Difference between revisions of "2018 AMC 10B Problems/Problem 23"

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<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>
 
<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>
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Let <math>x = lcm(a, b)</math>, and <math>y = gcd(a, b)</math>. Therefore, <math>a\cdot b = lcm(a, b)\cdot gcd(a, b) = x\cdot y</math>. Thus, the equation becomes
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<cmath>x\cdot y + 63 = 20x + 12y</cmath>,
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<cmath>x\cdot y - 20x - 12y + 63 = 0</cmath>.
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(awesomeag)

Revision as of 15:22, 16 February 2018

23. How many ordered pairs $(a, b)$ of positive integers satisfy the equation \[a\cdot b + 63 = 20\cdot \text{lcm}(a, b) + 12\cdot\text{gcd}(a,b),\] where $\text{gcd}(a,b)$ denotes the greatest common divisor of $a$ and $b$, and $\text{lcm}(a,b)$ denotes their least common multiple?

$\textbf{(A)} \text{ 0} \qquad \textbf{(B)} \text{ 2} \qquad \textbf{(C)} \text{ 4} \qquad \textbf{(D)} \text{ 6} \qquad \textbf{(E)} \text{ 8}$


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

\[x\cdot y + 63 = 20x + 12y\], \[x\cdot y - 20x - 12y + 63 = 0\].

(awesomeag)

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