Question

The local food bank is preparing take-home packages for people in need. They have 300 boxes of cereal, 550 cans of vegetables, and 480 bags of pasta. Find the greatest number of items that can be equally divided into take-home packages.(1 point) Responses 10 10 30 30 25 25 20

Answer 1

To find the greatest number of items that can be equally divided into take-home packages, we need to find the greatest common divisor (GCD) of the three quantities: 300 (boxes of cereal), 550 (cans of vegetables), and 480 (bags of pasta).

1. **Find the prime factorization of each number:**
- **300:**
\( 300 = 3 \times 100 = 3 \times 10^2 = 3 \times (2 \times 5)^2 = 3 \times 2^2 \times 5^2 \)
So, the prime factorization of 300 is \( 2^2 \times 3^1 \times 5^2 \).

- **550:**
\( 550 = 5 \times 110 = 5 \times (10 \times 11) = 5 \times 2 \times 5 \times 11 \)
So, the prime factorization of 550 is \( 2^1 \times 5^2 \times 11^1 \).

- **480:**
\( 480 = 48 \times 10 = (16 \times 3) \times (2 \times 5) = (2^4 \times 3) \times (2^1 \times 5) = 2^5 \times 3^1 \times 5^1 \)
So, the prime factorization of 480 is \( 2^5 \times 3^1 \times 5^1 \).

2. **Identify the lowest powers of the common prime factors:**
- For \(2\): The minimum power is \(2^1\) (from 550).
- For \(3\): The minimum power is \(3^0\) (there is no 3 in 550).
- For \(5\): The minimum power is \(5^1\) (from 480).

3. **Multiply the common prime factors raised to their lowest powers:**
\[
GCD = 2^1 \times 3^0 \times 5^1 = 2 \times 1 \times 5 = 10.
\]

Therefore, the greatest number of items that can be equally divided into take-home packages is **10**.