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).
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Find the prime factorization of each number:
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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 \( 2^5 \times 3^1 \times 5^1 \).
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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).
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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.
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