If $N$ is the smallest positive integer whose digits have a product of 1728 , then the sum of the digits of $N$ is
(A) 28
(B) 26
(C) 18
(D) 27
(E) 21
In order to find $N$, which is the smallest possible integer whose digits have a fixed product, we must first find the minimum possible number of digits with this product. (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 $N$ itself is formed by writing the digits in increasing order. (Given a fixed set of digits, the leading digit of $N$ will contribute
to the largest place value, and so should be the smallest digit. The next largest place value should get the next smallest digit, and so on.)

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

Also, the digits of $N$ cannot include 1, otherwise we could remove the 1s and obtain an integer with fewer digits (thus, a smaller integer) with the same product of digits.

Since the product of the digits of $N$ is 1728 , we find the prime factorization of 1728 to help us determine what the digits are:

$$
1728=9 \times 192=3^{2} \times 3 \times 64=3^{3} \times 2^{6}
$$

We must try to find a combination of the smallest number of possible digits whose product is 1728 .

Note that we cannot have 3 digits with a product of 1728 since the maximum possible product of 3 digits is $9 \times 9 \times 9=729$.

Let us suppose that we can have 4 digits with a product of 1728 .

In order for $N$ to be as small as possible, its leading digit (that is, its thousands digit) must be as small as possible.

From above, this digit cannot be 1 .

This digit also cannot be 2 , since otherwise the product of the remaining 3 digits would be 864 , which is larger than the product of 3 digits can be.

Can the thousands digit be 3 ? If so, the remaining 3 digits have a product of 576 .

Can 3 digits have a product of 576 ?

If one of these 3 digits were 7 or less, then the product of the 3 digits would be at most $7 \times 9 \times 9=567$, which is too small.

Therefore, if we have 3 digits with a product of 576 , then each digit is 8 or 9 .

Since the product is even, then at least one of the digits would have to be 8 , leaving the remaining two digits to have a product of $576 \div 8=72$.

These two digits would then have to be 8 and 9 .

Thus, we can have 3 digits with a product of 576 , and so we can have 4 digits with a product of 1728 with smallest digit 3 .

Therefore, the digits of $N$ must be $3,8,8,9$. The smallest possible number formed by these digits is when the digits are placed in increasing order, and so $N=3889$.

The sum of the digits of $N$ is $3+8+8+9=28$.

ANSWER: (A)