2019 AMC 10B Problems/Problem 14
The base-ten representation for is , where , , and denote digits that are not given. What is ?
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 . Since their divisibility rules gives us that is congruent to mod and that is congruent to mod . By inspection, we see that is a valid solution. Therefore the answer is , which is (C).
With investing just a little bit of time, we can manually calculate 19!. If we prime factorize 19!, it becomes . This looks complicated, but we can use elimination methods to make it simpler. , and . If we put these aside for a moment, we have left from the original 19!. , and . We have the 2's and 3's out of the way, and then we have . Now if we multiply all the values calculated, we get . Thus , and the answer , thus (C).
We know that and are both factors of . Furthermore, we know that H is 0 because ends in three zeroes. We can simply use the divisibility rules for and for this problem to find T and M. For to be divisible by , the sum of digits must be divisible by . Summing the digits, we get that T + M + must be divisible by . This leaves either A or C as our answer choice. Now we test for divisibility by . For a number to be divisible by eleven, the alternating sum must be divisible by 11(ex. , -+- = -11 so is divisible by ). Applying the alternating sum to this problem, we see that T-M-7 must be divisible by 11. By inspection, we can see that this holds if T is and M is . The sum is + + = 12 or (C). -- krishdhar
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