Comments on: School Lockers: Math Problem

By: bunny

bunny — Sat, 23 Apr 2011 20:12:53 +0000

Now I get it, thank you!

By: noncoolman

noncoolman — Sun, 25 Apr 2010 02:03:59 +0000

My school had to do this, but it was 150 lockers.
Try to find the answer for it.

I got 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, and 144 .

By: ahsergio

ahsergio — Fri, 27 Nov 2009 11:26:07 +0000

Good that u guys enjoyed it =D

By: Henkie

Henkie — Wed, 25 Nov 2009 17:00:53 +0000

door 1, 4, 9, 16 and 25 will be open after all students passed

By: Mashplum

Mashplum — Tue, 24 Nov 2009 16:47:02 +0000

1, 4, 9, 16, 25

All prime numbered doors will be opened by number 1 and closed by their own number. They will be otherwise untouched.

All composite numbers have pairs of factors (and therefore an even number of factors) except perfect squares (which have an odd number of factors.)

By: Frankros

Frankros — Tue, 24 Nov 2009 08:48:17 +0000

1,4,9,16 and 25

By: Ahmed

Ahmed — Tue, 24 Nov 2009 08:02:06 +0000

This will be easy if you can figure out the perfect squares from 1 to 30 which are:
1-4-9-16-25
so, these are the lockers that remain open

By: saeedsafasaeed

saeedsafasaeed — Tue, 24 Nov 2009 04:29:52 +0000

answer, 4,9,16,25 (the square numbers)

One can draw a table with all of the multiples, (3 times table, 4 times table etc) and can count the number of times each number appears, if the number is odd then that means that that door is closed, if the number is even then that means that the door is open

By: Bobo The Bear

Bobo The Bear — Mon, 23 Nov 2009 21:48:41 +0000

Lockers 1, 4, 9, 16, and 25 will remain open.

The only way a locker will remain open is if it is affected by an odd number of students. This only happens for the lockers that represent the perfect squares. Here’s why:

Each locker is affected only by the students whose numbers are factors of the locker number. For example, locker #10 if affected by students 1, 2, 5, and 10. This is an even number of factors, and so locker #10 will end up being closed. Because factors come in pairs (1×10; 2×5), most numbers have an even number of factors, meaning that most lockers will end up being closed.

The perfect squares are different. They are the result of a number multiplied by itself, but that number doesn’t get counted twice when writing a list of factors. Look at locker #9, for example. The number 9 can be factored as 1×9 or 3×3, but the list of factors is just 1, 3, and 9. So locker #9 is only affected by three students, and remains open at the end.

By: Shawn

Shawn — Mon, 23 Nov 2009 19:35:28 +0000

Well, if I programmed Excel correctly, it looks like only 5 doors remain open.

The open doors are on lockers whose numbers are a perfect square,

1
4
9
16
25

By: aaronlau

aaronlau — Mon, 23 Nov 2009 18:31:30 +0000

All the squares will be Open.
1, 4, 9, 16, 25.

Because prime numbers only once. Close
All other numbers with 2 different denominators will be past by twice, neutralizing the effect. Plus the number itself, a odd number of times. Close.
Left only the squared numbers that pass by an even number of times. Open.

By: joe

joe — Mon, 23 Nov 2009 16:09:09 +0000

Well I did it the long way, on a spreadsheet methodically changing it every time (although I know there is a quicker mathematical method)

I ended up with OPEN for 1, 4, 9, 16, 25.

So an easy pregressive series, which I am sure others will explain better…
Nice test thoguh, thanks.

By: Migrated

Migrated — Mon, 23 Nov 2009 13:40:38 +0000

Since there are 30 lockers and 30 students, all the lockers will be open after the first student. The first locker therefore will remain open. All the even numbers will be closed. When the third student comes around, the multiples of 3 will be changed and so on. We eventually find a pattern that all the square numbers will be remained open, so that is 1,4,9,16 and 25. This would work for any x number of lockers and x number of students.

By: Shofnite

Shofnite — Mon, 23 Nov 2009 13:29:30 +0000

My idea for this one…
The # of door which has even number of divisors will remain closed, and the one with odd number of divisors will remain closed.
For example,
Door # 4 will remain open because divisors of 4 are – 1,2,4, so the number of divisors is – 3.
Door # 7 will remain closed because number of divisors of 7 are 2 (1,7).

This clearly shows that Prime Numbered Doors will remain closed.

Remaining Numbers –
4,6,8,9,10,12,14,15,16,18,20,21,22,24,25,26,27,28,30.

Out of them, only 4,9,16,25 have odd number of divisors.
So only door numbered 4,9,16 and 25 will remain open.
Oh and yes, #1 will remain open as well.

So,
Door # 1,4,9,16 and 25 will remain open.
So, 5 doors will remain open.

The answer looks promising enough, though I’m not too sure.

One thing I noticed, all the numbers I obtained above are perfect squares. A little bit of googling and I found out that –
“Only Squares have odd number of divisors”.
I didn’t know about this before. So, a new fact learned today!

By: suineg

suineg — Mon, 23 Nov 2009 13:17:17 +0000

cool puzzle, one solution could be this:
there is a pattern 1 only is open once an no other touches again that locker, 1 has 1 factor, 2 is open and then close and has 2 factors so to know which lockerss remains open you have to see the factors of the numbered lockers, if is an odd number then is open if is an even number then is close:
1: 1 (1) open 16: 1,2,4,8,16 (5) open
2: 1,2 (2) close 17: 1,17 (2) close
3: 1,3 (2) close 18: 1,2,3,6,9,18 (6) close
4: 1,2,4 (3) open 19: 1,19 (2) close
5: 1,5 (2) open 20: 1,2,4,5,10,20 (6) close
6: 1,2,3,6 (4) close 21: 1,3,7,21 (4) close
7: 1,7 (2) close 22: 1,2,11,22 (4) close
8: 1,2,4,8 (4) close 23: 1,23 (2) close
9: 1,3,9 (3) open 24: 1,2,3,4,6,8,12,24 (8) close
10: 1,2,5,10 (4) close 25: 1,5,25 (3) open
11: 1,11 (2) close 26: 1,2,13,26 (4) close
12: 1,2,3,4,6,12 (6)close 27: 1,3,9,27 (4) close
13: 1,13 (2) close 28: 1,2,4,7,14,28 (6) close
14: 1,2,7,14 (4) close 29: 1,29 (2) close
15: 1,3,5,15 (4) close 30: 1,2,3,5,6,10,15,30 (8) close

So the 1,4,5,9,16 and 25 are open.