Difference between revisions of "2019 AMC 10B Problems/Problem 14"
(→Solution) |
(→Solution) |
||
| Line 6: | Line 6: | ||
- AZAZ12345 | - 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== | ==See Also== | ||
Revision as of 16:56, 14 February 2019
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 | |
| 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. 
