Every day, \(100\) students enter a school that has \(100\) lockers. All the lockers are closed when they arrive.

Student \(1\) opens every locker.

Student \(2\) closes every second locker.

Student \(3\) changes the state of every third locker i.e. he opens it if it is closed and closes it if its open.

Student \(4\) changes the state of every fourth locker and so on... so that student \(n\) changes the state of every \(nth\) locker.

One day several students are absent. Regardless, those present complete the procedure and simply skip the students who are absent. For e.g. if student \(3\) is absent, then nobody changes the state of every third locker.

Satyen Nabar discussed the problem when one door was opened and all others were closed.This inspired me and tried to interpret the situation when 2 doors are opened and all the others are closed.

I took two cases

\(Case 1:\) Here after 100th operation locker number \(1,2\) were opened and all others were closed.I found that \((2*3=6,2*5=10,2*7=14,2*11=22,..........,2*47=94)\)and \((8,16,32,64)\),\((9.27,81)\),\((25)\),\((49)\) student nos who were absent.So total \(14+4+3+1+1=23\) students were absent.

\(Case 2:\)In this case after 100th operation locker number \(2,4\) were opened and all other were closed.Here I found that \(26\) students were present.Student no \(2,6,10,14,18,.......,98\) and \(4\)(which is an exception).So in this case \(74\)students were absent.

Anyone please confirm these two answers and help me to generalize the problem.

## Comments

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TopNewestSorry to say that your answers are ...inaccurate. in Case 1 it should be 48 and in Case 2 it should be 75 (quite close, though).

I used a PC program to solve, and checked back to confirm. If you wanna the detail result (which student is absent and which isnot), I can provide for you.

I found out that, since the number of combination of lockers' states is \(2^n\) and the number of combination of student's states is also \(2^n\), no 2 combination of student's states lead to the same result. It's some kind of function. :) – Ngọc Bùi Tiến · 1 year, 12 months ago

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– Calvin Lin Staff · 1 year, 12 months ago

The last statement is quite simple to prove mathematically. Do you see how to do that?Log in to reply

Fortunately, I described the algorithm in response to Mr. Guha. ;) – Ngọc Bùi Tiến · 1 year, 12 months ago

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– Kalpok Guha · 1 year, 12 months ago

Thank you very much but please provide the the details.And which function do you used to solve it?Log in to reply

Let \(i\) be the counter from 1 to 100. If \(current_i\neq result_i\), then \(i^{th}\) student goes to school (so he can change his locker's state); also the state of every lockers \(j\), which are divisible by \(i\), should be altered (close to open and vice versa). If \(current_i=result_i\), \(i^{th}\) student is absent, since all students after him cannot change his locker's state.

The "states" of students can be stored in another array for recheck.

Here is result for Case 1 – Ngọc Bùi Tiến · 1 year, 12 months ago

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– Kalpok Guha · 1 year, 11 months ago

Thank you.Have you got any generalization of this problem?Log in to reply

– Ngọc Bùi Tiến · 1 year, 11 months ago

Not yet. I can only solve with specific cases.Log in to reply

It's great that you are looking at further extensions of this problem.

Note that locker 1 will always remain open, so case 2 cannot be true.

I believe that the answer will depend on which 2 lockers stay open, and you need to look at the factors/multiples of the other (non-1) locker. – Calvin Lin Staff · 2 years ago

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– Kalpok Guha · 2 years ago

I think if the No.1 student is absent,then locker 1 will not remain open.But I think you are right that I should look on the factors of lockers.Log in to reply

– Calvin Lin Staff · 2 years ago

oh yes indeed. Not sure why, but it didn't cross my mind that he could be absent.Log in to reply

– Kalpok Guha · 2 years ago

This is okay but can you help me generalize it?I think I have got line to think but I need the help from you people.Log in to reply