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

(Solution)
(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 = \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>
 
<cmath>x\cdot y + 63 = 20x + 12y</cmath>
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<cmath>(x - 12)(y - 20) = 177</cmath>
 
<cmath>(x - 12)(y - 20) = 177</cmath>
  
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>. Obviously, the first pair does not work. Assume <math>a>b</math>. We must have <math>a = 21 \cdot 8</math> and <math>b = 21</math>, and we could then have <math>a<b</math>, so there are <math>\boxed{2}</math> solutions.
+
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>. Obviously, 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.
 
(awesomeag)
 
(awesomeag)
 +
 +
Edited by IronicNinja~
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2018|ab=B|num-b=22|num-a=24}}
 
{{AMC10 box|year=2018|ab=B|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:37, 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 $x = \lcm(a, b)$ (Error compiling LaTeX. ! Undefined control sequence.), and $y = \gcd(a, b)$. Therefore, $a\cdot b = \lcm(a, b)\cdot gcd(a, b) = x\cdot y$ (Error compiling LaTeX. ! Undefined control sequence.). 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 (ProblemsAnswer KeyResources)
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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