2019 AMC 10B Problems/Problem 14
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 | |
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| All AMC 10 Problems and Solutions | ||
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