Difference between revisions of "2024 AMC 8 Problems/Problem 16"
Iwowowl253 (talk | contribs) (→Solution) |
Niuniumaths (talk | contribs) (→Video Solution 3 by SpreadTheMathLove) |
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==Video Solution 3 by SpreadTheMathLove== | ==Video Solution 3 by SpreadTheMathLove== | ||
https://www.youtube.com/watch?v=Svibu3nKB7E | 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🔥]== | == Video Solution by CosineMethod [🔥Fast and Easy🔥]== |
Revision as of 02:11, 27 January 2024
Contents
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
. 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
of our non-multiples, but if we do
rows and
columns,
will intersect. With our
being enough as we need only
non-multiples of
(
minus the
overlapped). We check to see if we can fill out one more row/column, and when that fails we conclude the final answer to be
-IwOwOwl253 ~andliu766(Minor edits)
(If someone would add a diagram that would be greatly appreciated)
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 (D) 11.
- SahanWijetunga
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
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.