## Monday, January 18, 2010

### The Locker Issue in SST

An impending headache to the administrator in planning the locker operation in SST. He seeks your advise on how to resolve this issue:

Here is the problem:

Task: In SST, there is a row of 100 closed lockers numbered 1 to 100. A student goes through the row and opens every locker. A second student goes through the row and for every second locker if it is closed, she opens it and if it is opened, she closes it. A third student does the same thing for every third, a fourth for every fourth locker and so on, all the way to the 100th locker. The goal of the problem is to determine which lockers will be open at the end of the process.

Sample Questions:

In words, explain your thinking to the following problems clearly. Be sure to use appropriate mathematical language and methods:

Which lockers remain open after the 100th student has passed?

If there were 500 students and lockers, which lockers remain opened after the 500tth student has passed?

Further Activity:

Identify another problem that uses the same method or technique to solve and post it to this blog

1. 1,4,5,6,7,8,11,13,16,17,18,19,20,23,25,29,31,32,35,37,,41,42,43,47,49,52,53,54,55,56,59,61,64,65,66,67,68,71,73,76,77,78,79,80,83,85,88,89,90,91,92,95,97100

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3. (a)Method 1:
1 of the lockers would remain open after all the opening and closing of all the lockers,which is actually the first locker.Lets imagine that the students are numbers.

1---2---3---4---5---6---7---8---9---10
| | | | | | | | | |
11--12--13--14--15--16--17--18--19--20
| | | | | | | | | |
21--22--23--24--25--26--27--28--29--30
| | | | | | | | | |
31--32--33--34--35--36--37--38--39--40
| | | | | | | | | |
41--42--43--44--45--46--47--48--49--50
| | | | | | | | | |
51--52--53--54--55--56--57--58--59--60
| | | | | | | | | |
61--62--63--64--65--66--67--68--69--70
| | | | | | | | | |
71--72--73--74--75--76--77--78--79--80
| | | | | | | | | |
81--82--83--84--85--86--87--88--89--90
| | | | | | | | | |
91--92--93--94--95--96--97--98--99--100

This diagram(square) display the rows of numbers(students).Lets imagine,all multiples of 2 less than 101 will be closed by the 2nd student.And the multiples of 3 lesser than 100 will be closed by the 3rd person and so on and so forth.But the first locker will not be locked due to the 1st person who opened all the locks.So through mathematical reasoning you will find out that all lockers would be locked except the 1st locker.

Method 2:

Solve using prime factorisation(more of reasoning)(Find the highest common factor)
1,2,3,4,5,6,7,8,9,10,11,12,13,14...100.
All numbers can be divisible by 1.So,all numbers from 1 to 100 have the highest common factor of 1.

Ans:_1__
(b)Again,Solve using prime factorisation(more of reasoning)(Find the highest common factor)
1,2,3,4,5,6,7,8,9,10,11,12,13,14...500.
All numbers can be divisible by 1.So,all numbers from 1 to 100 have the highest common factor of 1.

Ans:_1__

Example of question that requires prime factorisation to solve.

Find the lowest common multiple of 24,56 and 147.

You will get 2 to the power of 3,3 to the power of 2 and 7 to the power of 2.Using the calculator,you will find that the LCM of 24,56,147 is 3528

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5. Take each number up to 10 and find it's square.You should have all the lockers that are open.

1x1=1,2x2=4,3x3=94x4=16etc.
Ans:1,4,9,16,25,36,49,64,81,100

Ans:1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324,361,400,441,484

An alternative way is to calculate the difference between the last two numbers and add 2 to it and then add the result to the larger of the 2 numbers to form the next number. E.g 4-1=3, (3+2)+4=9

6. 1, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. which is all the prime numbers in 100.plus a few lockers that are opened and closed which is the multiple of either 2 and 3,2 and 4, 3 and 4

and these numbers are:

for 2 and 3: 6,18,30,42,54,66,78,90

for 2 and 4: 4,8,16,32,40,56,64,80,88,

for 3 and 4: none as three times four is 12.which is also a multiple of 2.

4 ,5, 6,7, 8 , 11, 13, 16 , 17,18, 19, 23, 29,30, 31, 32 , 37, 40 , 41, 42,43, 47, 53, 54,56 , 59 , 61, 64,66, 67, 71, 73,78, 79,80, 83,88, 89, 90,97.

7. redpoint records is ong bing juefrom 108 index number 16

8. I am impress by some of your answers and if you can get more of your classmates to give comments (not copy and paste) I will consider reducing your assignment.

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10. For a locker to be closed it has to have an odd numbers of factors which can only happen if the number is a square number E.g. 4=2x2,1x4 so there is only 3 different factor so the door will be closed.A locker with the number that has even factors will be open.
Ans Q1 : square numbers from 1 to 100. 1 , 4, , 9, 16, 25, 36, 49, 64, 81, 100.
Ans Q2: it is the same logic for queestion 2 so the answers are square numbers from 1 to 500. 1 , 4, , 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484.

an addition to his answer, instead of calculating the difference between last 2 numbers and adding it to the larger of 2. Just do 1+3+5+7+9.
The odd numbers in sequence

11. The method i do is square numbers:
1,4,9,16,25,36,49,64,81,100

12. 1)
1, 4, 9, 16, 25, 36, 49, 64, 81, 100.

2) Above numbers and 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484,

13. my answers are square numbers.
Q1.1,4,9,16,25,36,49,64,81,100
for 500 i use the same square numbers method.
Q2. 1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324,361,400,441,484

14. 1.I used square numbers to solve the problem.
eg. 1x1=1,2x2=4,3x3=9,etc.
Ans: 1,4,9,16,25,36,49,64,81,100

2. Continue to find the square numbers until I reached the biggest number before 500.
Ans: 1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324,361,400,441,484.
(23x23 is 529, bigger than 500)

15. sorry..there was a problem with my listing.
the lockers opened are square numbers.
(b): 1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324,361,400,441,484

16. I am going to solve this math problem by listing down the square numbers that are less than 100.
1x1=1
2x2=4
.
.
.
8x8=64
9x9=81
10x10=100
Therefore, all the lockers that are open will be:
1,4,9,16,25,36,49,64,81 and 100.

17. For this question, only square numbers will be left open as it will be the only numbers that would have an odd number of factors.

a) 1, 4, 9, 26, 25, 36, 49, 64, 81, 100
b) 1, 4, 9, 26, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484

18. Only the square numbers from 1 to 100 will be opened.
1]- 1, 4, 9, 16, 25, 36, 49, 64, 81, 100
The square numbers from 1 to 500
2]- 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484

19. By reducing the number of cabinets from 100 to 10. It is found that the first and fourth and the ninth cabinets will be left open therefore using "list the pattern" it is found that the squared number of numbers 1 to 100 will be opened.
It applies to the second question too.

Ans:
a)1, 4, 9, 16, 25, 36, 49, 64, 81, 100

b)1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484

20. Good observation - I see various methods in analysing the problems:
.1 establishing patterns
.2 reducing the problem to something simple and easy to analyse before making generalisation
.3 listing method

21. 1.1,4,9,16,25,36,49,64,81,100

2.1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324,361,400,441,484.

My method is too confusing to work out. I think i made a mistake i am not too sure.

22. The total factor need to be an odd number in order for the locker to be opened. Hence, only square numbered lockers will be open.
Eg. Number 50. 1x50, 2x25, 5x10. There are a total of 6 factors. Therefore it's closed. But for number 81. It's 1x81, 3x27, 9x9. There's only 5 factors. So it will be opened. Applies to all square numbers too.