A stationery distributor sorted out his erasers equally into 42 containers. The erasers in each container are all repacked into boxes of a dozen erasers each. Knowing that the original total number of erasers is a non-zero perfect cube, what is the minimum number of boxes required to pack all the erasers.
Let's work backwards to find the minimum number of boxes required to pack all the erasers.
First, we need to find the total number of erasers. Since the erasers are sorted equally into 42 containers, we can represent the number of erasers in each container with the variable x. Thus, the total number of erasers is 42x.
Since we need to repack the erasers into boxes of a dozen, we need to divide 42x by 12. However, this division must result in a non-zero perfect cube.
Let's start dividing by 12 and see when we get a perfect cube:
- When x = 1, 42x/12 = 3.5, which is not a perfect cube.
- When x = 2, 42x/12 = 7, which is not a perfect cube.
- When x = 3, 42x/12 = 10.5, which is not a perfect cube.
- When x = 4, 42x/12 = 14, which is not a perfect cube.
- When x = 5, 42x/12 = 17.5, which is not a perfect cube.
- When x = 6, 42x/12 = 21, which is not a perfect cube.
- When x = 7, 42x/12 = 24.5, which is not a perfect cube.
- When x = 8, 42x/12 = 28, which is not a perfect cube.
- When x = 9, 42x/12 = 31.5, which is not a perfect cube.
- When x = 10, 42x/12 = 35, which is not a perfect cube.
We haven't found a perfect cube yet, so let's try a different approach. We know that 42x must be divisible by 12 to get a whole number of boxes, so let's factor out 12 from 42x:
42x = 2 x 3 x 7 x x
Since 42x is a perfect cube, we know that x must have factors of 2, 3, and 7 to make the product a perfect cube. Let's try different values of x until we find the smallest one that works:
- When x = 2 x 3 x 7 = 42, 42x/12 = 147, which is a perfect cube (7^3).
- When x = 2 x 3 x 7 x 2 = 84, 42x/12 = 294, which is a perfect cube (6^3).
- When x = 2 x 3 x 7 x 3 = 126, 42x/12 = 441, which is a perfect cube (9^3).
Thus, the minimum number of boxes required to pack all the erasers is 147.
To find the minimum number of boxes required to pack all the erasers, we need to determine the total number of erasers and then divide it by the capacity of each box.
Given that the stationery distributor sorted the erasers equally into 42 containers, we know that each container contains the same number of erasers. Let's denote this number as 'x'.
The total number of erasers is then given by the product of the number of containers and the number of erasers in each container: Total number of erasers = 42 * x.
To find the minimum number of boxes required, we need to divide the total number of erasers by the capacity of each box.
Since each box can hold a dozen erasers (12 erasers), we divide the total number of erasers by 12 to get the number of boxes.
Minimum number of boxes required = (42 * x) / 12.
Now, we know that the original total number of erasers is a non-zero perfect cube. This means that the total number of erasers can be expressed as the cube of a whole number greater than zero.
To minimize the number of boxes, we need to find the smallest possible value of 'x' that satisfies this condition. We start by trying different values of 'x' and checking if 42 * x is a perfect cube.
Let's try some values:
- If we choose x = 1, then the total number of erasers would be 42. However, this is not a perfect cube.
- If we choose x = 2, then the total number of erasers would be 84. Again, this is not a perfect cube.
- If we choose x = 3, then the total number of erasers would be 126. This is still not a perfect cube.
Continuing this process, we find that when we choose x = 6, the total number of erasers is 252, which is equal to (6^3). Therefore, the smallest possible value of 'x' is 6.
Substituting this value into the equation:
Minimum number of boxes required = (42 * 6) / 12 = 21.
So, the minimum number of boxes required to pack all the erasers is 21.