HMMT 十一月 2024 · GEN 赛 · 第 6 题
HMMT November 2024 — GEN Round — Problem 6
题目详情
- A positive integer n is stacked if 2 n has the same number of digits as n and the digits of 2 n are multiples of the corresponding digits of n . For example, 1203 is stacked because 2 × 1203 = 2406 , and 2 , 4 , 0 , 6 are multiples of 1 , 2 , 0 , 3 , respectively. Compute the number of stacked integers less than 1000 . ◦
解析
- A positive integer n is stacked if 2 n has the same number of digits as n and the digits of 2 n are multiples of the corresponding digits of n . For example, 1203 is stacked because 2 × 1203 = 2406, and 2 , 4 , 0 , 6 are multiples of 1 , 2 , 0 , 3, respectively. Compute the number of stacked integers less than 1000. Proposed by: Srinivas Arun Answer: 135 Solution: We do casework on the number of digits of n. One digit. There are 4 one-digit stacked integers: 1, 2, 3, 4. Two digits. Suppose n = ab is a two-digit integer. If a < 5 and b < 5, then the digits of 2 n are double the respective digits of n , so n is stacked; there are 4 · 5 = 20 such n . Otherwise, since 2 n < 100, we still must have a < 5, so b ≥ 5. Then the last digit of 2 n is 2 b − 10, so b | 2 b − 10, which implies that b = 5 . Then the first digit of 2 n is 2 a + 1, which a must divide, so a = 1. Thus, the only stacked n with b ≥ 5 is 15. Adding that to the 20 stacked numbers with b < 5 gives us 21 two-digit stacked integers. Three digits. Suppose n = abc is a three-digit integer. If a , b , and c are all less than 5, then the digits of 2 n are double the respective digits of n , so n is stacked; there are 4 · 5 · 5 = 100 such n . Otherwise, since 2 n < 1000, we must have a < 5. We now casework on which of b and c are at least 5. • If b ≥ 5 and c ≥ 5, then the digits of 2 n are 2 a + 1, 2 b − 9, and 2 c − 10 in order. Thus, a | 2 a + 1, b | 2 b − 9, and c | 2 c − 10, which implies a = 1, b = 9, and c = 5. Thus 195 is the only stacked number in this case. • If c ≥ 5 only, then 2 n = 200 a + 2 bc has first digit 2 a and last two digits 2 bc , so n is stacked if and only if bc to be stacked. Since c ≥ 5, as proved before, the only such stacked bc is 15, so we get 4 stacked numbers in this case: 115, 215, 315, and 415. • If b ≥ 5 only, then 2 n has last digit 2 c and first two digits 2 ab , so n is stacked if and only if ab to be stacked. As b ≥ 5, similar to the previous case, the only such stacked ab is ab = 15, so we get 5 stacked numbers in this case: 150, 151, 152, 153, and 154. Summing over all cases, there are 100 + 1 + 4 + 5 = 110 three-digit stacked integers. Our final answer is 4 + 21 + 110 = 135 . ◦