house number = 50



-----



the work out:











Subtracting:

   n>m;  n,m  integers



28 is 1*28, 2*14, or 4*7



Case 1:

  n-m = 1

  n+m = 28



(adding): 2n = 29   non-integer solution, no good



Case 2:

  n-m = 2

  n+m = 14



   2n = 16  -->  n=8 --> m=6  this solution works



Case 3:

   n-m = 4

   n+m = 7



   2n = 11   non-integer solution, no good



Using case 2, and :

     -->  x = 50



Check using :

     (ok)  

Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!
 


Here is an unorthodox solution method that is fast if you understand it....


The house number is 14 less than one perfect square and 14 more than another, so it is halfway between two perfect squares whose difference is 28.


Now use the fact that the difference between consecutive perfect squares n^2 and (n+1)^2 is 2n+1.  That difference is clearly always odd.  So the two perfect squares in this problem are not consecutive perfect squares.


So what about squares of numbers that differ by 2?  28 = 13+15.  13 = 2(6)+1 is the difference between 6^2 and 7^2, and 15 = 2(7)+1 is the difference between 7^2 and 8^2.  And so 28 is the difference between 6^2 and 8^2.


So the two perfect squares in the problem are 6^2=36 and 8^2 = 64; the house number is halfway between 36 and 64, so it is 50. 

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