Difference between revisions of "2019 AMC 10B Problems/Problem 14"
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==Solution 2== | ==Solution 2== | ||
− | We can manually calculate 19!. If we prime factorize 19!, it becomes <math>2^16 \cdot 3^8 \cdot 5^3 \cdot 7^2 \cdot 11 \cdot 13 \cdot 17 \cdot 19</math>. This looks complicated, but we can use elimination methods to make it simpler. <math>2^3 \cdot 5^3 = 1000</math>, and <math>7 \cdot 11 \cdot 13 \cdot = 1001</math>. If we put these aside for a moment, we have <math>2^13 \cdot 3^8 \cdot 7 \cdot 17 \cdot 19</math>. <math>2^13 = 2^10 \cdot 2^3 = 1024 \cdot 8 = 8192</math>, and <math>3^8 = (3^4)^2 = 81^2 = 6561</math>. We have the 2's and 3's out of the way, and then we have <math>7 \cdot 17 \cdot 19 = 2261</math>. Now if we multiply all the values calculated, we get <math>1000 \cdot 1001 \cdot 8192 \cdot 6561 \cdot 2261 = 121,645,100,408,832,000</math>. Thus <math>T = 4, M = 8, H = 0</math>, and the answer <math>T + M + H = 12</math>, thus (C). | + | We can manually calculate 19!. If we prime factorize 19!, it becomes <math>2^{16} \cdot 3^8 \cdot 5^3 \cdot 7^2 \cdot 11 \cdot 13 \cdot 17 \cdot 19</math>. This looks complicated, but we can use elimination methods to make it simpler. <math>2^3 \cdot 5^3 = 1000</math>, and <math>7 \cdot 11 \cdot 13 \cdot = 1001</math>. If we put these aside for a moment, we have <math>2^{13} \cdot 3^8 \cdot 7 \cdot 17 \cdot 19</math>. <math>2^{13} = 2^10 \cdot 2^3 = 1024 \cdot 8 = 8192</math>, and <math>3^8 = (3^4)^2 = 81^2 = 6561</math>. We have the 2's and 3's out of the way, and then we have <math>7 \cdot 17 \cdot 19 = 2261</math>. Now if we multiply all the values calculated, we get <math>1000 \cdot 1001 \cdot 8192 \cdot 6561 \cdot 2261 = 121,645,100,408,832,000</math>. Thus <math>T = 4, M = 8, H = 0</math>, and the answer <math>T + M + H = 12</math>, thus (C). |
==See Also== | ==See Also== |
Revision as of 22:51, 14 February 2019
Contents
Problem
The base-ten representation for is , where , , and denote digits that are not given. What is ?
Solution
We can figure out by noticing that will end with zeroes, as there are three 's in its prime factorization. Next we use the fact that is a multiple of both and . Since their divisibility rules gives us that is congruent to mod and that is congruent to mod . By inspection, we see that is a valid solution. Therefore the answer is , which is (C).
Solution 2
We can manually calculate 19!. If we prime factorize 19!, it becomes . This looks complicated, but we can use elimination methods to make it simpler. , and . If we put these aside for a moment, we have . , and . We have the 2's and 3's out of the way, and then we have . Now if we multiply all the values calculated, we get . Thus , and the answer , thus (C).
See Also
2019 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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All AMC 10 Problems and Solutions |
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