Problem number 28 states, how many 5-digit positive numbers have the product of their digits equal to 1000? So here we have a 5-digit number, ABCDE, and we know the product of its digits is going to be equal to 1000. So let's break 1000 down into its prime factorization. Doing this, we get that it is equal to 3 copies of 2, 3 copies of 5. Unfortunately, this is 6 digits, and we can only have 5, so let's try combining some of these terms. We cannot combine 5 and 5 because that would be 25, which would be more than 1 digit. We also cannot combine 2 and 5 because that would be 10, which would also be more than 1 digit. We could, however, combine 2 of the 2s. This would give us a digit of 4. We could even combine 3 of the 2s, and this would give us a digit of 8. Let's keep in mind the restriction that we can only have 5 digits. The first combination doesn't work because there are 6 digits. The second combination does work because there are 5 digits, and the third combination doesn't work because there are only 4 digits. But we could just add times 1, and now we would have 5 digits. So this works. So now we know the digits that exist in our number are either 18555 or 24555. Now we just need to find all the possible combinations of these digits into a unique number. To do this, we will use the idea of permutations with repetition. Basically, permutations tell you how many ways a set of elements can be ordered where order matters. So the basic formula to find this is that if you have n unique elements, then they can be arranged in n factorial unique ways. The formula with repetition that we're going to use states that if we have n elements where k of them are identical, then they can be ordered in n factorial over k factorial unique ways. In this example, we are going to be arranging 5 digits. So n is going to be equal to 5, and 3 of them repeat, 3 5s. So in total, there are going to be 5 factorial over 3 ways to arrange one of the sets of digits. So let's compute this. 5 factorial is 5 times 4 times 3 times 2 times 1, and 3 factorial is 3 times 2 times 1. Many of the values in the numerator and the denominator cancel out, and ultimately we get 5 times 4, which is a value of 20. This means that there are 20 ways to make each of the combinations of digits for ABCDE. We have two of those, so let's multiply our number by 2 to get how many total unique combinations there exist. So 2 times 20, which is 40, meaning that there exist 45 digit numbers whose digits multiply up to 1000. So the question asked us, how many 5 digit positive numbers have the product of their digits equal to 1000? The answer is 40, letter D.

5-digit numbers

digit multiplication

prime factorization

permutations

unique numbers