ as possible numbers. $Y \not\in \{4, 8\}$ or we would get $Y * B = Y \pmod{10} \not= H$. The other three numbers result to an odd carry digit. However, $L * B + c_{5} = JO$ results to a contradiction, since $(L * even) + odd \not= even$.

Therefore, $O = 4$. This also means $D = 8$ and we are left with the numbers $\{2, 3, 5, 7, 9\}$ for $LY * B = JOH$. The only valid numbers for $Y * B = H \pmod{10}$ are $Y = 7, H = 2$, with a carry of 4. This leaves us with $L * 6 + carry = J4$, or $L * 6 = J0$, which finally allows us to conclude the only solution with $L = 5, J = 3$.

Answer 2

Solution

$570140 * 6 = 3420840$

Long way:

N can be:
0 - fits the patterns
5 if B is odd.
2,4,8 if B is 6.

Starting at the top:

N = 0.
The alphametic reduces to $LY0DO * B = JOH0SO$.
The same rules for N apply now for O except for $O = 0$.
Trying with $B = 6$ since is more likely statistically.
We get $LY0DO * 6 = JOH0SO$.
Trying for $O=2$.
$LY0D2 * 6 = J2H0S2$
From this we see that $D = 1$ otherwise we will not be able to get 0*6 + carriage to end up again to 0.
Now we have:
$LY012 * 6 = J2H072$
removing the last 3 digits from both parts because they don't affect the rest:
We get $LY * 6 = J2H$ with the available digits 3,4,5,8,9.
Y cannot 4 or 8 because it will result in H=Y.
The carriage from $6*Y$ must be an even number in order to get 2 as a result, but this does not happen for any of the digits 3,5,9.
Conclusions. This does not work.

Moving on to O=4.
$LY0D4 * 6 = J4H0S4$
Same as above, we see that $D=1$ otherwise we will not be able to get 0*6 + carriage to end up again to 0.
Now we have:
$LY014 * 6 = J4H074$
Again, we can remove the last 3 digits because we know them and they don't influence the rest.
$LY * 6 = J4H$ with the available digits: 2,3,5,7,9.
Obviously Y cannot be 9 because H would end up being 4.
Y cannot be 2 because H would be 2.
Y cannot be 5 because we end up with H = 0 and we already assumed N=0.
For Y = 3 we get:
$L3 * 6 = J48$.
$6*L+1$ must end with 4 and this is not possible because $6*L$ is even.
for $Y = 7$ we get:
$L7 *6 = J42$.
This works for $L = 5$ and $J = 3$.
Bingo! We got one.

Since the OP does not say that there are multiple answers I stopped after finding this.
If the question get's modified and I have to find more results I will go on with

$O=8$ and the combination $N = 5$ and B is odd.