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

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We know that if a row/column of numbers has a single multiple of <math>3</math>, that entire row/column will be divisible by <math>3</math>. Since there are <math>27</math> multiples of <math>3</math> from <math>1</math> to <math>81</math>, We need to find a way to place the <math>54</math> non-multiples of <math>3</math> such that they take up as many entire rows and columns as possible.
 
We know that if a row/column of numbers has a single multiple of <math>3</math>, that entire row/column will be divisible by <math>3</math>. Since there are <math>27</math> multiples of <math>3</math> from <math>1</math> to <math>81</math>, We need to find a way to place the <math>54</math> non-multiples of <math>3</math> such that they take up as many entire rows and columns as possible.
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If we naively put in non-multiples of <math>3</math> in <math>6</math> rows from the top, we get <math>9 - 6 = 3</math> rows that are multiples of <math>3</math>, along with all <math>9</math> columns! However, we can improve this number by making some rows and columns intersect so that some squares help fill out both rows and columns
+
If we naively put in non-multiples of <math>3</math> in <math>6</math> rows from the top, we get <math>9 - 6 = 3</math> rows that are multiples of <math>3</math>, along with all <math>9</math> columns!  
−
We see that filling <math>7</math> rows/columns would usually take <math>7 \times 9 = 63</math> of our non-multiples, but if we do <math>4</math> rows and <math>3</math> columns, <math>12</math> will intersect. With our <math>54</math> being enough as we need only <math>51</math> non-multiples of <math>3</math>(<math>63</math> minus the <math>12</math> 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 <math>18 - 7 = \boxed{\textbf{(D)} 11}</math>  -IwOwOwl253 ~andliu766(Minor edits) -c29ss1(Diagram)
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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 <math>7</math> rows/columns would usually take <math>7 \times 9 = 63</math> non-multiples of 3, but if we do <math>4</math> rows and <math>3</math> columns, <math>4 \times 3 = 12</math> of those would intersect. We now need only <math>63 - 12 = 51</math> non-multiples of <math>3</math>.
 +
This is the most efficient way to overlap squares, as <math>4</math> and <math>3</math> are close together and would be bigger than any of the other pairs like <math>5 \times 2</math> and <math>6 \times 1</math>.
 +
With only <math>54 - 51 = 3</math> of our non-multiples left, we probably can’t make another row/column and conclude the final answer to be <math>18 - 7 = \boxed{\textbf{(D)} 11}</math>  -IwOwOwl253 ~andliu766(Minor edits) -c29ss1(Diagram)
  
 
==Solution 2==
 
==Solution 2==

Revision as of 13:40, 3 August 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

[asy] unitsize(0.2cm); draw((9,18)--(-9,18)); draw((9,16)--(-9,16)); draw((9,14)--(-9,14)); draw((9,12)--(-9,12)); draw((9,10)--(-9,10)); draw((9,8)--(-9,8)); draw((9,6)--(-9,6)); draw((9,4)--(-9,4)); draw((9,2)--(-9,2)); draw((9,0)--(-9,0)); draw((9,18)--(9,0)); draw((7,18)--(7,0)); draw((5,18)--(5,0)); draw((3,18)--(3,0)); draw((1,18)--(1,0)); draw((-1,18)--(-1,0)); draw((-3,18)--(-3,0)); draw((-5,18)--(-5,0)); draw((-7,18)--(-7,0)); draw((-9,18)--(-9,0)); draw((-9,17)--(9,17), red); draw((-9,15)--(9,15), red); draw((-9,13)--(9,13), red); draw((-8,0)--(-8,18), red); draw((-6,0)--(-6,18), red); draw((-4,0)--(-4,18), red); draw((-2,0)--(-2,18), red); [/asy]

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 $9 - 6 = 3$ rows that are multiples of $3$, along with all $9$ columns! 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 $7 \times 9 = 63$ non-multiples of 3, but if we do $4$ rows and $3$ columns, $4 \times 3 = 12$ of those would intersect. We now need only $63 - 12 = 51$ non-multiples of $3$. This is the most efficient way to overlap squares, as $4$ and $3$ are close together and would be bigger than any of the other pairs like $5 \times 2$ and $6 \times 1$. With only $54 - 51 = 3$ of our non-multiples left, we probably can’t make another row/column and conclude the final answer to be $18 - 7 = \boxed{\textbf{(D)} 11}$ -IwOwOwl253 ~andliu766(Minor edits) -c29ss1(Diagram)

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 $\boxed{\textbf{(D)} 11}$.

- 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

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

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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