Question
Edwin has 12 commemorative plates and 44 commemorative spoons. He wants to display them in groups throughout his house, each with the same combination of plates and spoons, with none left over. What is the greatest number of groups Edwin can display?
Answers
GPT-4o mini
To find the greatest number of groups in which Edwin can display his commemorative plates and spoons, we need to determine the greatest common divisor (GCD) of the two quantities: 12 plates and 44 spoons.
The GCD can be found using the prime factorization method or the Euclidean algorithm.
First, we will perform the prime factorization for both numbers:
**Prime factorization of 12:**
- 12 can be factored as \( 2^2 \times 3^1 \).
**Prime factorization of 44:**
- 44 can be factored as \( 2^2 \times 11^1 \).
Now, we identify the common prime factors and their lowest powers:
- The only common prime factor between 12 and 44 is \( 2 \).
- The lowest power of \( 2 \) in both factorizations is \( 2^2 \).
Thus, the GCD is:
\[
GCD(12, 44) = 2^2 = 4.
\]
Now we can verify this using the Euclidean algorithm:
1. We start with the two numbers: \( 44 \) and \( 12 \).
2. We divide \( 44 \) by \( 12 \) to get the quotient and the remainder:
\[
44 \div 12 = 3 \quad \text{(quotient)} \quad \text{remainder} = 44 - (12 \times 3) = 44 - 36 = 8.
\]
3. Then, apply the algorithm again using \( 12 \) and the remainder \( 8 \):
\[
12 \div 8 = 1 \quad \text{(quotient)} \quad \text{remainder} = 12 - (8 \times 1) = 4.
\]
4. Now apply it again using \( 8 \) and \( 4 \):
\[
8 \div 4 = 2 \quad \text{(quotient)} \quad \text{remainder} = 8 - (4 \times 2) = 0.
\]
Since the last non-zero remainder is \( 4 \), this confirms that the GCD of \( 12 \) and \( 44 \) is indeed \( 4 \).
Thus, the greatest number of groups Edwin can display, with each group containing the same combination of plates and spoons without any left over, is:
\[
\boxed{4}.
\]
The GCD can be found using the prime factorization method or the Euclidean algorithm.
First, we will perform the prime factorization for both numbers:
**Prime factorization of 12:**
- 12 can be factored as \( 2^2 \times 3^1 \).
**Prime factorization of 44:**
- 44 can be factored as \( 2^2 \times 11^1 \).
Now, we identify the common prime factors and their lowest powers:
- The only common prime factor between 12 and 44 is \( 2 \).
- The lowest power of \( 2 \) in both factorizations is \( 2^2 \).
Thus, the GCD is:
\[
GCD(12, 44) = 2^2 = 4.
\]
Now we can verify this using the Euclidean algorithm:
1. We start with the two numbers: \( 44 \) and \( 12 \).
2. We divide \( 44 \) by \( 12 \) to get the quotient and the remainder:
\[
44 \div 12 = 3 \quad \text{(quotient)} \quad \text{remainder} = 44 - (12 \times 3) = 44 - 36 = 8.
\]
3. Then, apply the algorithm again using \( 12 \) and the remainder \( 8 \):
\[
12 \div 8 = 1 \quad \text{(quotient)} \quad \text{remainder} = 12 - (8 \times 1) = 4.
\]
4. Now apply it again using \( 8 \) and \( 4 \):
\[
8 \div 4 = 2 \quad \text{(quotient)} \quad \text{remainder} = 8 - (4 \times 2) = 0.
\]
Since the last non-zero remainder is \( 4 \), this confirms that the GCD of \( 12 \) and \( 44 \) is indeed \( 4 \).
Thus, the greatest number of groups Edwin can display, with each group containing the same combination of plates and spoons without any left over, is:
\[
\boxed{4}.
\]