2019 AMC 10B Problem 14

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 $\boxed{\textbf{(C)}\ 12}$.

A factorial is divisible by every small number, so the classical digit tests for $9$ and $11$ become equations in the unknown digits. One test alone leaves ambiguity; two independent tests intersect in a single solution. Nail down the easiest digit (trailing zeros) first to shrink the unknowns.

• Applying only the divisibility-by-9 test, which leaves T + M equal to 3 or 12, and guessing between choices (A) and (C).
• Miscounting positions in the alternating sum for $11$ (the number has $18$ digits), which flips the sign of $M - T$.