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2019 AMC 10B Problems/Problem 14

Revision as of 15:14, 15 February 2019 by Krishdhar (talk | contribs)

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

With 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 from the original 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).

Solution 3

We know that $9$ and $11$ are both factors of $19!$. Furthermore, we know that H is 0 because $19!$ ends in three zeroes. 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 be divisible by $9$. Summing the digits, we get that T + M + $33$ must be divisible by $9$. This leaves either A or C as our answer choice. Now we test for divisibility by $11$. For a number to be divisible by eleven, the alternating sum must be divisible by 11(ex. $2728$, $2$-$7$+$2$-$8$ = -11 so $2728$ is divisible by $11$). 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 $4$ and M is $8$. The sum is $8$ + $4$ + $0$ = 12 or (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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