2019 AMC 10B Problems/Problem 14

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 $9$ . Their divisibility rules 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

By investing just a little bit of time, 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 $2^{13} \cdot 3^8 \cdot 7 \cdot 17 \cdot 19$ left. $2^{13} = 2^{10} \cdot 2^3 = 1024 \cdot 8 = 8192$ , 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 is $T + M + H = \boxed{\textbf{(C) }12}$ .

Solution 3 (similar to Solution 1)

We know that $9$ and $11$ are both factors of $19!$ . Furthermore, we know that $H = 0$ , because $19!$ ends in three zeroes (see Solution 1). 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 $8 + 4 + 0 = \boxed{\textbf{(C) }12}$ .

2019 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2019 AMC 10B Problems/Problem 14

Contents

Problem

Solution 1

Solution 2

Solution 3 (similar to Solution 1)

See Also