Difference between revisions of "2024 AMC 8 Problems/Problem 16"

Latest revision as of 21:33, 1 December 2024

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 $1$ 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 3

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

~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

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

Video Solution by Dr. David

https://youtu.be/0kp2LdaCWYw

Video Solution by WhyMath

https://youtu.be/xtiBWCVHIHY

@@ Line 1: / Line 1: @@
-Four pigs are piggling. Mommy pig also wants to piggle. How many ways can mommy piggle?
+==Problem 16 ==
+Minh enters the numbers <math>1</math> through <math>81</math> into the cells of a <math>9 \times 9</math> 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 <math>3</math>?
+<math>\textbf{(A) } 8\qquad\textbf{(B) } 9\qquad\textbf{(C) } 10\qquad\textbf{(D) } 11\qquad\textbf{(E) } 12</math>
+==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 <math>a \times b</math> rectangle. This has <math>ab</math> area and <math>a+b</math> rows and columns divisible by <math>3</math>. We want <math>ab\ge 27</math> and <math>a+b</math> minimized.
+If <math>ab=27</math>, we achieve minimum with <math>a+b=9+3=12</math>.
+If <math>ab=28</math>,our best is <math>a+b=7+4=11</math>. Note if <math>a+b=10</math>, then <math>ab\le 25</math>, and hence there is no smaller answer, and we get <math>\boxed{\textbf{(D)} 11}</math>.
+- SahanWijetunga
+~vockey(minor edits)
+==Solution 2==
+For a row or column to have a product divisible by <math>3</math>, there must be a multiple of <math>3</math> in the row or column. To create the least amount of rows and columns with multiples of <math>3</math>, we must find a way to keep them all together, to minimize the total number of rows and columns. From <math>1</math> to <math>81</math>, there are <math>27</math> multiples of <math>3</math> (<math>81/3</math>). So we have to fill <math>27</math> cells with numbers that are multiples of <math>3</math>. If we put <math>25</math> of these numbers in a <math>5 x 5</math> grid, there would be <math>5</math> rows and <math>5</math> columns (<math>10</math> in total), with products divisible by <math>3</math>. However, we have <math>27</math> numbers, so <math>2</math> numbers remain to put in the <math>9 x 9</math> grid. If we put both numbers in the <math>6</math>th column, but one in the first row, and one in the second row, (next to the <math>5 x 5</math> already filled), we would have a total of <math>6</math> columns now, and still <math>5</math> rows with products that are multiples of <math>3</math>. So the answer is <math>\boxed{\textbf{(D)} 11}</math>
+~goofytaipan
+==Solution 3==
+In the numbers <math>1</math> to <math>81</math>, 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 <math>25</math>, meaning there are two multiples of three remaining. However, you can place these multiples right above the 5x5 square, meaning the answer is <math>\boxed{\textbf {(D)} 11}</math>
+~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==
+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
+==Video Solution by Dr. David==
+https://youtu.be/0kp2LdaCWYw
+==Video Solution by WhyMath==
+https://youtu.be/xtiBWCVHIHY
+==See Also==
+{{AMC8 box|year=2024|num-b=15|num-a=17}}
+{{MAA Notice}}

2024 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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