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

(Original had 21*8 = 189 which is not true. Also, fixed the latex formatting.)
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==Solution==
 
==Solution==
−
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 =</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
  
 
<cmath>x\cdot y + 63 = 20x + 12y</cmath>
 
<cmath>x\cdot y + 63 = 20x + 12y</cmath>

Revision as of 14:41, 25 September 2018

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

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 $(189, 21)$. Obviously, 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 $\boxed{2}$ solutions. (awesomeag)

Edited by IronicNinja~

See Also

2018 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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