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

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==Solution 2 (similar to Solution 1)==
 
==Solution 2 (similar to Solution 1)==
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We know that <math>9</math> and <math>11</math> are both factors of <math>19!</math>. Furthermore, we know that <math>H = 0 </math>, because <math>19!</math> ends in three zeroes (see Solution 1). We can simply use the divisibility rules for <math>9</math> and <math>11</math> for this problem to find <math>T</math> and <math>M</math>. For <math>19!</math> to be divisible by <math>9</math>, the sum of digits must simply be divisible by <math>9</math>. Summing the digits, we get that <math>T + M + 33</math> must be divisible by <math>9</math>. This leaves either <math>\text{A}</math> or <math>\text{C}</math> as our answer choice. Now we test for divisibility by <math>11</math>. For a number to be divisible by <math>11</math>, the alternating sum must be divisible by <math>11</math> (for example, with the number <math>2728</math>, <math>2-7+2-8 = -11</math>, so <math>2728</math> is divisible by <math>11</math>). Applying the alternating sum test to this problem, we see that <math>T - M - 7</math> must be divisible by 11. By inspection, we can see that this holds if <math>T=4</math> and <math>M=8</math>. The sum is <math>8 + 4 + 0 = \boxed{\textbf{(C) }12}</math>.
+
We know that <math>H = 0 </math>, because <math>19!</math> ends in three zeroes (see Solution 1). Furthermore, we know that <math>9</math> and <math>11</math> are both factors of <math>19!</math>. We can simply use the divisibility rules for <math>9</math> and <math>11</math> for this problem to find <math>T</math> and <math>M</math>. For <math>19!</math> to be divisible by <math>9</math>, the sum of digits must simply be divisible by <math>9</math>. Summing the digits, we get that <math>T + M + 33</math> must be divisible by <math>9</math>. This leaves either <math>\text{A}</math> or <math>\text{C}</math> as our answer choice. Now we test for divisibility by <math>11</math>. For a number to be divisible by <math>11</math>, the alternating sum must be divisible by <math>11</math> (for example, with the number <math>2728</math>, <math>2-7+2-8 = -11</math>, so <math>2728</math> is divisible by <math>11</math>). Applying the alternating sum test to this problem, we see that <math>T - M - 7</math> must be divisible by 11. By inspection, we can see that this holds if <math>T=4</math> and <math>M=8</math>. The sum is <math>8 + 4 + 0 = \boxed{\textbf{(C) }12}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 15:54, 14 September 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 $T+M+H$?

$\textbf{(A) }3 \qquad\textbf{(B) }8 \qquad\textbf{(C) }12 \qquad\textbf{(D) }14 \qquad\textbf{(E) } 17$

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 (see Solution 2) 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 (similar to Solution 1)

We know that $H = 0$, because $19!$ ends in three zeroes (see Solution 1). Furthermore, we know that $9$ and $11$ are both factors of $19!$. 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 (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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