Difference between revisions of "2019 AMC 10B Problems/Problem 14"

Revision as of 16:57, 14 February 2019

Problem

The base-ten representation for $19!$ is $121,6T5,100,40M,832,H00$ , where $T$ , $M$ , and $H$ denote digits that are not given. What is $T+M+H$ ?

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

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.

@@ Line 6: / Line 6: @@
 - 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==

Page

Toolbox

Search

Difference between revisions of "2019 AMC 10B Problems/Problem 14"

Revision as of 16:57, 14 February 2019

Problem

Solution

See Also