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

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==Solution==
 
==Solution==
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'''These are just left here for future conveniency.'''
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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.
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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.
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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>18 - 6 = 12</math> rows that are multiples of <math>3</math>. However, we can improve this number by making some rows and columns intersect so that some squares help fill out both rows and columns
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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
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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)
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We see that filling 7 rows/columns would usually take 7 x 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 = (D) 11  -IwOwOwl253
 
 
==Solution 2==
 
==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 <math>a \times b</math> rectangle. This has <math>ab</math> are and <math>a+b</math> rows and columns divisible by <math>3</math>. We want <math>ab\ge 27</math> and <math>ab</math> minimized.  
 
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> are and <math>a+b</math> rows and columns divisible by <math>3</math>. We want <math>ab\ge 27</math> and <math>ab</math> minimized.  

Revision as of 17:16, 26 January 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 $3$?

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)

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 $ab$ are and $a+b$ rows and columns divisible by $3$. We want $ab\ge 27$ and $ab$ 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 (D) 11.

- SahanWijetunga

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 CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=DLzFB4EplKk

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