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Topic: A 2006 Puzzle (Read 558 times) |
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K Sengupta
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A 2006 Puzzle
« on: Jun 25th, 2006, 7:54pm » |
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Determine the total number of positive integer solutions of this equation:
2/x + 3/y = 1/2006.
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jollytall
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Re: A 2006 Puzzle
« Reply #1 on: Jun 26th, 2006, 4:45am » |
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Well, not the complete solution, simply some equation manipulation gave the way to all solutitions.
2/x + 3/y = 1/2006
2*17*59*(2y+3x) = x*y
X and y shall have 2, 17, and 59 as dividers. There are seven options:
X=2*17*59*x' => y has no restrictions
x=2*17*x', y=59*y'
x=17*59*x', y=2*y'
x=59*2*x', y=17*y'
x=2*x', y=17*59*y'
x=17*x', y=59*2*y'
x=59*x', y=2*17*y'
x is free, y=2*17*59*y'
If we try them all:
First equation:
2*y+3*2*17*59*x'=x'*y
y=(3*2006*x')/(x'-2) | X'>2
There are a number of solutions, where x'-2 is dividing 2*3*17*59 (x' does not count, since there are only two numbers, 3 and 4, where x'-2 divides x' too, but those are covered anyway:
x'=3, x=3*2006, y=9*2006
x'=4, x=4*2006, y=6*2006
x'=5, x=5*2006, y=5*2006
etc.
Second equation:
2*59*y'+3*2*17*x'=x'*y'
y'=(3*2*17*x')/(x'-2*59)
Again there are many solutions x'>2*59:
x'=119, x=4046, y'=12138, y=716142
etc.
For all equations we can do it.
The last one:
y'=3*x/(x-2*2006)
The first few solutions:
x=4013, y'=12039, y=24150234
x=4014, y'=6021, y=12078126
x=4015, y'=4015, y=8054090
x=4016, y'=3012, y=6042072
etc.
There might be a more elegant solution, but I still don't have it.
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Icarus
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Re: A 2006 Puzzle
« Reply #2 on: Jun 26th, 2006, 3:32pm » |
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Try the substitutions: r = x - 4012, s = y - 6018.
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| « Last Edit: Jun 26th, 2006, 3:37pm by Icarus » |
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I wonder: Which is larger
When their digits are reversed? " - Anonymous
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K Sengupta
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Re: A 2006 Puzzle
« Reply #3 on: Jun 26th, 2006, 11:08pm » |
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On Today at 3:32pm , Icarus wrote:
Quote:| Try the substitutions: r = x - 4012, s = y - 6018. |
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I would like to thank Icarus for his invaluable guidance in the matter.
Substituting, r = x - 4012, s = y - 6018., we obtain:
r*s = 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 = 72.
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| « Last Edit: Jun 27th, 2006, 9:17pm by K Sengupta » |
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Barukh
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Re: A 2006 Puzzle
« Reply #4 on: Jun 26th, 2006, 11:29pm » |
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on Jun 26th, 2006, 11:08pm, K Sengupta wrote:| However, it would be observed that this number only corresponds to the number of solutions whenever x> 4012 and y>6018, and does not take into account the pairs (x,y) satisfying 1<=x<= 4012 and 1<=y<=6018. |
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If x <= 4012, then 2/x >= 1/2006, and so no positive y yields the required equality.
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| « Last Edit: Jun 26th, 2006, 11:39pm by Barukh » |
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K Sengupta
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Re: A 2006 Puzzle
« Reply #5 on: Jun 27th, 2006, 12:48am » |
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On, Jun 26th, 2006, 11:29pm Barukh wrote:
Quote:| If x <= 4012, then 2/x >= 1/2006, and so no positive y yields the required equality. |
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I wish to thank Barukh for obviating one of my doubts regarding solution to the problem under reference.
In light of the foregoing, I confirm having suitably amended my previous post (Jun 26th, 2006, 11:08pm) to reflect the precise solution to the given problem.
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| « Last Edit: Jun 27th, 2006, 12:50am by K Sengupta » |
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