2024 AMC 8
Complete problem set with solutions and individual problem pages
Minh enters the numbers through into the cells of a grid in some order. She calculates the product of the numbers in each row and column. What is the least number of rows and columns that could have a product divisible by ?
- A.
- B.
- C.
- D.
- E.
Solution 1
Note you can swap/rotate any configuration of rows, such that all the rows and columns that have a product of 3 are in the top left. Hence the points are bounded by a rectangle. This has area and rows and columns divisible by . We want and minimized.
If , we achieve minimum with .
If ,our best is . Note if , . Because , there is no smaller answer, and we get .
Solution 2
For a row or column to have a product divisible by , there must be a multiple of in the row or column. To create the least amount of rows and columns with multiples of , we must find a way to keep them all together, to minimize the total number of rows and columns with multiples of threes in it. From to , there are multiples of . So we have to fill cells with numbers that are multiples of . If we put of these numbers in a grid, there would be rows and columns ( in total), with products divisible by . However, we have numbers, so numbers still need to be put in the grid. If we put both numbers in the th column, but one in the first row, and one in the second row, (next to the by grid already filled), we would have a total of columns now, and still rows with products that are multiples of . Since , the answer is
