Problem 17. The number n is the largest 6-digit number with the product of all its digits equal to 180. What is the sum of the digits of n? So to find the largest 6-digit number with this property, we want to maximize the digits in order. So first it's important to make sure the first digit is as large as possible. Then, given that, we'll try to make the second digit as large as possible, and so on for the rest of the digits. So first let's look at the prime factorization of 180. 180 is 2 squared times 3 squared times 5, and we want to pick the largest possible first digit, such that the product of all the digits can be 180. So that means the first digit has to be a divisor of 180, and since 9, which is the largest digit, is a divisor of 180, we will make the first digit 9. Now the product of the remaining digits is 20, which is 180 divided by 9. Now, given that the product of the remaining 5 digits is 20, we want to maximize the second digit. So the second digit is going to be the largest digit that is a divisor of 20, which is 5. Then the product of the remaining digits is 4, and so the third digit is going to be the largest digit that's a divisor of 4, which is just 4. And then the product of the remaining digits is going to have to be 1, so the last three digits of the six digit number are all going to be 1. So now the digits in order are 9, 5, 4, 1, 1, 1, and so their sum is 21, and the answer is A.

largest number

6-digit

product of digits

factorization

180