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Math Puzzle

eddybob123

Active member
Aug 18, 2013
76
1) Find all solutions (a,b) to the equation

\(\displaystyle \frac{1}{a}+\frac{1}{b}=\frac{1}{8}\).

2) Generalize to

\(\displaystyle \frac{1}{a}+\frac{1}{b}=\frac{1}{n}\).

If no one gets the answer in long time I'll post hints.
 

agentmulder

Active member
Feb 9, 2012
33
Hi there eddybob123 , welcome to MHB !

-agentredlum (Cool)
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
1) Find all solutions (a,b) to the equation

\(\displaystyle \frac{1}{a}+\frac{1}{b}=\frac{1}{8}\).

2) Generalize to

\(\displaystyle \frac{1}{a}+\frac{1}{b}=\frac{1}{n}\).

If no one gets the answer in long time I'll post hints.
For the generalized part put $a=n-x$ and $b=n-y$. One gets $n^2=xy$.
 

Albert

Well-known member
Jan 25, 2013
1,225
1) Find all solutions (a,b) to the equation

\(\displaystyle \frac{1}{a}+\frac{1}{b}=\frac{1}{8}\).

2) Generalize to

\(\displaystyle \frac{1}{a}+\frac{1}{b}=\frac{1}{n}\).

If no one gets the answer in long time I'll post hints.
\(\displaystyle \frac{1}{a}+\frac{1}{b}=\frac{1}{n}\).
$\dfrac{1}{n}=\dfrac {k}{nk}=\dfrac {1+(k-1)}{nk}=\dfrac {1}{nk}+\dfrac {k-1}{nk}$ (k=2,3,----)
here nk must be a multiple of (k-1)
take n=8
$\dfrac {1}{8}=\dfrac {1}{16}+\dfrac{1}{16}=\dfrac {1}{24}+\dfrac{1}{12}=
\dfrac {1}{40}+\dfrac{1}{10}=\dfrac {1}{72}+\dfrac{1}{9}$
I found four pairs of (a,b)
 
Last edited:

Albert

Well-known member
Jan 25, 2013
1,225
$n=xy ,\,\, x\geq 1,\,\, y\geq 1$
$\dfrac {1}{n}=\dfrac {k}{nk}=\dfrac {1+(k-1)}{nk}=\dfrac{1}{nk}+\dfrac{k-1}{nk}$
here nk mod (k-1)=0-----(1)
$k=x,y,(x+y),---$
$if \,\, n=8=1\times 8=2\times 4$
$\therefore k=2,4,(2+4),(1+8),(2+4+1+8)----(2)$
compare (1) and (2) and choose suitable k,and get corresponding a and b
 
Last edited:

eddybob123

Active member
Aug 18, 2013
76