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

Revision as of 23:47, 11 September 2022

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) (The most illogical solution)

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

Do not do this in a real contest.

Solution 4 (1001?)

H = 0 because 19! will end with 3 zeroes, as there are three factors of 5 in its prime factorization.

7, 11, 13 are < 19 and 1001 = 7 * 11 * 13. Check the alternating sum of block 3: 000 - 832 + 40M - 100 + 6T5 - 121 and it is divisible by 1001. TM + 5 - 53 = 0 (mod 1001) => TM = 48.

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

Do this in a real contest.

Video Solution

https://youtu.be/mXvetCMMzpU

~IceMatrix

@@ Line 24: / Line 24: @@
 , 11, 13 are < 19 and 1001 = 7 * 11 * 13. Check the alternating sum of block 3: 000 - 832 + 40M - 100 + 6T5 - 121 and it is divisible by 1001. TM + 5 - 53 = 0 (mod 1001) => TM = 48.
-The answer is <math>4 + 8 + 0 = \boxed{\textbf{(C) }12}</math>.
+The answer is <math>4 + 8 + 0 = \boxed{\textbf{(C) }12}</math>. ~ AliciaWu
 Do this in a real contest.

2019 AMC 10B (Problems • Answer Key • Resources)
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