# Problem of the Week Problem D and Solution Digits Multiplied

## Problem

The digits of any positive integer can be multiplied together to give the digit product for the integer. For example, $$345$$ has the digit product of $$3 \times 4 \times 5 = 60$$. There are many other positive integers that have $$60$$ as a digit product. For example, $$2532$$ and $$14\,153$$ both have a digit product of $$60$$. Note that $$256$$ is the smallest positive integer with a digit product of $$60$$.

There are also many positive integers that have a digit product of $$2160$$. Determine the smallest such integer.

## Solution

Let $$N$$ be the smallest positive integer whose digit product is $$2160$$.

In order to find $$N$$, we must find the minimum possible number of digits whose product is $$2160$$. This is because if the integer $$a$$ has more digits than the integer $$b$$, then $$a>b$$. Once we have determined the digits that form $$N$$, then the integer $$N$$ is formed by writing those digits in increasing order.

Note that the digits of $$N$$ cannot include $$0$$, or else the digit product of $$N$$ would be $$0$$.

Also, the digits of $$N$$ cannot include $$1$$, otherwise we could remove the $$1$$ and obtain an integer with fewer digits (and thus, a smaller integer) with the same digit product. Therefore, the digits of $$N$$ will be between $$2$$ and $$9$$, inclusive.

Since digits of $$N$$ multiply to $$2160$$, we can use the prime factorization of $$2160$$ to help determine the digits of $$N$$: $2160 = 2^4 \times 3^3 \times 5$

In order for the digit product of $$N$$ to have a factor of $$5$$, one of the digits of $$N$$ must equal $$5$$.

The digit product of $$N$$ must also have a factor of $$3^3= 27$$. We cannot find one digit whose product is $$27$$ but we can find two digits whose product is $$27$$. In particular, $$27 = 3 \times 9$$. Therefore, $$N$$ could also have the digits $$3$$ and $$9$$.

Then the remaining digits of $$N$$ must have a product of $$2^4 = 16$$. We need to find a combination of the smallest number of digits whose product is $$16$$. We cannot have one digit whose product is $$16$$, but we can have two digits whose product is $$16$$. In particular, $$16 = 2\times 8$$ and $$16 = 4\times 4$$.

Therefore, it is possible for $$N$$ to have $$5$$ digits. We have seen that this can happen when the digits of $$N$$ are $$5$$, $$3$$, $$9$$, $$2$$, $$8$$ or $$5$$, $$3$$, $$9$$, $$4$$, $$4$$.

However, notice that the product of $$2$$ and $$3$$ is $$6$$. Therefore, rather than using the digits $$5$$, $$3$$, $$9$$, $$2$$, $$8$$, we can replace the two digits $$2$$ and $$3$$ with the single digit $$6$$. We now have the digits $$6$$, $$5$$, $$8$$, and $$9$$. The smallest integer using these digits is $$5689$$.

It is possible that we can take a factor of $$2$$ from the $$8$$ and a factor of $$3$$ from the 9 to make another $$2 \times 3 = 6$$. However, the digits will be now be $$5$$, $$6$$, $$6$$, $$4$$, and $$3$$. This means we will have a five-digit number which is larger than than the four-digit number $$5689$$.

Therefore, the smallest possible integer with a digit product of $$2160$$ is $$5689$$.