Difference between revisions of "2019 AMC 10B Problems/Problem 14"
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(→Solution 3 (Brute force) (why for the life of you would you do this)) |
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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>. | 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) ( | + | ==Solution 3 (Brute force) (The most illogical solution)== |
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>. | 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>. | ||
Revision as of 14:36, 21 June 2022
Contents
[hide]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) (The most illogical solution)
Multiplying it out, we get . Evidently, , , and . The sum is .
why would you do this
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 |
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