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

Revision as of 15:28, 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

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

Using Simon's Favorite Factoring Trick, we rewrite this equation as

$(x - 12)(y - 20) - 240 + 63 = 0$ $(x - 12)(y - 20) = 177$ . (awesomeag)

@@ Line 11: / Line 11: @@
 <cmath>x\cdot y - 20x - 12y + 63 = 0</cmath>.
+Using Simon's Favorite Factoring Trick, we rewrite this equation as
+<cmath>(x - 12)(y - 20) - 240 + 63 = 0</cmath>
+<cmath>(x - 12)(y - 20) = 177</cmath>.
 (awesomeag)

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

Revision as of 15:28, 16 February 2018