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avatar+208 

\( Given\ positive\ integers\ x\ and\ y\ such\ that\ x \not = y\ and\ \frac {1}{x} + \frac {1}{y} = \frac {1}{12},\\ what\ is\ the\ smallest\ possible\ positive\ value\ for\ x + y?\)

I know this is on the website, but, you got 24+24... but x cant be the same as y

 Jun 22, 2019
 #1
avatar+7711 
+1

\(\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{12}\\ \dfrac{x+y}{xy} = \dfrac{1}{12}\\ \text{Let }x + y = k, xy = 12k.\\ y(k-y) = 12k\\ -y^2 +ky = 12k\\ y^2 -ky+12k = 0\\ y = \dfrac{k+\sqrt{k^2 -48k}}{2}\\ \implies k^2 - 48k \text{ is a perfect square.}\\ k^2 - 48k \ge 0\\ k^2 - 48k = t^2\\ k \le 0\text{(rejected)} \vee k\ge48\\ k \ge 48\\ k \text{ cannot be 48. If } k = 48\text{, } x = y = 24.\\ k \ge 49\\ k = 49 \text{ satisfies the condition that } k^2 - 48k \text{ is a perfect square.}\\ \min(k) = 49\\ \boxed{\min(x + y) = \color{red}49}\)

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 Jun 22, 2019
 #6
avatar+208 
+2

Great explantation with latex coding! (howd you do that... ?)

NoobGuest  Jun 22, 2019
 #8
avatar+7711 
0

https://web2.0calc.com/questions/latex-coding

^ You can learn LaTeX with the links in this post. 

MaxWong  Jun 22, 2019
 #2
avatar
0

a=1; b=1;c= 1/a + 1/b; if(c==1/12, goto4, goto5);printa, b; a++;if(a<200, goto2, 0);a=1;b++;if(b<200, goto2, discard=0;

a         b

156    13
84      14
60      15
48      16
36      18
30      20
28      21
24      24
21      28
20      30
18      36
16      48
15      60
14      84
13      156

 Jun 22, 2019
 #3
avatar+7711 
0

Nice. Is this a linear search on the set of integers?

MaxWong  Jun 22, 2019
 #7
avatar+208 
0

Wow... That must have taken a while!

NoobGuest  Jun 22, 2019
 #4
avatar
+1

Exactly!.

 Jun 22, 2019
 #5
avatar+208 
0

OHHHH wow... I just checked here and BOOOOMMMM THX GUYS!

 Jun 22, 2019

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