To find the minimum number of pears that both Philip and Grayson must pack, we need to find the least common multiple (LCM) of the number of pears each crate can hold, which are 7 and 6.
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Find the prime factorization:
- The prime factorization of 7 is \(7^1\).
- The prime factorization of 6 is \(2^1 \times 3^1\).
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Determine the LCM:
- The LCM takes the highest power of each prime factor present in the factorizations.
- For 7, the highest power is \(7^1\).
- For 2, the highest power is \(2^1\).
- For 3, the highest power is \(3^1\).
Therefore, the LCM is calculated as follows: \[ \text{LCM} = 7^1 \times 2^1 \times 3^1 = 7 \times 2 \times 3 \]
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Perform the multiplication: \[ 7 \times 2 = 14 \] \[ 14 \times 3 = 42 \]
Thus, the least common multiple of 7 and 6 is 42.
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Calculate how many crates each person packs:
- Philip's crates (7 pears each): \[ \frac{42}{7} = 6 \text{ crates} \]
- Grayson's crates (6 pears each): \[ \frac{42}{6} = 7 \text{ crates} \]
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Minimum number of pears packed: Both Philip and Grayson must pack a total of 42 pears.
Therefore, the answer is:
\[ \boxed{42} \text{ pears} \]
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