2019 AMC 10B Problems/Problem 14
Problem
The base-ten representation for 
is 
, where 
, 
, and 
denote digits that are not given. What is 
?
Solution
We can figure out H = 0 by noticing that 19! will end with 3 zeroes, as there are three 5's in its prime factorization. Next we use the fact that 19! is a multiple of both 11 and 9. sing their divisibility rules gives us that T+M is congruent to 3 mod 9 and T-M is congruent to 7 mod 11. By inspection, we see that T = 4, M = 8 is a valid solution. Therefore the answer is 4+8+0 = 12. C
- AZAZ12345
Immediately we know that H is 0 (Since there are 3 5's which contribute to 3 0's in the product) Notice the 19! is divisible by both 3 and 11. Using the divisibility rules, we find that the sum and alternating sum of the digits are 33 and -7 respectively. Thus we can eliminate all the choices that are not divisible by 3. This leaves 12 and 3. Since T-M must be 7 (so the number can be divisible by 11) 3 cannot be an answer choice. This leaves 12 and we are done.
- Pascal
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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