2019 AMC 10B Problems/Problem 14

Revision as of 17:39, 14 February 2019 by AZAZ12345 (talk | contribs) (Solution)

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

- AZAZ12345

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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