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

(Work in progress of my answer to this question.)
(This is my solution to the problem.)
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==Solution 1==
 
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
 
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
  
<cmath>x\cdot y + 63 = 20x + 12y</cmath>,
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<cmath>x\cdot y + 63 = 20x + 12y</cmath>
<cmath>x\cdot y - 20x - 12y + 63 = 0</cmath>.
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<cmath>x\cdot y - 20x - 12y + 63 = 0</cmath>
  
 
Using Simon's Favorite Factoring Trick, we rewrite this equation as
 
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) - 240 + 63 = 0</cmath>
<cmath>(x - 12)(y - 20) = 177</cmath>.
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<cmath>(x - 12)(y - 20) = 177</cmath>
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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>, which can be translated back to two solution for <math>a</math> and <math>b</math>. Thus, the answer is <math>\boxed{2}</math>.
 
(awesomeag)
 
(awesomeag)

Revision as of 15:34, 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?

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


Solution 1

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

\[(x - 12)(y - 20) - 240 + 63 = 0\] \[(x - 12)(y - 20) = 177\]

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 $(189, 21)$, which can be translated back to two solution for $a$ and $b$. Thus, the answer is $\boxed{2}$. (awesomeag)

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