# Bulb Problem

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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?

vishal Scholar Asked on 27th July 2015 in

yes it is possible….
position of the light bulb which is in on state is not given so assume that they are adjacent as…

b1(on) b2(on) b3(off) b4(off) b5(off)……………………………. b2014(off) (assume it is in circular path)

now as two bulb is in on state and remaining 2012 bulbs are in off state
clearly all 2012 bulbs which is in off state are adjacent ,now we will operate on these 2012 bulbs as….

we can choose any bulb and able to make two bulb adjacent to it in on state.now choose anyone of two bulb to which we just turn in on state and made two more bulb in on state.
so clearly we can make four adjacent bulbs in on state.

(ex- at starting all four bulbs are in off state as
1(0ff) 2(0ff) 3(0ff) 4(0ff)
choose 2nd bulb and turn 1 and 3 in on state…as
1(0n) 2(0ff) 3(0n) 4(0ff)
now choose 3rd bulb and turn 2 and 4 in on state…as
1(0n) 2(0n) 3(0n) 4(0n)
)

by repeating these steps we can turn all 2012 bulbs in on state (4*53=2012)
so now all 2014 bulbs are in on state…………………….

satyen Guru Answered on 28th July 2015.

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