Question
Greg has 42 glasses and 28 mugs. He would like to set them out in identical groups, with none left over, in preparation for a dinner party. What is the greatest number of groups Greg can set out?
Answers
GPT-4o mini
To find the greatest number of identical groups that Greg can form using his glasses and mugs, we need to determine the greatest common divisor (GCD) of the two amounts: 42 glasses and 28 mugs.
We can start by finding the GCD using the prime factorization method.
First, we factor each number:
- For \( 42 \):
\[
42 = 2 \times 3 \times 7
\]
- For \( 28 \):
\[
28 = 2^2 \times 7
\]
Next, we identify the common prime factors. The common prime factors of 42 and 28 are \( 2 \) and \( 7 \).
Next, we take the lowest power of each common factor:
- For \( 2 \): the lowest power is \( 2^1 \) (because \( 42 \) has \( 2^1 \) and \( 28 \) has \( 2^2 \)).
- For \( 7 \): the lowest power is \( 7^1 \) (both have \( 7^1 \)).
Now, we multiply these together to find the GCD:
\[
\text{GCD} = 2^1 \times 7^1 = 2 \times 7 = 14
\]
Thus, the greatest number of groups that Greg can set out, with none left over, is \( \boxed{14} \).
We can start by finding the GCD using the prime factorization method.
First, we factor each number:
- For \( 42 \):
\[
42 = 2 \times 3 \times 7
\]
- For \( 28 \):
\[
28 = 2^2 \times 7
\]
Next, we identify the common prime factors. The common prime factors of 42 and 28 are \( 2 \) and \( 7 \).
Next, we take the lowest power of each common factor:
- For \( 2 \): the lowest power is \( 2^1 \) (because \( 42 \) has \( 2^1 \) and \( 28 \) has \( 2^2 \)).
- For \( 7 \): the lowest power is \( 7^1 \) (both have \( 7^1 \)).
Now, we multiply these together to find the GCD:
\[
\text{GCD} = 2^1 \times 7^1 = 2 \times 7 = 14
\]
Thus, the greatest number of groups that Greg can set out, with none left over, is \( \boxed{14} \).