Difference between revisions of "2019 AMC 10B Problems/Problem 14"
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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 | |
| − | + | \qquad\textbf{(B) }8 | |
| + | \qquad\textbf{(C) }12 | ||
| + | \qquad\textbf{(D) }14 | ||
| + | \qquad\textbf{(E) } 17 </math> | ||
| − | - | + | ==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, 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)== | ||
| + | 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>. | ||
| + | |||
| + | ==Solution 3 (Brute Force)== | ||
| + | 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. | ||
| + | |||
| + | ==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 <math>4 + 8 + 0 = \boxed{\textbf{(C) }12}</math>. | ||
| + | |||
| + | ~ 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== | ==See Also== | ||
Latest revision as of 12:47, 14 July 2024
Contents
Problem
The base-ten representation for 
is 
, where 
, 
, and 
denote digits that are not given. What is 
?
Solution 1
We can figure out 
by noticing that will end with 
zeroes, as there are three factors of 
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 is a multiple of both 
and 
. Their divisibility rules (see Solution 2) tell us that 
and that 
. By guess and checking, we see that 
is a valid solution. Therefore the answer is 
.
Solution 2 (similar to Solution 1)
We know that , because ends in three zeroes (see Solution 1). Furthermore, we know that and are both factors of . We can simply use the divisibility rules for and for this problem to find and . For to be divisible by , the sum of digits must simply be divisible by . Summing the digits, we get that 
must be divisible by . This leaves either 
or 
as our answer choice. Now we test for divisibility by . For a number to be divisible by , the alternating sum must be divisible by (for example, with the number 
, 
, so is divisible by ). Applying the alternating sum test to this problem, we see that 
must be divisible by 11. By inspection, we can see that this holds if 
and 
. The sum is 
.
Solution 3 (Brute Force)
Multiplying it out, we get 
. Evidently, 
, 
, and . The sum is .
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 .
~ AliciaWu
Video Solution by OmegaLearn
https://youtu.be/p5f1u44-pvQ?t=760
~ pi_is_3.14
Video Solution
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
