There r 100 doors and 100 students. intially all doors r closed. first student no 1 comes n toggles(open if close/close if open) the door no 1 and its multiples, next comes student no 2 and toggles door no 2 and its multiples i.e. door no 2,4,6,8........,student no 3 toggles door no 3 and its multiples i.e.3,6,9,12........., and so on till 100 students complete. at the end how many r open n how many r closed.?

i assume 51 doors will be opened and 49 closed. please let me know the correct answer.

I guess there will be 84 doors closed and 16 doors opened. I have not done it through formula, but tried manually. Have a check on the attached excel and tell if the answer is correct.

Attached Files

Puzz.zip (24.9 KB, 10 views)

Hi Friends,

Here is the answer for your question...

Opened 10 doors
Closed 90 doors

All those doors with door number equal to a perfect square will remain open, i.e 1, 4, 9, 16, 25, 36, 49, 64, 81, 100

For a door to remain open it has to be toggled an odd number of times.

e.g
open or open-close-open or open-close-open-close-open and so on.

Since the perfect squares have odd number of factors, the door number equal to perfect square will remain open.

Like door number 36.
On first whistle it will get open.
On second whistle it will get closed.
On third whistle it will get open.
On fourth whistle it will get close.
On sixth whistle it will get open.
On ninth whistle it will get closed.
On twelth whistle it will get open.
On eighteenth whistle it will get closed.
On thirtysixth whistle it will get open.

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suresh

answer given by psureh1982 is exactly correct..

hi, psuresh1982 I just couldn't get ur answer . please clarify it.