Difference between revisions of "2021 AMC 10B Problems/Problem 11"
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Scrabbler94 (talk | contribs) (solution 2 is exactly the same as solution 1 except x,y are used instead of m,n. Added more detail to solution 1.) |
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<math>\textbf{(A)} ~24 \qquad\textbf{(B)} ~30 \qquad\textbf{(C)} ~48 \qquad\textbf{(D)} ~60 \qquad\textbf{(E)} ~64</math> | <math>\textbf{(A)} ~24 \qquad\textbf{(B)} ~30 \qquad\textbf{(C)} ~48 \qquad\textbf{(D)} ~60 \qquad\textbf{(E)} ~64</math> | ||
− | ==Solution 1 | + | ==Solution 1== |
− | Let the side lengths of | + | Let the side lengths of the rectangular pan be <math>m</math> and <math>n</math>. It follows that <math>(m-2)(n-2) = \frac{mn}{2}</math>, since half of the brownie pieces are in the interior. This gives <math>2(m-2)(n-2) = mn \iff mn - 2m - 2n - 4 = 0</math>. Adding 8 to both sides and applying [[Simon's Favorite Factoring Trick]], we obtain <math>(m-2)(n-2) = 8</math>. Since <math>m</math> and <math>n</math> are both positive, we obtain <math>(m, n) = (5, 12), (6, 8)</math> (up to ordering). By inspection, <math>5 \cdot 12 = \boxed{\textbf{(D) }60}</math> maximizes the number of brownies. |
~ ike.chen | ~ ike.chen | ||
− | ==Solution 2 | + | ==Solution 2== |
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Obviously, no side of the rectangular pan can have less than <math>5</math> brownies beside it. We let one side of the pan have <math>5</math> brownies, and let the number of brownies on its adjacent side be <math>x</math>. Therefore, <math>5x=2\cdot3(x-2)</math>, and solving yields <math>x=12</math> and there are <math>5\cdot12=60</math> brownies in the pan. <math>64</math> is the only choice larger than <math>60</math>, but it cannot be the answer since the only way to fit <math>64</math> brownies in a pan without letting a side of it have less than <math>5</math> brownies beside it is by forming a square of <math>8</math> brownies on each side, which does not meet the requirement. Thus the answer is <math>\boxed{\textbf{(D) }60}</math>. | Obviously, no side of the rectangular pan can have less than <math>5</math> brownies beside it. We let one side of the pan have <math>5</math> brownies, and let the number of brownies on its adjacent side be <math>x</math>. Therefore, <math>5x=2\cdot3(x-2)</math>, and solving yields <math>x=12</math> and there are <math>5\cdot12=60</math> brownies in the pan. <math>64</math> is the only choice larger than <math>60</math>, but it cannot be the answer since the only way to fit <math>64</math> brownies in a pan without letting a side of it have less than <math>5</math> brownies beside it is by forming a square of <math>8</math> brownies on each side, which does not meet the requirement. Thus the answer is <math>\boxed{\textbf{(D) }60}</math>. | ||
Revision as of 23:13, 13 February 2021
Contents
Problem
Grandma has just finished baking a large rectangular pan of brownies. She is planning to make rectangular pieces of equal size and shape, with straight cuts parallel to the sides of the pan. Each cut must be made entirely across the pan. Grandma wants to make the same number of interior pieces as pieces along the perimeter of the pan. What is the greatest possible number of brownies she can produce?
Solution 1
Let the side lengths of the rectangular pan be and . It follows that , since half of the brownie pieces are in the interior. This gives . Adding 8 to both sides and applying Simon's Favorite Factoring Trick, we obtain . Since and are both positive, we obtain (up to ordering). By inspection, maximizes the number of brownies.
~ ike.chen
Solution 2
Obviously, no side of the rectangular pan can have less than brownies beside it. We let one side of the pan have brownies, and let the number of brownies on its adjacent side be . Therefore, , and solving yields and there are brownies in the pan. is the only choice larger than , but it cannot be the answer since the only way to fit brownies in a pan without letting a side of it have less than brownies beside it is by forming a square of brownies on each side, which does not meet the requirement. Thus the answer is .
-SmileKat32
Video Solution by OmegaLearn (Simon's Favorite Factoring Trick)
~ pi_is_3.14
See Also
2021 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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 AMC 10 Problems and Solutions |
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