Difference between revisions of "2019 AMC 10B Problems/Problem 14"

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==Solution==
 
==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. sing their divisibility rules gives us that T+M is congruent to 3 mod 9 and 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. C
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We can figure out <math>H = 0</math> by noticing that <math>19!</math> will end with <math>3</math> zeroes, as there are three <math>5</math>'s in its prime factorization. Next we use the fact that <math>19!</math> is a multiple of both <math>11</math> and <math>9</math>. Since their divisibility rules gives us that <math>T + M</math> is congruent to <math>3</math> mod <math>9</math> and that <math>T - M</math> is congruent to <math>7</math> mod <math>11</math>. By inspection, we see that <math>T = 4, M = 8</math> is a valid solution. Therefore the answer is <math>4 + 8 + 0 = 12</math>, which is (C).
  
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Edited (LaTeXed) by greersc
  
 
==See Also==
 
==See Also==

Revision as of 17:11, 14 February 2019

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

See Also

2019 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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