Junior 2009/7

[Eesha]

Express $\frac{7}{26}$ as $\frac{1}{a}+\frac{1}{b}$ (a and b, both are positive integers)

[Labib]

First, assume WLOG,

$\frac 1 a \geq \frac 1 b$
$\Rightarrow \frac 7 {52} \leq \frac 1 a \leq \frac 7 {26}$
$\Rightarrow \frac 1 7 \leq a\leq \frac 1 4$
$\Rightarrow 7\geq a\geq 4$

Now, using trial & error, we get,
$a=4, b=52.$

Thus,
$ \frac 7 {26}= \frac 1 {4}+ \frac 1 {52}$.

[nafistiham]

if it at last trial and error, why not doing it from the beginning ?
here $26$ has just $2$ factors.we can find $a,b$ just trying thrice

[Labib]

logic দেখানোটা জরুরি, তাই...
নাহলে তো সব ইচ্ছামত করা যাইতো...
তুমি ভালো সমাধান পারলে সেটা দাও...

[MATHPRITOM]

I have a solution....... I usually use this way for solving this... it is also trial & error.....
$ \frac{1}{a}+\frac{1}{b}=\frac{7}{26} $
or,$\frac{a+b}{ab}=\frac{7}{26}$
or,$ 26a+26b-7ab=0$
or,$ a(26-7b)-\frac{26}{7}(26-7b)$+$ \frac{(26^2)}{7}=0 $
or,$ 7a(26-7b)-26(26-7b)+26^2=0 $
or,$ (7a-26)(7b-26)=26^2=676 $

Now,676=1*676=2*338=4*169=13*52=26*26.

so,possible conditions are,
7a-26=1 & 7b-26=676...1
or,7a-26=2 & 7b-26=338...2
or,7a-26=4 & 7b-26=169...3
or,7a-26=13 & 7b-26=52...4
or,7a-26=26 & 7b-26=26...5

by solving the equation ,we just get integer value only for the equation ...2 (a,b)=(4,52) again (a,b)=(52,4)..

[Eesha]

hmm বুঝলাম