2024 AMC 8 Problems/Problem 16

1 Problem 16
2 Solution 1
3 Solution 2
4 Solution 4
5 Video Solution by Math-X (Apply this simple strategy that works every time!!!)
6 Video Solution 1 (easy to digest) by Power Solve
7 Video Solution 2 by OmegaLearn.org
8 Video Solution 3 by SpreadTheMathLove
9 Video Solution by NiuniuMaths (Easy to understand!)
10 Video Solution by CosineMethod [🔥Fast and Easy🔥]
11 Video Solution by Interstigation
12 See Also

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 $3$ ?

$\textbf{(A) } 8\qquad\textbf{(B) } 9\qquad\textbf{(C) } 10\qquad\textbf{(D) } 11\qquad\textbf{(E) } 12$

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 $a \times b$ 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 $\boxed{\textbf{(D)} 11}$ .

- SahanWijetunga ~vockey(minor edits)

Solution 2

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

Solution 4

In the numbers $1$ to $81$ , 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 $25$ , meaning there are two multiples of three remaining. However, you can place these multiples right above the 5x5 square, meaning the answer is $\boxed{\textbf {(D)} 11}$ ~ 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

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

Video Solution by Interstigation

https://youtu.be/ktzijuZtDas&t=1709

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

Page

Toolbox

Search