phillip1882 Posted April 4, 2012 Posted April 4, 2012 (edited) out of a set of 12 lights, 3 are burned out. you have a set of 12 light switches in front of you numbered 1-12, with the 12 lights behind a glass wall. each light switch will toggle the state of 2 lights. the 2 lights it toggles will be 2*n %13, and 3*n %13. for example, light switch 7 will toggle light 1, and light 8. is it possible that: no matter what three lights are burned out, you can light up the remaining 9 lights? Edited April 4, 2012 by phillip1882
John Cuthber Posted April 4, 2012 Posted April 4, 2012 I suspect that this hasn't been interpreted by the forum software in the way you expected. the 3 lights it toggles will be 1*n %13, 2*n %13, and 3*n %13.
imatfaal Posted April 4, 2012 Posted April 4, 2012 I guess the question should be "Switch n will toggle 3 lights - (1*n) mod 13, (2*n) mod 13, (3*n) mod 13"
phillip1882 Posted April 4, 2012 Author Posted April 4, 2012 bah you guys are too quick. i modified the original question because i quickly realized it was too easy.
imatfaal Posted April 4, 2012 Posted April 4, 2012 Surely if you throw all the switches then all the lights will be toggled three times ( ie on, off again, back on). With mod 13 you will get 1 through 12 with the n*1 (in order), you will also get 1-12 with n*2 (going up by 2 each time) and also with n*3 (going up by 3)
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