In a jail with 100 rooms, all locked initially, 100 rioters break in and disturb the rooms in the following way. First one stops at all rooms and opens them all. Second rioters stops at rooms numbered 2, 4, 6, . . . and locks open rooms, and leaving the other rooms as they were. The third rioter stops at rooms numbered 3, 6, 9, . . . and again opens a locked room and locks an open room, leaving others undisturbed. And this process continues. After all the 100 rioters have left which rooms would be open?
i dont know! :P any good reasons ?
the ones that will be open are the squares (i.e. 4, 9, 16, 25, etc)
the ones that will be closed are the not squares
@lgbasallote I think room 1 will be surely opened..
yes it will
1 is a perfect square isnt it
1 opens 4 2 closes 4 4 opens 4
100 th room
for 9 1 opens 9 3 closes 9 9 opens 9
for 16 1 opens 16 2 closes 16 4 opens 16 8 closes 16 16 opens 16
for 25 1 opens 25 5 closes 25 25 opens 25
for 36 1 opens 36 3 closes 36 6 opens 36 12 closes 36 36 opens 36
for 49 1 opens 49 7 closes 49 49 opens 49
I guess the number which has odd number of factors will be open
in 36 you missed some in between but result remains open,,hmm i see maybe you're correct..
And the room number which has even number of factors will be closed
for 64 1 opens 64 2 closes 64 4 opens 64 8 closes 64 16 opens 64 32 closes 64 64 opens 64
what comesafter 64 again?
81 1o 3c 9o 27c 81o
yes that
finally 100 1o 2c 4o 5c 10o 20c 25o 50c 100o nice.. thanks..
welcome
can there be some reason why its always sq. no ? or its just a pattern ? ! ?
it's a tedious proof... lol
i just know the trial and error proof of it
I guess only sq. no have odd number of factors
aha..yes it seems so..lol..might make up another good ques that only sq no. have odd no. of factors! :D
damn i closed the quess,,wait i shall tag,,
@mukushla @eliassaab
Oh..... got it....... I think I can prove it
IF its not an sq. number.... for example 12 then the factors are 1,2,3,4,6,12
right?
yes..
|dw:1345558048314:dw|Now,
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