The product of a list of (not necessarily distinct) positive integers is 120. What is the least possible sum of the numbers in the list?

Answer 1

To find the least possible sum of a list of positive integers whose product is 120, we can start by determining the prime factorization of 120:

\[ 120 = 2^3 \times 3^1 \times 5^1 \]

The problem involves splitting the number 120 into factors that minimize the sum of those factors. A good strategy is to use numbers that are as close together as possible and utilize smaller integers when possible, especially since smaller integers generally lead to smaller sums.

First, let’s list the different ways to factor 120 while keeping an eye on the sum of the factors.

1. Using 2's and combinations with 3 and 5:

   Start with all factors:

   • \( 1 \times 120 \): sum = \( 1 + 120 = 121 \)
   • \( 2 \times 60 \): sum = \( 2 + 60 = 62 \)
   • \( 3 \times 40 \): sum = \( 3 + 40 = 43 \)
   • \( 4 \times 30 \): sum = \( 4 + 30 = 34 \)
   • \( 5 \times 24 \): sum = \( 5 + 24 = 29 \)
   • \( 6 \times 20 \): sum = \( 6 + 20 = 26 \)
   • \( 8 \times 15 \): sum = \( 8 + 15 = 23 \)
   • \( 10 \times 12 \): sum = \( 10 + 12 = 22 \)

2. Using combinations of smaller factors:

   Next, we can examine other combinations using more factors:

   • \( 2 \times 2 \times 30 \): sum = \( 2 + 2 + 30 = 34 \)
   • \( 2 \times 3 \times 20 \): sum = \( 2 + 3 + 20 = 25 \)
   • \( 2 \times 4 \times 15 \): sum = \( 2 + 4 + 15 = 21 \)
   • \( 2 \times 5 \times 12 \): sum = \( 2 + 5 + 12 = 19 \)
   • \( 2 \times 6 \times 10 \): sum = \( 2 + 6 + 10 = 18 \)
   • \( 3 \times 4 \times 10 \): sum = \( 3 + 4 + 10 = 17 \)
   • \( 3 \times 5 \times 8 \): sum = \( 3 + 5 + 8 = 16 \)

3. Finding combinations leading to the smallest sum:

   We can break down further into as many factors of \( 2, 3, 4, 5 \) etc., as possible. Let’s breakdown \( 120 \) into \( 2 \times 3 \times 4 \times 5 \):

   • \( 2 \times 3 \times 4 \times 5 \): sum = \( 2 + 3 + 4 + 5 = 14 \)

   This combination works and gives us a total sum of \( 14 \).

4. Check common combinations for being distinct or with repetitions:

   • \( 2 + 2 + 2 + 3 + 5 \) from \( 2^3, 3, 5 = 14 \)
   • Or, combinations like \( 2 + 2 + 4 + 5 = 13 \), but we have \( 2 \times 2 \times 4 \times 5 \) giving \( 80 \neq 120 \).

Given these combinations, the least sum we have discovered that meets the criteria of a product of \( 120 \) is through various strategies:

\[ \text{The least possible sum of the numbers in the list is } \boxed{14}. \]

This optimal breakdown achieves a nice balance of minimalizing the total numerically while yielding the correct product.