Problem D and Solution

Digits Multiplied

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.

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 \(5689\).