2024 AMC 8 Problems/Problem 16
Contents
- 1 Problem 16
- 2 Solution
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4
- 6 Video Solution by Math-X (Apply this simple strategy that works every time!!!)
- 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 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
We know that if a row/column of numbers has a single multiple of , that entire row/column will be divisible by
. Since there are
multiples of
from
to
, We need to find a way to place the
non-multiples of
such that they take up as many entire rows and columns as possible.
If we naively put in non-multiples of in
rows from the top, we get
rows that are multiples of
, along with all
columns!
However, we can improve this number by making some rows and columns intersect so that some squares help fill out both rows and columns.
We see that filling rows/columns would usually take
non-multiples of 3, but if we do
rows and
columns,
of those would intersect. We now need only
non-multiples of
.
This is the most efficient way to overlap squares, as and
are numbers that are close together and that would make
bigger than any of the other pairs like
and
.
With only
of our non-multiples left, we can’t make another row/column and conclude the final answer to be
-IwOwOwl253 ~andliu766(Minor edits) -c29ss1(Diagram)
Solution 2
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 3
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 4
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___
Video Solution by Math-X (Apply this simple strategy that works every time!!!)
https://youtu.be/BaE00H2SHQM?si=Z4Y7xHZEdRfDR-Bb&t=3952
~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
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.