Difference between revisions of "2024 AMC 8 Problems/Problem 16"
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==Solution== | ==Solution== | ||
− | + | 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 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 <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 |
− | 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 <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) |
− | We see that filling 7 rows/columns would usually take 7 | ||
==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
Contents
[hide]Problem 16
Minh enters the numbers through into the cells of a 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 ?
Solution
We know that if a row/column of numbers has a single multiple of , that entire row/column will be divisible by . Since there are multiples of from to , We need to find a way to place the non-multiples of such that they take up as many entire rows and columns as possible. If we naively put in non-multiples of in rows from the top, we get rows that are multiples of . 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 rows/columns would usually take of our non-multiples, but if we do rows and columns, will intersect. With our being enough as we need only non-multiples of ( minus the 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 -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 rectangle. This has are and rows and columns divisible by . We want and minimized.
If , we achieve minimum with .
If ,our best is . Note if , then , and hence there is no smaller answer, and we get (D) 11.
- SahanWijetunga
Video Solution 1 (easy to digest) by Power Solve
Video Solution 2 by OmegaLearn.org
Video Solution 3 by SpreadTheMathLove
https://www.youtube.com/watch?v=Svibu3nKB7E