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When more than one solution is provided, this is done to illustrate a significant contrast in methods. These solutions are by no means the only ones possible, nor are they superior to others the reader may devise.  \nTeachers are encouraged to share copies of the problem booklet and official solutions with their students for educational purposes. All problems should be credited to the MAA AMC (for example,“2017 AMC 12 B, Problem \\#21”). The publication, reproduction, or communication of the competition’s problems or solutions for revenue-generating purposes requires written permission from the Mathematical Association of America (MAA) .  \nQuestions and comments about this competition should be sent to:  \n[amcinfo@maa.org](amcinfo@maa.org)  \nor  \nMAA American Mathematics Competitions  \nPO Box 471  \nAnnapolis Junction, MD 20701  \nThe problems and solutions for this AMC 8 were prepared by the  \nMAA AMC 8 Editorial Board under the direction of:  \nSilva Chang and Zsuzsanna Szaniszlo  \n© 2022 Mathematical Association of America  \n2 2022 AMC 8 Solutions  \n1. The Math Team designed a logo shaped like a multiplication symbol, shown below on a grid of 1-inch squares. What is the area of the logo in square inches?  \n\n|  |  |  |  |  |  |\n| --- | --- | --- | --- | --- | --- |\n|  |  |  |  |  |  |\n|  |  | Math |  |  |  |\n|  |  | Te | am |  |  |\n|  |  |  |  |  |  |\n|  |  |  |  |  |  |\n\n(A) 10 (B) 12 (C) 13 (D) 14 (E) 15  \nAnswer (A): The multiplication symbol consists of 4 squares, each measuring 1 inch by 1 inch, surrounded by 12 right triangles, each exactly half of a square. Thus the total area of the logo is 4 · 1 + 12 · ~~1~~2 = 4 + 6 = 10 square inches.  \nOR  \nBy the Pythagorean Theorem, each 1-by-1-inch square has a diagonal length of √12 + 12 = √2. The multiplication symbol consists of 5 congruent squares of side length √ 2 . Therefore the total area of the logo is 5 􀀀√2􀀁 2 = 5 · 2 = 10 square inches.  \nOR  \n2022 AMC 8 Solutions 3  \nThe logo can be enclosed in a large square of side length 4. There are 12 right triangles, each with area ~~1~~2 square inch, that are outside the logo and inside that square. The area of the logo equals the area of the large square minus the area of the triangles: 42 − 12 · ~~1~~2 = 16 − 6 = 10 square inches.  \n2. Consider these two operations:  \na♦ b = a2 − b2  \na⋆ b = (a − b)2  \nWhat is the value of 􀀀5♦ 3 􀀁 ⋆ 6 ?  \n(A) −20 (B) 4 (C) 16 (D) 100 (E) 220  \nAnswer (D): The value of the expression in parentheses is  \n5♦ 3 = 52 − 32 = 25 − 9 = 16 .  \nSubstituting that value into the larger expression gives  \n􀀀5♦ 3 􀀁 ⋆ 6 = 16⋆ 6 = (16 − 6)2 = 102 = 100 .  \n3. When three positive integers a, b, and c are multiplied together, their product is 100. Suppose a \u003C b \u003C c. In how many ways can the numbers be chosen?  \n(A) 0 (B) 1 (C) 2 (D) 3 (E) 4  \nAnswer (E): The factors of 100 are 1, 2, 4, 5, 10, 20, 25, 50, and 100. Consider each possible value of a and determine whether there are any solutions given that a · b · c = 100 and a \u003C b \u003C c.  \nIf a = 1, then b · c = 100 . There are 3 solutions: 1 · 2 · 50, 1 · 4 · 25, and 1 · 5 · 20.  \nIf a = 2, then b · c = 50 . There is 1 solution: 2 · 5 · 10.  \nIf a ≥ 4, then b · c ≤ 25. There are no solutions given that a \u003C b \u003C c.  \nIn summary, there are 4 ways to choose the numbers a, b, and c:  \n100 = 1 · 2 · 50 = 1 · 4 · 25 = 1 · 5 · 20 = 2 · 5 · 10.  \n4 2022 AMC 8 Solutions  \n4. The letter M in the figure below is first reflected over the line q and then reflected over the line p. What is the resulting image?  \n(A)  \n(D)  \n(B)  \n(E)  \n(C)  \nAnswer (E): Reflecting the given figure over the line q results in a sideways M located below th","cbCaiv7Xt5RB595L","https://ap.wps.com/l/cbCaiv7Xt5RB595L","pdf",817130,"English","# Introduction\n## Usage and Permissions\n# Solutions\n## Problem 1\n## Problem 2\n## Problem 3\n## Problem 4\n## Problem 5\n## Problem 6","[{\"question\":\"Do the official AMC 8 solutions require a calculator?\",\"answer\":\"No. The solutions state that all problems can be solved without the use of a calculator.\"},{\"question\":\"Why are multiple solutions sometimes included for the same problem?\",\"answer\":\"When more than one solution is provided, it is intended to illustrate a significant contrast in methods.\"},{\"question\":\"What should educators and students do regarding sharing and credit?\",\"answer\":\"Teachers are encouraged to share copies of the problem booklet and official solutions for educational purposes, and all problems should be credited to the MAA AMC.\"}]","2022 AMC 8 Solutions - Official Solutions | PDF",1790670287]