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

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~vockey(minor edits)
 
~vockey(minor edits)
  
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==Solution 3==
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==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>
 
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>
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~goofytaipan
 
~goofytaipan
  
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==Solution 4==
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==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>
 
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>
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==Video Solution by Math-X (Apply this simple strategy that works every time!!!)==
 
==Video Solution by Math-X (Apply this simple strategy that works every time!!!)==
 
https://youtu.be/BaE00H2SHQM?si=Z4Y7xHZEdRfDR-Bb&t=3952
 
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
 
~Math-X
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https://www.youtube.com/watch?v=DLzFB4EplKk
 
https://www.youtube.com/watch?v=DLzFB4EplKk
 +
 
==Video Solution by Interstigation==
 
==Video Solution by Interstigation==
 
https://youtu.be/ktzijuZtDas&t=1709
 
https://youtu.be/ktzijuZtDas&t=1709
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 +
==Video Solution by Dr. David==
 +
https://youtu.be/0kp2LdaCWYw
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2024|num-b=15|num-a=17}}
 
{{AMC8 box|year=2024|num-b=15|num-a=17}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 00:54, 24 September 2024

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___

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

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

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