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Another 2006 Problem (Posted on 2006-09-30) Difficulty: 2 of 5
Determine the total number of positive integer solutions of:

2/x + 3/y = 1/2006

  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.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
No SubjectBennsan2021-11-22 12:00:58
re(2): A Computer Assisted look ???brianjn2006-10-01 18:55:23
re: A Computer Assisted look ???Charlie2006-10-01 10:17:41
A Computer Assisted look ???brianjn2006-10-01 02:15:23
Some Thoughtsagree with xdogDennis2006-09-30 17:28:20
re(2): SolutionCharlie2006-09-30 13:49:50
re: Solutionxdog2006-09-30 12:50:54
Questionre: SolutionK Sengupta2006-09-30 10:46:05
SolutionDej Mar2006-09-30 07:53:54
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