GIVEN   :   (1/a)+(1/b) +(1/c) +(1/d)=1  with condition that a

To FIND : The NUMBER of positive integral solutions (a, b, c, d) satisfying

Solution:

(1/a)+(1/b) +(1/c) +(1/d)=1

1/1 = 1  hence a can not be 1

1/2

1/3

1/4

1/5

1/6

if we leave  1/2

then max sum is 1/3 + 1/4 + 1/5 + 1/6  = ( 20 + 15 + 12 + 10) /60 = 57/60

Hence  not possible

so a must be  2

a = 2

=> 1/b + 1/c  + 1/d  =  1 - 1/2

=> 1/b + 1/c + 1/d  = 1/2

Assume b = 3

then 1/c  + 1/d  = 1/2 - 1/3  => 1/c  + 1/d  =  1/6

=> c < 6    taking c as 7 , 8 , 9 so on find d

1/6  = 1/7  + 1/42

1/6  = 1/8  + 1/24

1/6 =  1/9 + 1/18

1/6  = 1/10  +  1/15

1/6  =  1/12  + 1/12  ( not possible as c and  d would be same )

Till now a = 2 , b = 3   c , d = ( 7 , 42) , ( 8 , 24) , ( 9 , 18) , (10 , 15)

now taking b = 4

=> 1/c + 1/d  =  1/2 - 1/4   =  1/4    

c  < 4

=> 1/4  = 1/5  + 1/20

   1/4  =  1/6  +  1/12

    1/4 = 1/8 + 1/8   not possible

a = 2 , b = 4   c , d = ( 5 , 20) , ( 6 ,12)  

now taking b = 5

=> 1/c + 1/d  =  1/2 - 1/5   = 3/10

c < 5

=>    no  feasible

now taking b = 6

=> 1/c + 1/d  =  1/2 - 1/6   =  1/3

1/7 + 1/8  <  1/3   as that is max possible sum  , Hence no further solution possible

a = 2 , b = 3   c , d = ( 7 , 42) , ( 8 , 24) , ( 9 , 18) , (10 , 15)

 a = 2 , b = 4   c , d = ( 5 , 20) , ( 6 ,12)  

( 2  , 3 , 7 , 42)  , ( 2 , 3 , 8 , 24) , ( 2 , 3 , 9 , 18) , ( 2 , 3 , 10 , 15) ,

(2 , 4 , 5 , 20) , ( 2 , 4 , 6 , 12)

The number of positive integral solutions (a, b, c, d)

satisfying = (1/a)+(1/b) +(1/c) +(1/d)=1   are  6

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