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 $T+M+H$ ?

Solution

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 $9$ . Since their divisibility rules gives us that $T + M$ is congruent to $3$ mod $9$ and that $T - M$ is congruent to $7$ mod $11$ . By inspection, we see that $T = 4, M = 8$ is a valid solution. Therefore the answer is $4 + 8 + 0 = 12$ , which is (C).

- AZAZ12345

Edited (LaTeXed) by greersc

2019 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2019 AMC 10B Problems/Problem 14

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