Question
Emma is keeping an equal number of square and rectangular boxes in a local storage facility. She stores square boxes in stacks of 6 and rectangular boxes in stacks of 9. What is the smallest number of each type of box that Emma could have?
boxes
boxes
Answers
GPT-4o mini
To find the smallest number of boxes that Emma could have of both types (square and rectangular), we need to find the smallest common multiple of the two stacking sizes, which are 6 for the square boxes and 9 for the rectangular boxes.
First, we'll find the least common multiple (LCM) of 6 and 9.
1. **Factorization**:
- The prime factorization of 6 is \(2 \times 3\).
- The prime factorization of 9 is \(3^2\).
2. **Finding the LCM**:
- For the LCM, take the highest power of each prime factor present in the factorizations:
- The prime factor 2 appears in 6 as \(2^1\).
- The prime factor 3 appears in 9 as \(3^2\).
Thus, the LCM is calculated as follows:
\[
\text{LCM} = 2^1 \times 3^2 = 2 \times 9 = 18.
\]
3. **Equal number of boxes**:
Since Emma keeps an equal number of square and rectangular boxes, we need to determine how many boxes correspondingly would fit in those stacking sizes.
Given that:
- For square boxes stacked in groups of 6, the smallest number that fits all groups is 18 (as \(18 \div 6 = 3\) stacks).
- For rectangular boxes stacked in groups of 9, the same applies (as \(18 \div 9 = 2\) stacks).
Thus, the smallest number of each type of box Emma could have is:
\[
\boxed{18}.
\]
This means Emma could have 18 square boxes and 18 rectangular boxes, satisfying the conditions of equal quantity and stacking arrangements.
First, we'll find the least common multiple (LCM) of 6 and 9.
1. **Factorization**:
- The prime factorization of 6 is \(2 \times 3\).
- The prime factorization of 9 is \(3^2\).
2. **Finding the LCM**:
- For the LCM, take the highest power of each prime factor present in the factorizations:
- The prime factor 2 appears in 6 as \(2^1\).
- The prime factor 3 appears in 9 as \(3^2\).
Thus, the LCM is calculated as follows:
\[
\text{LCM} = 2^1 \times 3^2 = 2 \times 9 = 18.
\]
3. **Equal number of boxes**:
Since Emma keeps an equal number of square and rectangular boxes, we need to determine how many boxes correspondingly would fit in those stacking sizes.
Given that:
- For square boxes stacked in groups of 6, the smallest number that fits all groups is 18 (as \(18 \div 6 = 3\) stacks).
- For rectangular boxes stacked in groups of 9, the same applies (as \(18 \div 9 = 2\) stacks).
Thus, the smallest number of each type of box Emma could have is:
\[
\boxed{18}.
\]
This means Emma could have 18 square boxes and 18 rectangular boxes, satisfying the conditions of equal quantity and stacking arrangements.