Question
Clayton is posting 28 flyers for the science club and 42 for the music club. He wants to make all the locations identical, with the same combination of science club flyers and music club flyers. In addition, he wants to make sure that no flyers are left over. What is the greatest number of locations that Clayton can post at?
Answers
Answered by
GPT-4o mini
To determine the greatest number of locations where Clayton can post the flyers without leaving any flyers left over, we need to find the greatest common divisor (GCD) of the number of science club flyers (28) and music club flyers (42).
We will use the prime factorization method to find the GCD.
1. **Find the prime factorization of 28:**
\[
28 = 2 \times 14 = 2 \times 2 \times 7 = 2^2 \times 7^1
\]
2. **Find the prime factorization of 42:**
\[
42 = 2 \times 21 = 2 \times 3 \times 7 = 2^1 \times 3^1 \times 7^1
\]
3. **Identify the common prime factors and their minimum exponents:**
- The common prime factors are \(2\) and \(7\).
- For \(2\), the minimum exponent is \(1\) (from \(2^1\) in \(42\)).
- For \(7\), the minimum exponent is \(1\) (common in both).
4. **Calculate the GCD by taking the product of these common factors:**
\[
\text{GCD} = 2^1 \times 7^1 = 2 \times 7 = 14
\]
Thus, the greatest number of locations that Clayton can post at without leaving any flyers left over is \( \boxed{14} \).
To verify this, we divide both the number of science club flyers and music club flyers by the GCD:
- For science club flyers:
\[
\frac{28}{14} = 2
\]
- For music club flyers:
\[
\frac{42}{14} = 3
\]
This means that at each of the 14 locations, Clayton can post 2 science club flyers and 3 music club flyers, confirming that there are no flyers left over.
We will use the prime factorization method to find the GCD.
1. **Find the prime factorization of 28:**
\[
28 = 2 \times 14 = 2 \times 2 \times 7 = 2^2 \times 7^1
\]
2. **Find the prime factorization of 42:**
\[
42 = 2 \times 21 = 2 \times 3 \times 7 = 2^1 \times 3^1 \times 7^1
\]
3. **Identify the common prime factors and their minimum exponents:**
- The common prime factors are \(2\) and \(7\).
- For \(2\), the minimum exponent is \(1\) (from \(2^1\) in \(42\)).
- For \(7\), the minimum exponent is \(1\) (common in both).
4. **Calculate the GCD by taking the product of these common factors:**
\[
\text{GCD} = 2^1 \times 7^1 = 2 \times 7 = 14
\]
Thus, the greatest number of locations that Clayton can post at without leaving any flyers left over is \( \boxed{14} \).
To verify this, we divide both the number of science club flyers and music club flyers by the GCD:
- For science club flyers:
\[
\frac{28}{14} = 2
\]
- For music club flyers:
\[
\frac{42}{14} = 3
\]
This means that at each of the 14 locations, Clayton can post 2 science club flyers and 3 music club flyers, confirming that there are no flyers left over.
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