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

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==Solution 1==
 
==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 factors of <math>5</math> 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>. 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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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>.
  
 
==Solution 2 (similar to Solution 1)==
 
==Solution 2 (similar to Solution 1)==

Revision as of 13:20, 5 September 2021

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

Video Solution

https://youtu.be/mXvetCMMzpU

~IceMatrix

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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