2024 AMC 8 Problems/Problem 16

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

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. If we naively put in non-multiples of $3$ in $6$ rows from the top, we get $18 - 6 = 12$ rows that are multiples of $3$ . 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 $7$ rows/columns would usually take $7 \times 9 = 63$ of our non-multiples, but if we do $4$ rows and $3$ columns, $12$ will intersect. With our $54$ being enough as we need only $51$ non-multiples of $3$ ( $63$ minus the $12$ 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 $18 - 7 = \boxed{\textbf{(D)} 11}$ -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 $a \times b$ rectangle. This has $ab$ area and $a+b$ rows and columns divisible by $3$ . We want $ab\ge 27$ and $a+b$ minimized.

If $ab=27$ , we achieve minimum with $a+b=9+3=12$ .

If $ab=28$ ,our best is $a+b=7+4=11$ . Note if $a+b=10$ , then $ab\le 25$ , and hence there is no smaller answer, and we get $\boxed{(D) 11}$ .

- SahanWijetunga ~vockey(minor edits)

Solution 3

For a row or column to have a product divisible by $3$ , there must be a multiple of $3$ in the row or column. To create the least amount of rows and columns with multiples of $3$ , we must find a way to keep them all together, to minimize the total number of rows and columns. From $1$ to $81$ , there are $27$ multiples of $3$ ( $81/3$ ). So we have to fill $27$ cells with numbers that are multiples of $3$ . If we put $25$ of these numbers in a $5 x 5$ grid, there would be $5$ rows and $5$ columns ( $10$ in total), with products divisible by $3$ . However, we have $27$ numbers, so $2$ numbers remain to put in the $9 x 9$ grid. If we put both numbers in the $6$ th column, but one in the first row, and one in the second row, (next to the $5 x 5$ already filled), we would have a total of $6$ columns now, and still $5$ rows with products that are multiples of $3$ . So the answer is $\boxed{\textbf{(D)} 11}$

~goofytaipan

Video Solution 1 (easy to digest) by Power Solve

https://youtu.be/zxkL4c316vg

Video Solution 2 by OmegaLearn.org

https://youtu.be/xfiPVmuMiXs

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

2024 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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All AJHSME/AMC 8 Problems and Solutions

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