2019 AMC 10B Problems/Problem 14

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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 $T+M+H$?

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 tell us that $T + M \equiv 3 \;(\bmod\; 9)$ and that $T - M \equiv 7 \;(\bmod\; 11)$. By inspection, we see that $T = 4, M = 8$ is a valid solution. Therefore the answer is $4 + 8 + 0 = \boxed{\textbf{(C) }12}$.

Solution 2

By investing just a little bit of time, we can manually calculate $19!$. If we prime factorize $19!$, it becomes $2^{16} \cdot 3^8 \cdot 5^3 \cdot 7^2 \cdot 11 \cdot 13 \cdot 17 \cdot 19$. This looks complicated, but we can use elimination methods to make it simpler. $2^3 \cdot 5^3 = 1000$, and $7 \cdot 11 \cdot 13 \cdot = 1001$. If we put these aside for a moment, we have $2^{13} \cdot 3^8 \cdot 7 \cdot 17 \cdot 19$ left. $2^{13} = 2^{10} \cdot 2^3 = 1024 \cdot 8 = 8192$, and $3^8 = (3^4)^2 = 81^2 = 6561$. We have the $2$s and $3$s out of the way, and then we have $7 \cdot 17 \cdot 19 = 2261$. Now if we multiply all the values calculated, we get $1000 \cdot 1001 \cdot 8192 \cdot 6561 \cdot 2261 = 121,645,100,408,832,000$. Thus $T = 4, M = 8, H = 0$, and the answer is $T + M + H = \boxed{\textbf{(C) }12}$.

Solution 3 (similar to Solution 1)

We know that $9$ and $11$ are both factors of $19!$. Furthermore, we know that $H = 0$, because $19!$ ends in three zeroes (see Solution 1). 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 $8 + 4 + 0 = \boxed{\textbf{(C) }12}$.

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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