Difference between revisions of "2019 AMC 10B Problems/Problem 14"
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==Solution== | ==Solution== | ||
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 <math>5</math>'s in its prime factorization. Next we use the fact that <math>19!</math> is a multiple of both <math>11</math> and <math>9</math>. Since their divisibility rules gives us that <math>T + M</math> is congruent to <math>3</math> mod <math>9</math> and that <math>T - M</math> is congruent to <math>7</math> mod <math>11</math>. By inspection, we see that <math>T = 4, M = 8</math> is a valid solution. Therefore the answer is <math>4 + 8 + 0 = 12</math>, which is (C). | 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 <math>5</math>'s in its prime factorization. Next we use the fact that <math>19!</math> is a multiple of both <math>11</math> and <math>9</math>. Since their divisibility rules gives us that <math>T + M</math> is congruent to <math>3</math> mod <math>9</math> and that <math>T - M</math> is congruent to <math>7</math> mod <math>11</math>. By inspection, we see that <math>T = 4, M = 8</math> is a valid solution. Therefore the answer is <math>4 + 8 + 0 = 12</math>, which is (C). | ||
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| + | ==Solution 2== | ||
| + | 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 \cdot. 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. 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). | ||
==See Also== | ==See Also== | ||
Revision as of 22:47, 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 2^16 \cdot 3^8 \cdot 5^3 \cdot 7^2 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot. 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. 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).
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. 
