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

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==Problem==
 
==Problem==
 
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>?
 +
 +
<math>\textbf{(A) }3
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\qquad\textbf{(B) }8
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\qquad\textbf{(C) }12
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\qquad\textbf{(D) }14
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\qquad\textbf{(E) } 17 </math>
  
 
==Solution 1==
 
==Solution 1==
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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 <math>\boxed{\textbf{(C) }12}</math>
+
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 factors of <math>5</math> in its prime factorization, so there would be 3 powers of 10 meaning it will end in 3 zeros. Next, we use the fact that <math>19!</math> is a multiple of both <math>11</math> and <math>9</math>. Their divisibility rules (see Solution 2) tell us that <math>T + M \equiv 3 \;(\bmod\; 9)</math> and that <math>T - M \equiv 7 \;(\bmod\; 11)</math>. By guess and checking, we see that <math>T = 4, M = 8</math> is a valid solution. Therefore the answer is <math>4 + 8 + 0 = \boxed{\textbf{(C) }12}</math>.
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 +
==Solution 2 (similar to Solution 1)==
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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>.
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 +
==Solution 3 (Brute Force)==
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Multiplying it out, we get <math>19! = 121,645,100,408,832,000</math>. Evidently, <math>T = 4</math>, <math>M = 8</math>, and <math>H = 0</math>. The sum is <math>8 + 4 + 0 = \boxed{\textbf{(C) }12}</math>.
 +
 
 +
NEVER do this in a real contest unless you decide to devote most of your time to this problem.
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 +
==Solution 4 (1001?) ==
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 +
7, 11, 13 are < 19 and 1001 = 7 * 11 * 13. Check the alternating sum of block 3: H00 - 832 + 40M - 100 + 6T5 - 121 and it is divisible by 1001. HTM + 5 - 53 = 0 (mod 1001) => HTM = 48.
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 +
The answer is <math>4 + 8 + 0 = \boxed{\textbf{(C) }12}</math>.
 +
 
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~ AliciaWu
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== Video Solution by OmegaLearn ==
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https://youtu.be/p5f1u44-pvQ?t=760
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 +
~ pi_is_3.14
  
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==Solution 2==
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==Video Solution==
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With investing just a little bit of time, we can manually calculate 19!. If we prime factorize 19!, it becomes <math>2^{16} \cdot 3^8 \cdot 5^3 \cdot 7^2 \cdot 11 \cdot 13 \cdot 17 \cdot 19</math>. This looks complicated, but we can use elimination methods to make it simpler. <math>2^3 \cdot 5^3 = 1000</math>, and <math>7 \cdot 11 \cdot 13 \cdot = 1001</math>. If we put these aside for a moment, we have <math>2^{13} \cdot 3^8 \cdot 7 \cdot 17 \cdot 19</math> left from the original 19!. <math>2^{13} = 2^{10} \cdot 2^3 = 1024 \cdot 8 = 8192</math>, and <math>3^8 = (3^4)^2 = 81^2 = 6561</math>. We have the 2's and 3's out of the way, and then we have <math>7 \cdot 17 \cdot 19 = 2261</math>. Now if we multiply all the values calculated, we get <math>1000 \cdot 1001 \cdot 8192 \cdot 6561 \cdot 2261 = 121,645,100,408,832,000</math>. Thus <math>T = 4, M = 8, H = 0</math>, and the answer <math>T + M + H = 12</math>, thus <math>\boxed{\textbf{(C) }12}</math>.
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https://youtu.be/mXvetCMMzpU
  
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==Solution 3==
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~IceMatrix
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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 H is 0 because <math>19!</math> ends in three zeroes. We can simply use the divisibility rules for <math>9</math> and <math>11</math> for this problem to find T and M. For <math>19!</math> to be divisible by <math>9</math>, the sum of digits must be divisible by <math>9</math>. Summing the digits, we get that T + M + <math>33</math> must be divisible by <math>9</math>. This leaves either A or C as our answer choice. Now we test for divisibility by <math>11</math>. For a number to be divisible by eleven, the alternating sum must be divisible by 11(ex. <math>2728</math>, <math>2</math>-<math>7</math>+<math>2</math>-<math>8</math> = -11 so <math>2728</math> is divisible by <math>11</math>). Applying the alternating sum to this problem, we see that T-M-7 must be divisible by 11. By inspection, we can see that this holds if T is <math>4</math> and M is <math>8</math>. The sum is <math>8</math> + <math>4</math> + <math>0</math> = 12 or (C). -- krishdhar
 
  
 
==See Also==
 
==See Also==
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{{AMC10 box|year=2019|ab=B|num-b=13|num-a=15}}
 
{{AMC10 box|year=2019|ab=B|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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SUB2PEWDS
 

Latest revision as of 12:47, 14 July 2024

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 factors of $5$ in its prime factorization, so there would be 3 powers of 10 meaning it will end in 3 zeros. 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 guess and checking, 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 $8 + 4 + 0 = \boxed{\textbf{(C) }12}$.

Solution 3 (Brute Force)

Multiplying it out, we get $19! = 121,645,100,408,832,000$. Evidently, $T = 4$, $M = 8$, and $H = 0$. The sum is $8 + 4 + 0 = \boxed{\textbf{(C) }12}$.

NEVER do this in a real contest unless you decide to devote most of your time to this problem.

Solution 4 (1001?)

7, 11, 13 are < 19 and 1001 = 7 * 11 * 13. Check the alternating sum of block 3: H00 - 832 + 40M - 100 + 6T5 - 121 and it is divisible by 1001. HTM + 5 - 53 = 0 (mod 1001) => HTM = 48.

The answer is $4 + 8 + 0 = \boxed{\textbf{(C) }12}$.

~ AliciaWu

Video Solution by OmegaLearn

https://youtu.be/p5f1u44-pvQ?t=760

~ pi_is_3.14

Video Solution

https://youtu.be/mXvetCMMzpU

~IceMatrix

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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