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Light Bulbs PUZZLE/MATH?
there are 10 lights. you turn them ON (0) and OFF (X) ten times. here's the thing, you turn each light by it's sequence of numbers. such as. 2,4,6 or 3,6,8. (ALL lights are off when you start, but you flight the light at 1 making the lights turn ON (0)
here's how i'd set it up
-1,2,3,4,5,6,7,8,9,10 (the lights)
1
2
3
4
5
6
7
8
9
10 (the number of times you flick the lights (vertically)
here's an example
-..1,2,3,4,5,6,7,8,9,10 (the lights)
1.0,0,0,0,0,0,0,0,0,,0
2..,X....X....X....X.....X
3......X.......0......X
4.........0.........0
5
6
7
8
9
10
and so on.
WHAT LIGHTS REMAIN ON AT THE END?
WHAT IS AN EQUATION YOU MAY USE, TO FIND THIS?
1 Answer
- J. J..Lv 71 decade agoFavorite Answer
The lights that remain on at the end are numbers 1, 4 and 9.
This relates to square numbers. The square of 1 is 1, the square of 2 is 4 and thesquare of 3 = 9.
Every number has a number of factors. Each factor has a reciprocal which when multiplied by becomes the original number. Therefore there will always be an EVEN number of factors EXCEPT with square numbers which will have an ODD number of factors,as the reciprocal is the same number again.
eg 24 the pairs of factors are
1 and 24, 2 and 12, 3 and 8, 4 and 6. ie there are 4 pairs = 8 factors = EVEN
now a square number say 36
1 and 36, 2 and 18, 3 and 12, 4 and 9, 6 and itself (ie 6 and 6). = 9 factors only = ODD
Therefore for all numbers apart from square numbers the first factor switches the light on and the reciprocal switches it off. Therefore at the end the light is OFF. With square numbers with one factor that does not have a reciprocal, the light will remain ON.