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January 25th, 2014, 02:12 AM  #1 
Newbie Joined: Jan 2014 Posts: 4 Thanks: 0  Two Lighthouses. When do they disappear from view together?
Two lighthouses can be seen from the sea. Both lights turn on and off in regular repeating patterns. One is 7 secs on and 16 secs off. The other is 8 secs on and 23 secs off. 15secs ago they became visible at precisely the same moment. a) In how many seconds will they become visible together again? b) In how many seconds will they disappear from view together? __________________________________________________ ______________ a) is easy. Lighthouse 1) 7+16=23sec cycle Lighthouse 2) 8+23=31sec cycle Apply LCM 23*31=713 sec. They will become visible at the same moment every 713 sec. Subtract when they were visible at the same moment for the last time: 71315sec=698sec. They will become visible together again in 698sec. b) This is actually a different problem IS THERE AN MATHEMATICAL APPROACH? Making a drawing the answer is 62815=613sec. But I want to know whether there's a way to do it mathematically. 
January 25th, 2014, 05:50 AM  #2 
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: Two Lighthouses. When do they disappear from view togeth
If they both become visible 15 seconds ago, one "disappeared" 15 7= 8 seconds ago and the other 15 8= 7 seconds ago. The two cycles have, as you say, periods 23 and 31 seconds. So the first "disappears" at 23n 8 seconds for some integer n and the other at 31m 7 seconds for some integer m. You want to solve the "Diophantine equation" 23n 8= 31m 7 which is the same as 23n 31m= 1. (A "Diophantine equation" is an equation in which the solutions must be integers.) Here is how I would solve that: 23 divides into 31 once with remainder 8. That is, 31 23= 8. 8 divides into 23 twice with remainder 7. That is, 23 2(= 7. 7 divides into 8 once with remainder 1. That is, 8 7= 1. Replace the 7 in that last equation with 23 2(: 8 (23 2()= 3( 23= 1. Replace that 8 with 31 23: 3(31 23) 23= 3(31) 4(23)= 23(4) 31(3)= 1. That tells us that one solution is n= 4 and m= 3. But it is easy to see that m= 3+ 23i and n= 4+ 31i is also a solution for any integer i: 31(3+ 23i) 23(4+ 31i)= 93+ (31)(23i) (92) (23)(31i)= 1 for all i. We need to take i= 1, so that m= 20 and n= 27 is the smallest positive solution. After 27 cycles, the first light will disappear in 23(27) 8= 613 seconds and, after 20 cycles, the second light will disappear in 31(20) 7= 613 seconds. 
January 25th, 2014, 01:02 PM  #3  
Newbie Joined: Jan 2014 Posts: 4 Thanks: 0  Re: Two Lighthouses. When do they disappear from view togeth
Thank you! very helpful indeed! Just one thing: Is there a stepbystep method to solve the diophantine equation or you just see how you could solve it without a method? I refer to this part: "Here is how I would solve that: 23 divides into 31 once with remainder 8. That is, 31 23= 8. 8 divides into 23 twice with remainder 7. That is, 23 2(= 7. 7 divides into 8 once with remainder 1. That is, 8 7= 1. Replace the 7 in that last equation with 23 2(: 8 (23 2()= 3( 23= 1. Replace that 8 with 31 23: 3(31 23) 23= 3(31) 4(23)= 23(4) 31(3)= 1. That tells us that one solution is n= 4 and m= 3." Quote:
 
January 25th, 2014, 02:08 PM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,935 Thanks: 1129 Math Focus: Elementary mathematics and beyond  Re: Two Lighthouses. When do they disappear from view togeth
23n  31m = 1 31(n  m)  8n = 1 31 = (1 + 8n)/(n  m) 31 is a factor of 1 + 8n 31 * 1  1 = 30 31 * 3  1 = 92 31 * 5  1 = 154 31 * 7  1 = 216. 216/8 = 27. n = 27. (1 + 8 * 27)/31 = 7. n  m = 7. m = 20. 
January 25th, 2014, 04:27 PM  #5  
Newbie Joined: Jan 2014 Posts: 4 Thanks: 0  Re: Two Lighthouses. When do they disappear from view togeth
Thanks greg1313. I see it for this example. But I am trying with a different set of numbers to see if I can make it work on a different example, and can't. I give the details: DATA: A) 5'' on 17''off B) 13''on 23" off 15" started to shine at the same time. ______________________________________________ A) 5+17= 22 sec cycle B) 13+28= 41 sec cycle A)155> 10 #15"ago was on for 5'' (then off) B)1513> 2 #15'' ago was on for 13" (then off) Diphantine Equation 22n10=41m2; 22n41m=8 41(nm)19n=8 41=(8+19n)/(nm) 41 is a factor of 8+19n 41*18 = 33 /19=1,7 41*38 = 115/19=6,05 41*58 = 197/19=10,36 (...) 41*478=1919/19=101 >integer So n=101. 22(101)41m=8; m=1026/41 = 25,02 (not an integer :S) Drawing a diagram, I know the answer for the question: "when will they disappear from view together the next time?" is 423 secs after they became visible at the same moment, but I can't prove it mathematically... :S not even using a diphantine solver I get this result http://www.math.uwaterloo.ca/~snburris/ ... inear.html Am I doing something wrong? Quote:
 
January 25th, 2014, 04:52 PM  #6 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,935 Thanks: 1129 Math Focus: Elementary mathematics and beyond  Re: Two Lighthouses. When do they disappear from view togeth
41 * 9  8 = 361, n = 19, m = 19  9 = 10.

January 26th, 2014, 01:32 AM  #7  
Newbie Joined: Jan 2014 Posts: 4 Thanks: 0  Re: Two Lighthouses. When do they disappear from view togeth
Thank you! hence I do not need to use prime numbers for the multiplications, but any number? Quote:
 
January 27th, 2014, 02:41 PM  #8  
Global Moderator Joined: Dec 2006 Posts: 20,629 Thanks: 2077  Quote:
 

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