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

Revision as of 20:09, 21 August 2024

1 Problem 16
2 Solution 2
3 Solution 3
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 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

@@ Line 4: / Line 4: @@
 <math>\textbf{(A) } 8\qquad\textbf{(B) } 9\qquad\textbf{(C) } 10\qquad\textbf{(D) } 11\qquad\textbf{(E) } 12</math>
-==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 <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.
-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.
-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 numbers that are close together and that would make <math>4 \times 3</math> 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 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==

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