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![[parent]](http://images.planetmath.org:8080/images/uparrow.png) Viewing Message
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| ``Re: Exciting Problem''
by lynx on 2008-07-07 13:55:06 |
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| I had no patience to compute the exact value, but I tell you what I did.
We extend the triangle to the lattice generated by the vectors u:=C-->A and v:=C-->B. We color the lattice with the letters A, B and C in an proper way. Our solution correspond to a line segment that links the zero vector to another vector w colored by C and with positive coordinates (i.e. w=au+bv, a,b>0). This vector should satisfies the following conditions:
(1). a-b=0 mod 3. This means that w=au+bv is colored by C.
(2). a and b are coprimes. This means that the laser doesn't get out before w.
(3). (a-1)+(b-1)+(a+b-1) = 2a+2b-3 = 12017639147. This means that the laser hits the walls 12017639147 times before get out.
Now the condition (3) tells that
(4) a+b=6008819575 (=1mod3).
Thus by (1) we have a=b=2mod3
By (4) we have that a and b are coprimes iff a and a and 6008819575 are coprimes (i.e. a is a multiple of a prime factor of 6008819575).
6008819575=5*5*11*17*23*29*41*47
Now all we have to do is to count the number of the positive integers congruent with 2 mod3 less than 6008819575 and not multiple of 5, 11, 17, 23, 29, 41 or 47.
It is not difficult but it is a lot of calculus for a guy that is good in computer programing.
The best I can do is an estimation
(1/3)*6008819575*(1-1/5)*(1-1/11)*...*(1-1/47)=1209002666.(6)
I guess it's less than that.
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