Difference between revisions of "2024 AMC 8 Problems/Problem 16"
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Latest revision as of 21:33, 1 December 2024
Contents
- 1 Problem 16
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Video Solution by Math-X (Apply this simple strategy that works every time!!!)
- 6 Video Solution (A Clever Explanation You’ll Get Instantly)
- 7 Video Solution 1 (easy to digest) by Power Solve
- 8 Video Solution 2 by OmegaLearn.org
- 9 Video Solution 3 by SpreadTheMathLove
- 10 Video Solution by NiuniuMaths (Easy to understand!)
- 11 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 12 Video Solution by Interstigation
- 13 Video Solution by Dr. David
- 14 Video Solution by WhyMath
- 15 See Also
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 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 
, then 
, and hence there is no smaller answer, and we get 
.
- SahanWijetunga ~vockey(minor edits)
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. 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 remain to 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 already filled), we would have a total of columns now, and still rows with products that are multiples of . So the answer is
~goofytaipan
Solution 3
In the numbers to , there are 27 multiples of three. In order to minimize the rows and columns, the best way is to make a square. However, the closest square is , meaning there are two multiples of three remaining. However, you can place these multiples right above the 5x5 square, meaning the answer is 
~e
~Shriyans Chowdhury (minor configuration error)
Video Solution by Math-X (Apply this simple strategy that works every time!!!)
https://youtu.be/BaE00H2SHQM?si=Z4Y7xHZEdRfDR-Bb&t=3952
Video Solution (A Clever Explanation You’ll Get Instantly)
https://youtu.be/5ZIFnqymdDQ?si=EKPTJZxRQUL6PAoS&t=2017
~hsnacademy
~Math-X
Video Solution 1 (easy to digest) by Power Solve
Video Solution 2 by OmegaLearn.org
Video Solution 3 by SpreadTheMathLove
https://www.youtube.com/watch?v=Svibu3nKB7E
Video Solution by NiuniuMaths (Easy to understand!)
https://www.youtube.com/watch?v=V-xN8Njd_Lc
~NiuniuMaths
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=DLzFB4EplKk
Video Solution by Interstigation
https://youtu.be/ktzijuZtDas&t=1709
Video Solution by Dr. David
Video Solution by WhyMath
See Also
| 2024 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
