AMC10 2019 B

AMC10 2019 B · Q14

AMC10 2019 B · Q14. It mainly tests Primes & prime factorization, Digit properties (sum of digits, divisibility tests).

The base-ten representation for 19! is 121,675,100,40M,832,H00, where T, M, and H denote digits that are not given. What is $T + M + H$?

19!的十进制表示是121,675,100,40M,832,H00，其中T、M和H表示未给出的数字。$T + M + H$是多少？

(A) 3 3

(B) 8 8

(D) 14 14

(E) 17 17

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Because 5, 10, and 15 all have a single factor of 5 in their prime factorization, 19! ends with 3 0s. Hence $H = 0$. To determine $T$ and $M$, divisibility tests for 9 and 11 can be used. Because 19! is divisible by 9, its digit sum, $T + M + 33$, must also be divisible by 9, which implies $T + M = 3$ or $T + M = 12$. Similarly, because 19! is divisible by 11, its alternating digit sum, $(T + 13) - (M + 20) = T - M - 7$, must also be divisible by 11. This implies that $T - M = -4$ or $T - M = 7$. Combining these constraints results in only one solution in which $T$ and $M$ are digits, namely $T = 4$ and $M = 8$. Hence $T + M + H = 4 + 8 + 0 = 12$.

答案（C）：因为 5、10 和 15 在其质因数分解中都只含有一个因子 5，$19!$ 末尾有 3 个 0。因此 $H = 0$。为确定 $T$ 和 $M$，可以使用对 9 和 11 的整除判别法。由于 $19!$ 能被 9 整除，其各位数字之和 $T + M + 33$ 也必须能被 9 整除，这推出 $T + M = 3$ 或 $T + M = 12$。同样地，由于 $19!$ 能被 11 整除，其交错数字和 $(T + 13) - (M + 20) = T - M - 7$ 也必须能被 11 整除。这推出 $T - M = -4$ 或 $T - M = 7$。结合这些约束，只得到一个满足 $T$ 和 $M$ 都是数字的解，即 $T = 4$ 且 $M = 8$。因此 $T + M + H = 4 + 8 + 0 = 12$。

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