2015/056) How many ordered pairs of integers $(a,b)$ satisfy $\dfrac{1}{a} +\dfrac{1}{b}= \dfrac{1}{6}$

 $\dfrac{1}{a} +\dfrac{1}{b}= \dfrac{1}{6}$

hence $\dfrac{a+b}{ab}= \dfrac{1}{6}$

or $6 * (b + a) = ab$
or $6b + 6a = ab$

or $ab – 6b – 6a + 36 = 36$

or $(a-6)(b-6) = 36$
So  b - 6 must be a divisor of 36

Divisors:

-36 , -18 , -12 , -9 , -6 , -4 , -3 , -2 , -1 , 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36

except that b cannot be 0 as ( or b- 6 cannot be -6) as b is in denominator

secondly we have ordered pair so (a,b) and (b,a) are different

so number of ordered pairs = 17 ( one for each b -6 from -36 , -18 , -12 , -9 , -6 , -4 , -3 , -2 , -1 , 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36 except b-6 = -6 )
This I have solved at https://in.answers.yahoo.com/question/index?qid=20130530212220AA8Jqq3