Difference between revisions of "2024 AMC 8 Problems/Problem 16"

(Problem 16)
(Solution)
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==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 13:48, 26 January 2024

Problem 16

Minh enters the numbers $1$ through $81$ into the cells of a $9 \times 9$ 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 $3$?

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.”

Video Solution 1 (easy to digest) by Power Solve

https://youtu.be/zxkL4c316vg