AMC 8 2018 Problem 14
AMC8 2018 年第 14 题
整除与因数★★★☆☆
Let be the greatest five-digit number whose digits have a product of 120 . What is the sum of the digits of ?
表示各个位上数字之积为 120 的最大的五位数。那么 N 的各个位上数字之和是多少？
- A.15
- B.16
- C.17
- D.18
- E.20
正确答案 · Correct Answer
D
解析 · Solution
Solution 1
If we start off with the first digit, we know that it can't be 9 since 9 is not a factor of 120. We go down to the digit 8, which does work since it is a factor of 120. Now, we have to know what digits will take up the remaining four spots. To find this result, just divide . The next place can be 5, as it is the largest factor, aside from 15. Consequently, our next three values will be 3,1 and 1 if we use the same logic. Therefore, our five-digit number is 85311, so the sum is . 120 is 5!, so we have (5)(4)(3)(2)(1) = 120. (Alternatively, you could identify the prime factors (5)(3)(2)(2)(2) = 120.) Now look for the largest digit you can create by combining these factors. Use this largest digit for the ten-thousands place: 8 _, _ _ _ Next you use the 5 and the 3 for the next places: 85,3 _ _ (You can't use because the 2 was used to make 8.) Fill the remaining places with 1: 85,311 = (5)(8)(3)(1)(1) =120, 8+5+3+1+1=18.
继续练习