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2019 AMC 10B Problems/Problem 14: Difference between revisions

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


==Solution==
==Solution 1==
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).
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).


Revision as of 01:21, 15 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 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 $9$. 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).

Solution 2

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 $2^{13} \cdot 3^8 \cdot 7 \cdot 17 \cdot 19$. $2^{13} = 2^{10} \cdot 2^3 = 1024 \cdot 8 = 8192$, 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 $T + M + H = 12$, thus (C).

See Also

2019 AMC 10B (Problems • Answer Key • Resources)
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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