Maths Olympiad Prep

Track / Stage 3 / 84 of 260 #84 of 1964

Problem 84

AMC 10/12, early questions
Number theory Difficulty 3.4 Find the answer

Let aa, bb, cc, dd, and ee be distinct integers such that
(6a)(6b)(6c)(6d)(6e)=45(6-a)(6-b)(6-c)(6-d)(6-e)=45
What is a+b+c+d+ea+b+c+d+e?

Pick one

Official solution

If 4545 is expressed as a product of five distinct integer factors, the absolute value of the product of any four is at least (3)(1)(1)(3)=9|(-3)(-1)(1)(3)|=9, so no factor can have an absolute value greater than 55. Thus the factors of the given expression are five of the integers ±3,±1,±5\pm 3, \pm 1, \pm 5. The product of all six of these is 225=(5)(45)-225=(-5)(45), so the factors are 3,1,1,3,-3, -1, 1, 3, and 5.5. The corresponding values of a,b,c,d,a, b, c, d, and ee are 9,7,5,3,9, 7, 5, 3, and 1,1, and their sum is 25 (C)\fbox{25 (C)}

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.