2018 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 23 Hard

How many ordered pairs (a,b) of positive integers satisfy the equation a\cdot b+63=20\cdot lcm(a,b)+12\cdot gcd(a,b), where gcd(a,b) denotes the greatest common divisor of a and b, and lcm(a,b) denotes their least common multiple? (2018 AMC 10B Problem, Question#23)

  • A.

    0

  • B.

    2

  • C.

    4

  • D.

    6

  • E.

    8

Answer:B

Let x=lcm(a,b), and y=gcd(a,b). Therefore, a\cdot b= lcm(a,b)\cdot gcd(a,b)=x\cdot y.Thus, the equation becomes

x\cdot y+63=20x+12yx\cdot y-20x-12y+63=0, Using Simon's Favorite Factoring Trick, we rewrite this equation as (x-12)(y-20)-240+63=0, (x-12)(y-20)=177, From here we can already see that this is a quadratic, and thus must have 2 solutions. But, let's continue, to see if one of the solutions is extraneous.

Since 177=3\cdot 59 and x>y, we have x- 12=59 and y-20=3, or x-12=177 and y - 20=1. This gives us the solutions (71,23) and (189, 21). Since the Greatest Common Denominator must be a divisor of the Lowest Common Multiple, the first pair does not work. Assume a>b. We must have a=21\cdot 9 and b=21, and we could then have a<b, so  there are 2 solutions.

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