Difference between revisions of "2024 AMC 8 Problems/Problem 16"
(→Problem 16) |
Iwowowl253 (talk | contribs) (→Solution) |
||
| Line 5: | Line 5: | ||
==Solution== | ==Solution== | ||
'''These are just left here for future conveniency.''' | '''These are just left here for future conveniency.''' | ||
| + | “We know that if a row/column of numbers has a single multiple of 3, that entire row/column will be divisible by 3. Since there are 27 multiples of 3 from 1 to 81, We need to find a way to place the 54 non-multiples of 3 such that they take up as many entire rows and columns as possible.” | ||
==Video Solution 1 (easy to digest) by Power Solve== | ==Video Solution 1 (easy to digest) by Power Solve== | ||
https://youtu.be/zxkL4c316vg | https://youtu.be/zxkL4c316vg | ||
Revision as of 12:48, 26 January 2024
Problem 16
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 
?
Solution
These are just left here for future conveniency. “We know that if a row/column of numbers has a single multiple of 3, that entire row/column will be divisible by 3. Since there are 27 multiples of 3 from 1 to 81, We need to find a way to place the 54 non-multiples of 3 such that they take up as many entire rows and columns as possible.”
