The product of a set of positive integers is 1989. What is their least possible sum?

To find the least possible sum of a set of positive integers whose product is 1989, we can proceed with the following analysis:

1. Break Down the Problem

First, we need to factor the number 1989 into its prime factors, which will help us in determining the integers that multiply to give 1989.

2. Relevant Concepts

To find the prime factors of 1989:

• Start by dividing 1989 by the smallest prime numbers to find factors:
  1. Is 1989 divisible by $2$? No, because it is odd.
  2. Is 1989 divisible by $3$? Yes, $1989 \div 3 = 663$.
  3. Now, factor 663: Is it divisible by $3$? Yes, $663 \div 3 = 221$.
  4. Finally, factor 221: It is divisible by $13$ (since $221 \div 13 = 17$, both $13$ and $17$ are primes).

Thus, the prime factorization of 1989 is: $1989 = 3^2 \times 13 \times 17$

3. Analysis and Detail

To minimize the sum of these factors, we need to combine factors in such a way that produces larger numbers. This can reduce the number of integers and, thus, their sum.

• The prime factors are: $3, 3, 13, 17$.
• We can combine $3 \times 3 = 9$ to create the integer $9$.

Now we can consider the integers: $9, 13, 17$

Calculating the sum: $9 + 13 + 17 = 39$

4. Verify and Summarize

Now we verify that the product of $9$, $13$, and $17$ returns to $1989$: $9 \times 13 = 117,$ $117 \times 17 = 1989.$

Thus the combination works.

Final Answer

The least possible sum of the set of positive integers whose product is 1989 is: $\boxed{39}$