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 factors of in its prime factorization, so there would be 3 powers of 10 meaning it will end in 3 zeros. 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 (Brute force)
Multiplying it out, we get . Evidently, , , and . The sum is .
Do not do this in a real contest.
Solution 4 (1001?)
7, 11, 13 are < 19 and 1001 = 7 * 11 * 13. Check the alternating sum of block 3: H00 - 832 + 40M - 100 + 6T5 - 121 and it is divisible by 1001. HTM + 5 - 53 = 0 (mod 1001) => HTM = 48.
The answer is . ~ AliciaWu
Do this in a real contest.
Video Solution by OmegaLearn
https://youtu.be/p5f1u44-pvQ?t=760
~ pi_is_3.14
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.