To finite sets have m and n elements. The total no. of subsets of first set is 60 more than the total number of subsets of second sets. Find the value of m and n.
(Need explanation too!!)

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no!! i solved it till the step u wrote but what to do next??
can u pls solve that step??
i dont know next step..


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Given m and n are integers. They are cardinalities of two sets.

The total number of sub sets of first set:  2^m   (I hope the null set is a subset of any set.).    total number of subsets of second set: 2^n.
       Hence,               2^m = 2^n + 60

The above equation is not satisfied by m =0 or 1,   or n = 0 or 1.
 So m and n are >= 2.     Hence, 2^(m-2) and 2^(n-2) are also integers.

   Now divide the above equation by 4.    we get:
                     2^(m-2) = 2^(n-2) + 15

LHS is even as it is a power of 2.  On  RHS 15 is an odd number.  So only way this equation can be solved is if  2^(n-2) is an odd number.   The only possibility is n-2 = 0 for which it is equal to 1 and becomes an odd number.   Hence,
        n = 2.            So   m -2 = 4  and   so   m = 6

 if you want a simple solution, 

         2^m = 2^n + 60
     try n =0, 1, 2, 3,..  try to get  RHS as a power of 2.
     for n =2,  RHS = 64 , it is a power of 2. and 2^6.  Hence, m = 6.

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