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 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 inspection, we see that 
is a valid solution. Therefore the answer is 
.
Solution 2 (similar to Solution 1)
We know that and are both factors of . Furthermore, we know that , because ends in three zeroes (see Solution 1). 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(answer choices)
Because 9 is a factor 19! is congruent to 0 mod 9. Doing mod 9 we have 33+T+M+H is congruent to 0 mod 9, since 33 is congruent to -3 mod 9, T+M+H is congruent to 3 mod 9. The only answer choice that is 3 mod 9 is 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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