Difference between revisions of "2019 AMC 10B Problems/Problem 14"
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==Solution 3 (pure bash)== | ==Solution 3 (pure bash)== | ||
| − | Computing <math>19!</math>, we get <math>121,645,100,408,832,000</math>, so <math>T = 4, M = 8, H = 0</math>. <math>4 + 8 + 0 = \boxed{\textbf{(C) }12}</math> | + | Computing <math>19!</math>, we get <math>121,645,100,408,832,000</math>, so <math>T = 4, M = 8, H = 0</math>. |
| + | <math>4 + 8 + 0 = \boxed{\textbf{(C) }12}</math> | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 14:24, 4 September 2020
Contents
[hide]Problem
The base-ten representation for 
is 
, where 
, 
, and 
denote digits that are not given. What is 
?
Solution 1
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 
. Their divisibility rules (see Solution 2) tell us that 
and that 
. By guess and checking, we see that 
is a valid solution. Therefore the answer is 
.
Solution 2 (similar to Solution 1)
We know that , because ends in three zeroes (see Solution 1). Furthermore, we know that and are both factors of . We can simply use the divisibility rules for and for this problem to find and . For to be divisible by , the sum of digits must simply be divisible by . Summing the digits, we get that 
must be divisible by . This leaves either 
or 
as our answer choice. Now we test for divisibility by . For a number to be divisible by , the alternating sum must be divisible by (for example, with the number 
, 
, so is divisible by ). Applying the alternating sum test to this problem, we see that 
must be divisible by 11. By inspection, we can see that this holds if 
and 
. The sum is 
.
Solution 3 (pure bash)
Computing , we get 
, so 
.
Video Solution
~IceMatrix
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. 
