O-Level Additional Math

[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).

[OK Lor]

Yo Sir,


Please help:
Find the number of positive integers k<100 such that 2[3^(6n)] + k[2^(3n+1)] - 1 is divisible by 7 for any positive integer n.

Thanks

[OK Lor]

Hi Sir,


The following are my workings, please comment:

2[3^(6n)] + k[2^(3n+1)] - 1
=2[9^(3n)] + 2k(8^n) - 1

Let 7 = x - 1
=> f(x) = 2[(x+1)^(3n)] + 2k(x^n) - 1
=> f(1) = 2^(3n+1) + 2k - 1 = 0
k = 1/2 - 2^(3n) -> -ve? :?