Another 2006 Problem (Posted on 2006-09-30)

	Submitted by K Sengupta
Rating: 5.0000 (1 votes)
Solution:	(Hide)
Substituting, p = x - 4012, q = y - 6018 in the given relationship- we obtain: p*q = 6*20062 = 3*23*172*592 Accordingly, the required number of solutions = The total number of factors of 6*20062 = 2*4*3*3 =72. This corresponds to the case when x> 4012 and y>6018, and does not take into account the pairs (x,y) satisfying 1<=x<= 4012 and 1<=y<=6018. But when 1<=x<=4012, we obtain 2/x>= 2/4012 = 1/2006, which is a contradiction, since no positive value of y is then feasible in conformity with tenets corresponding to the problem under reference. Similarly, when 1<=y<=6018, we obtain 3/y>= 3/6018 = 1/2006, which is a contradiction, since no positive value of x is then possible for obvious reasons. Combining all the available cases, we must conclude that the required number of positive integer solutions to the given problem is equal to 72.

	Subject	Author	Date
	No Subject	Bennsan	2021-11-22 12:00:58
	re(2): A Computer Assisted look ???	brianjn	2006-10-01 18:55:23
	re: A Computer Assisted look ???	Charlie	2006-10-01 10:17:41
	A Computer Assisted look ???	brianjn	2006-10-01 02:15:23
	agree with xdog	Dennis	2006-09-30 17:28:20
	re(2): Solution	Charlie	2006-09-30 13:49:50
	re: Solution	xdog	2006-09-30 12:50:54
	re: Solution	K Sengupta	2006-09-30 10:46:05
	Solution	Dej Mar	2006-09-30 07:53:54