https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Ironinja186&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T10:23:47ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_23&diff=1017412018 AMC 10B Problems/Problem 232019-02-10T23:22:53Z<p>Ironinja186: /* Solution */</p>
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<div>How many ordered pairs <math>(a, b)</math> of positive integers satisfy the equation <br />
<cmath>a\cdot b + 63 = 20\cdot \text{lcm}(a, b) + 12\cdot\text{gcd}(a,b),</cmath><br />
where <math>\text{gcd}(a,b)</math> denotes the greatest common divisor of <math>a</math> and <math>b</math>, and <math>\text{lcm}(a,b)</math> denotes their least common multiple?<br />
<br />
<math>\textbf{(A)} \text{ 0} \qquad \textbf{(B)} \text{ 2} \qquad \textbf{(C)} \text{ 4} \qquad \textbf{(D)} \text{ 6} \qquad \textbf{(E)} \text{ 8}</math><br />
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==Solution==<br />
Let <math>x =</math> lcm<math>(a, b)</math>, and <math>y = </math>gcd<math>(a, b)</math>. Therefore, <math>a\cdot b = </math>lcm<math>(a, b)\cdot </math>gcd<math>(a, b) = x\cdot y</math>. Thus, the equation becomes<br />
<br />
<cmath>x\cdot y + 63 = 20x + 12y</cmath><br />
<cmath>x\cdot y - 20x - 12y + 63 = 0</cmath><br />
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Using Simon's Favorite Factoring Trick, we rewrite this equation as<br />
<br />
<cmath>(x - 12)(y - 20) - 240 + 63 = 0</cmath><br />
<cmath>(x - 12)(y - 20) = 177</cmath><br />
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Since <math>177 = 3\cdot 59</math> and <math>x > y</math>, we have <math>x - 12 = 59</math> and <math>y - 20 = 3</math>, or <math>x - 12 = 177</math> and <math>y - 20 = 1</math>. This gives us the solutions <math>(71, 23)</math> and <math>(189, 21)</math>. Since the Greatest Common Denominator must be a divisor of the Lowest Common Multiple, the first pair does not work. Assume <math>a>b</math>. We must have <math>a = 21 \cdot 9</math> and <math>b = 21</math>, and we could then have <math>a<b</math>, so there are <math>\boxed{2}</math> solutions.<br />
(awesomeag)<br />
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Edited by IronicNinja~<br />
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==See Also==<br />
{{AMC10 box|year=2018|ab=B|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Ironinja186