O-Level Additional Math

[OK Lor]

Hi CoffeeCat,


paiseh, from the questions asked, 不值一提 :oops:.

[ckh]

Hi, all,


This is an assignment question for my son in Sec 1. I would like to poll your understanding in interpreting this question, especially the last sentence:

Beginning of question:

Imagine a school with 1000 numbered, closed lockers, one for each of its students. Suppose all the students line up and go through the locker area to perform a specific task.

The first student will open all the locker doors.

The second student will close all the locker doors with even numbers.

The third student will change all the locker doors with numbers that are multiples of three. (Change means closing lockers that are open and opening locker doors that are closed.)

The fourth student will change the position of all the locker doors numbered with multiples of four, the fifth student will change the position of all the locker doors that are multiples of fives, and so on.

Every student goes through, reversing the lockers which correspond to his or her position in line until the 1000th student has passed through.

End of question.

To keep things short, I would like to identify the first 5 lockers' positions (whether opened or closed).

We would like to double check our understanding of the question

Thks and best rgds.

[corneyAmber]

2,3,5 Open

1 and 4 Closed

Correct?

[CoffeeCat]

ckh:

Every student goes through, reversing the lockers which correspond to his or her position in line until the 1000th student has passed through.

The last sentence indeed sound funny, but its valid for students whose numbers are very big.

I will say locker 1 and 4 are opened,
2, 3, 5 closed.

If you and your son has different interpretations it will be interesting to share with us.

[ckh]

Hi,


Thanks for your reply. This was what we got also.

[ckh]

Hi, dear CoffeeCat and ksi,


Indeed the last sentence sounds funny, and this is the reason we like to double check our interpretation. There are lots of solutions on the Internet that provided the same answers as yours, i.e. locker 1 and 4 are opened, 2, 3, 5 closed.

But because of the action on the last sentence, "Every student goes through, reversing the lockers which correspond to his or her position in line until the 1000th student has passed through", we felt this answer will get inverted, i.e. 2, 3, 5 Open, 1 and 4 Closed, which ksi had provided, and this was what we got also. Does this "inversion" make sense ?

Incidentally, most of the solutions on the Internet do not have this last sentence. I think the school added this sentence to introduce some variance to the question.

What do you all think ?

[CoffeeCat]

The funny sentence will sound funny if one is trying to apply it to small numbers.

Because the sentences above is describing 3 instances to give an idea,i suppose the funny sentence is just a concluding statement describing the whole process. Perhaps it was phrased carelessly as an afterthought by some teacher.
But the funny sentence is true for numbers from 501 to 1000. Those students are merely reversing lockers corresponding to their position.
Maybe your child ought to bug the teacher regarding the interpretation.

If the teacher had meant a 2-process thing, he/she is likely to include a "after this process is finished" clause.

[Guan Hui]

Ok lor for the latest question about partial fraction. I found out if you take the numerator divided by any of the denominator it will get the same remainder(using long division). But still thinking a way to give you the answer.

[ckh]

Yeap, thanks for your comments. I certailnly thought that it was a poorly worded sentence too. We will definitely clarify with the teacher.

[OK Lor]

Hi Sir,


Thanks. I think the remainders have to be the same, so that the quotient is common, in this case is x (qoute from the 高人 (CoffeCat) from the previous question: If the number has the same remainder 1 upon division by 3 and 5, it will leave the same remainder 1 upon division by 15 -> 15x+1 ).
By long division or algebraic juggling, (1-11x-4x^2+3x^3-2x^4)/(6-4x+3x^2-2x^3) = x + (1-17x)/(6-4x+3x^2-2x^3).
(1-17x)/(6-4x+3x^2-2x^3) = (1-17x)/[(3-2x )(x^2 + 2)] = A/(3-2x) + (Bx+C)(x^2 + 2).