2019 AMC 10B Problem 14

We cannot compute $19!$ by hand quickly, but we know a lot of numbers it divides by. Use that.

Step 1: find $H$. Trailing zeros come from factors of $10 = 2 \cdot 5$. Among $1, 2, \ldots, 19$, the multiples of $5$ are $5, 10, 15$, giving three $5$s, and there are plenty of $2$s. So $19!$ ends in exactly three zeros. The digits end in $H00$, so $H = 0$.

Step 2: test for $9$. $19!$ is a multiple of $9$, so its digit sum is a multiple of $9$. Add the known digits of $1216T5100408M832000$:
$$1 + 2 + 1 + 6 + 5 + 1 + 0 + 0 + 4 + 0 + 8 + 3 + 2 = 33,$$
so the digit sum is $33 + T + M$. The next multiples of $9$ are $36$ and $45$, so $T + M$ is $3$ or $12$. (It cannot be $21$, since two digits add to at most $18$.)

Step 3: test for $11$. $19!$ is a multiple of $11$. The test: starting from the right, add every other digit, add the remaining digits, and the two totals must differ by a multiple of $11$. The $18$ digits from the right are
$$0, 0, 0, 2, 3, 8, M, 0, 4, 0, 0, 1, 5, T, 6, 1, 2, 1.$$
- Positions $1, 3, 5, \ldots$: $0 + 0 + 3 + M + 4 + 0 + 5 + 6 + 2 = 20 + M$.
- Positions $2, 4, 6, \ldots$: $0 + 2 + 8 + 0 + 0 + 1 + T + 1 + 1 = 13 + T$.

The difference is $(20 + M) - (13 + T) = 7 + M - T$. For digits this lies between $-2$ and $16$, so it must be $0$ or $11$. That gives $M - T = -7$ or $M - T = 4$.

Step 4: combine.
- If $T + M = 3$: $M - T = 4$ would give $M = 3.5$, and $M - T = -7$ would give $M = -2$. Neither is a digit.
- If $T + M = 12$: $M - T = 4$ gives $M = 8$, $T = 4$. (And $M - T = -7$ gives $M = 2.5$, not a digit.)

So $T = 4$, $M = 8$, $H = 0$, and $T + M + H = 12$. (Indeed $19! = 121{,}645{,}100{,}408{,}832{,}000$.)

The answer is $\boxed{\textbf{(C)}\ 12}$.

A factorial divides evenly by every small number, so the divisibility tests you know become equations for the missing digits. One test alone left two possibilities; a second, independent test settled it. Start with the easiest digit (the trailing zeros) to cut down the number of unknowns.

Find $H$. The factors $5, 10, 15$ contribute three $5$s to $19!$, and there are plenty of $2$s, so $1000 \mid 19!$ and the number ends in three zeros: $H = 0$.

Divisibility by 9. Since $9 \mid 19!$, the digit sum is a multiple of $9$:
$$1 + 2 + 1 + 6 + T + 5 + 1 + 0 + 0 + 4 + 0 + M + 8 + 3 + 2 + 0 + 0 + 0 = 33 + T + M.$$
So $T + M \equiv 3 \pmod 9$, giving $T + M = 3$ or $12$.

Divisibility by 11. Since $11 \mid 19!$, the alternating digit sum is a multiple of $11$. Reading the $18$ digits from the right, the odd positions hold $0, 0, 3, M, 4, 0, 5, 6, 2$ (sum $20 + M$) and the even positions hold $0, 2, 8, 0, 0, 1, T, 1, 1$ (sum $13 + T$). Their difference $7 + M - T$ must be a multiple of $11$, so $M - T = 4$ or $M - T = -7$.

Combine. With $T + M = 12$: $M - T = 4$ gives $M = 8$, $T = 4$. With $T + M = 3$, neither $M - T = 4$ nor $M - T = -7$ yields digits. So $T = 4$, $M = 8$, $H = 0$, and $19! = 121{,}645{,}100{,}408{,}832{,}000$.

$T + M + H = 12$. The answer is $\boxed{\textbf{(C)}\ 12}$.