2019 AMC 10B Problems/Problem 14

Problem

The base-ten representation for $19!$ is $121,6T5,100,40M,832,H00$, where $T$, $M$, and $H$ denote digits that are not given. What is $T+M+H$?

$\textbf{(A) }3 \qquad\textbf{(B) }8 \qquad\textbf{(C) }12 \qquad\textbf{(D) }14 \qquad\textbf{(E) } 17$

Solution 1

We can figure out $H = 0$ by noticing that $19!$ will end with $3$ zeroes, as there are three $5$s in its prime factorization. Next, we use the fact that $19!$ is a multiple of both $11$ and $9$. Their divisibility rules (see Solution 2) tell us that $T + M \equiv 3 \;(\bmod\; 9)$ and that $T - M \equiv 7 \;(\bmod\; 11)$. By guess and checking, we see that $T = 4, M = 8$ is a valid solution. Therefore the answer is $4 + 8 + 0 = \boxed{\textbf{(C) }12}$.

Solution 2 (similar to Solution 1)

We know that $H = 0$, because $19!$ ends in three zeroes (see Solution 1). Furthermore, we know that $9$ and $11$ are both factors of $19!$. We can simply use the divisibility rules for $9$ and $11$ for this problem to find $T$ and $M$. For $19!$ to be divisible by $9$, the sum of digits must simply be divisible by $9$. Summing the digits, we get that $T + M + 33$ must be divisible by $9$. This leaves either $\text{A}$ or $\text{C}$ as our answer choice. Now we test for divisibility by $11$. For a number to be divisible by $11$, the alternating sum must be divisible by $11$ (for example, with the number $2728$, $2-7+2-8 = -11$, so $2728$ is divisible by $11$). Applying the alternating sum test to this problem, we see that $T - M - 7$ must be divisible by 11. By inspection, we can see that this holds if $T=4$ and $M=8$. The sum is $8 + 4 + 0 = \boxed{\textbf{(C) }12}$.

Solution 3 (pure bash)

Computing $19!$, we get $121,645,100,408,832,000$, so $T = 4, M = 8, H = 0$. \[4 + 8 + 0 = \boxed{\textbf{(C) }12}\]

Video Solution

https://youtu.be/mXvetCMMzpU

~IceMatrix

See Also

2019 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AMC 10 Problems and Solutions

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