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 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 | |
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