HMMT 十一月 2010 · 冲刺赛 · 第 8 题

HMMT November 2010 — Guts Round — Problem 8

[ 7 ] What is the sum of all four-digit numbers that are equal to the cube of the sum of their digits (leading zeros are not allowed)?

解析

[ 7 ] What is the sum of all four-digit numbers that are equal to the cube of the sum of their digits (leading zeros are not allowed)? Answer: 10745 We want to find all integers x between 1000 and 9999 that are the cube of the sum of their digits. Of course, our search is only restricted to perfect cubes. The smallest such cube is 3 3 10 = 1000 and the largest such cube is 21 = 9261. This means we only have to check 12 different cubes, which is quite doable, but we can reduce the search even further with a little number theory. Suppose we write our number as x = 1000 a + 100 b + 10 c + d , where a , b , c , and d are the decimal digits of x . Then we have 3 ( a + b + c + d ) ≡ 1000 a + 100 b + 10 c + d ≡ a + b + c + d (mod 9) 3 If we let k = a + b + c + d , then k must be a solution to the modular equation k ≡ k (mod 9). A quick check of the values 0 through 8 shows that the only solutions are 0, 1, and 8. Now, in our search, we only have to check values that are the cube of a number which is either 0, 1, or 8 mod 9. 3 10 = 1000, but 1 + 0 + 0 + 0 6 = 10. 3 17 = 4913, and 4 + 9 + 1 + 3 = 17. 3 18 = 5832, and 5 + 8 + 3 + 2 = 18. 3 19 = 6859, but 6 + 8 + 5 + 9 6 = 19. So the only solutions are 4913 and 5832, which sum to 10745. Guts Round