Re: jnu question

Posted by Amit Goyal on Apr 08, 2009; 2:31pm
URL: http://discussion-forum.276.s1.nabble.com/jnu-question-tp2592523p2605625.html

Hi Priyal,

For getting house number of A, we will first use the following information:
Sum of house numbers of 3 individuals is perfect square multiplied by two.
Given that B's House Number is 81, C's House Number is 64, let A's house number be x. Also we know that 100 > x > 50. Now, x+81+64 is a perfect square multiplied by 2. This implies that (x+81+64)/2 = (x + 145)/2 is a perfect square. Since 100 > x > 50 this implies that
245/2 > (x + 145)/2 > 195/2. Now we look for a perfect square in the interval (195/2, 245/2)
=(97.5, 122.5)
Two Possibilities are 100, 121
Solving,  (x + 145)/2  = 100 give us x = 55
and (x + 145)/2  = 121 give us x = 97
Now using the condition that A's house number is less than house number of both B and C we rule out x = 97
Hence, A's house number is 55.