Product of digits in an 11-digit integer

We first note that the maximum number of digits that are not $7$ s is $1+2+2+4=9$ , which is less than $11$ . This means that we could freely choose how many $1$ s, $2$ s, $3$ s, and $5$ s we have without caring for the restriction on the maximum possible number of total digits. There is no minimum number of numbers that are not $7$ s, because the number of digits in the number we are constructing is $11$ , and we could have up to $11$ $7$ s.

We could have $0$ or $1$ $1$ s, from $0$ to $2$ $2$ s, from $0$ to $2$ $3$ s, and from $0$ to $4$ $5$ s. The rest of the numbers would be $7$ s.

Hence we have $2$ choices for the $1$ s, $3$ choices for the $2$ s, $3$ choices for the $3$ , and $5$ choices for the $5$ s. Since the choices are independent as said above, there are $(2)(3)(3)(5)=90$ possible choices for the set of numbers used.

Note that if the set of numbers used is different, then the product of the digits would be different since all of the numbers $1, 2, 3, 5, 7$ are pairwise relatively prime.

Hence, there are $\boxed{90}$ . possibilities for the product of the digits of the number that Martin is making.

Since we are not asked to find the possible number of combinations/permutations of these numbers, the question becomes much simpler. The number of disting possibilities there are for products of the digts of the number depends entirely on the combination of numbers you choose. This also means that the order of the numbers does not matter, as the order in which you multiply digits does not matter.

If we think about it a certain way, it makes much more sense. If we think of the 7s as placeholders, we only have to worry about how many ways you can arrange the 1s, 2s, 3s, and 5s. If we use all of them, we still use 2 7s, but if we use none then we have enough 7s to still make an 11 digit number.

First we look at 1. Since there can be a maximum of 1, there are two possibilities in respect to the number: We can either have the number 1 or we can not have the number 1. We do this with every other number in regards to their maximum possibilities: For number 2, there are 3 possibilities: Zero 2s, One 2, and Two 2s.

With every number we get the final number of possibilities to be 2 x 3 x 3 x 5, which represent one, two, three and five respectively. We can ignore the sevens as we are simply thinking of them as placeholders. The final result of this multiplication is 90, therefore there are 90 distinct possibilities of the digits numbers in the 11 digit integer.

The first thing to notice in this problem is that since we are looking for distinct products of the digits, by the commutative property of multiplication, the order of the digits does not matter e.g. 12235777777 will yield the same product as 71777237752. The second thing to notice is that there is essentially no limit to the number of 7's we can choose (we can have all 7's if we'd like). Also, we know that regardless of our choice of 1's, 2's, 3's and 5's, there will always be remaining digits that need to be filled by 7's, because we have only one + two + two + four = nine digits to choose from that are not 7's, and we need to get to eleven digits. Therefore, we need only count the number of ways we can choose 1's, 2's, 3's and 5's from among one 1, two 2's, two 3's, and four 5's. The remaining digits are are restricted to being 7's.

Since we have the option of either choosing a number (1,2,3,5) to be included in our product or not be included in our product, we have two choices of how many 1's to choose (we can choose zero or one 1's), we have three choices of how many 2's to choose (we can choose zero, one or two 2's), we have three choices of how many 3's to choose (we can choose zero, one or two 3's), and we have five choices of how many 5's to choose (we can choose zero, one, two, three, or four 5's). We know that each combination of 1's, 2's, 3's, 5's and subsequently remaining 7's will yield a unique product because 2, 3, 5 and 7 are all coprime. Therefore the total number of distinct combinations is

2 * 3 * 3 * 5 = 90