Product Is 60

We have $60 = 2^2\cdot 3\cdot5$ . We also note that a digit of a number is a single digit number.

As for the minimum value of the number:

• It cannot be one digit, as that digit would have to be 60, which is a two-digit number.

• It cannot be two digits, as one of those digits must be a two digit number (ex. 6 and 10, or 5 and 12).

• However, it can be a three digit number. To minimize this number, we start by minimizing the hundred's digit. This digit cannot be 1, as that would require the number to be at least four digits. Our next choice is 2, which is optimal as it is part of the prime factorization of 60. This leaves $30 = 2 \cdot 3 \cdot 5$ . The ten's digit can be neither 2 nor 3, as that would make the last digit a two-digit number. Therefore, the second digit must be 5. That makes the last digit 6.

We therefore have the answer, $\fbox{256}$ .

Factors of $60: 1,\;2,\;3,\;4,\;5,\;6,\;10,\;12,\;15,\;20,\;30,\;60$

Since we are looking for digits, we can only use factors less than $10: 1,\;2,\;3,\;4,\;5,\;6$

We see that no two digits can give a product of $60$ , so we need at least $3$ digits.

If we use $1$ , we cannot get a product of $60$ with $3$ digits.

If we use $2$ , there is one set of $3$ digits that give a product of $60: 2\times5\times6$

Arranging it in ascending order gives us the smallest possible number. Therefore, the positive integer we are looking for is $\boxed{256}$

One way to attack this problem is by prime factorization. We have

$60=2^2 \cdot3 \cdot 5=2 \cdot 2\cdot 3 \cdot 5$

Then arrange the factors. Some thing like this:

$60=3(2^2)(5)=3(4)(5)$

or

$60=2(5)(2\cdot 3)=2(5)(6)$

And so on ...

The desired answer is $256$ .

60=1×60=10×6=2×5×6
=2×30=2×5×6
=3×20=3×5×4
=4×15=4×5×3
=10×6=2×5×6
=5×12=5×3×4
Therefore=256