On a circle there are 2014 light bulbs, 2 are ON, and 2012 are OFF. You can choose any bulb and change the neighbor’s state from ON to OFF or from OFF to ON. Doing so, can we get all 2014 light bulbs on ? If yes, How?
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Another Bulb Problem
(12 votes, average: 3.08 out of 5)
Just switch off any of these 2 bulbs. that way both of them will be switched off and a third one will go on. Now there is 1 bulb on and 2013 bulbs off. 2013 is a multiple of 3. IF 3 bulbs are off then we can switch on the middle one and 2 surrounding will go on. In this way we can turn on 3 at a time. since 2013 of the bulbs are off, we have 2013/3 that is 671 middle bulbs. we can switch all of them on and all the 2014 bulbs will be on!
easily get 2014 bulbs on … if 2 initial ON bulbs are set together and start the operate..like ON ON OFF OFF,……… start with second ON bulbs third bulb is ON , fourth bulb is ON … hence each and every bulbs in circle are ON
We can do this iff the no of bulbs between 2 ON bulbs is odd (no of OFF bulbs).
yes we can,
not a clear question?
Yup..choose the bulb whch is off n turn on it.. so remaining bulbs will be on..
Lets say, bulbs are b1, b2, b3, b4, b5, ……., b2012, b2013, b2014; and b2013, b2014 are ON
Select the bulb b2, now b1 and b3 are ON,
Select b3, now b2 and b4 are ON
So in two select we are able ON 4 bulbs.
Total: ((2014-2)/4)*2 = 1006.
What if the 2 initial ON bulbs are not together? (in your solution you have assumed that)
You can ON 2014 bulbs only if the number of bulbs between 2 initial ON bulbs is even.
in a circle ter is common switch,for all bulbs,2 bulbs on,2012 bulb off,on toggle of the switch,2 bulb off,2012 bulb on.