Given :   two digit number is such that sum of its digits is 3 time the difference of digits .  number is  neither prime nor a perfect square.

To find :  how many such two digit numbers are possible

Solution:

Let say number is AB or BA  where A > B

A + B  = 3( A - B  )

=> A + B = 3A  - 3B

=> 2A  = 4B

=> A  = 2B

B can be  1 , 2 , 3 , 4

A can be  2 , 4 , 6 , 8

Possible numbers are

12  , 24  , 36 , 48      , 21 , 42  , 63  , 84  

12 , 21 , 24 , 36 , 42 , 48 , 63 , 84

none of the  Numbers is prime  

36 is only perfect Square  so

Number left are

12 , 21 , 24 , 42 , 48 , 63 , 84

7 such two digit  Numbers are possible

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