O-Level Additional Math

[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? :?

[Guan Hui]

seems correct to me=D

[CoffeeCat]

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? :?

Your x is not a variable, it's actually a constant 8, so you can't treat it as a function.

2[3^(6n)] + k[2^(3n+1)] - 1
= 2 (729^n) + 2k ( 8^n) - 1
= 2 (728 + 1)^n + 2k ( 7+1)^n - 1

If you expand the above expression out and get rid of terms you know are divisible by 7 you will get
2 + 2k - 1 = 2k + 1 (since 1^n = 1)
So the k you are looking for is 3, 10, 17, ...

For questions like this, sometimes must expect that every number is there for a trick reason. Like 2^3 = 8 = (7+1) and 3^6 = 729 = (728 +1).
If you can't see why (ax + 1)^n will leave a remainder of 1 upon division by x i suggest you try small values of n.

lol oklor y you disable pm?

[OK Lor]

Hi Sir, thanks.


Hi Coffecat,
Got it, nice solution, thanks.
Just got to know what was the problem to my working,
=> f(1) = 2^(3n+1) + 2k - 1 = 0 (wrong, as the remainder can be 0 or multiple of 7).
=> f(1) = 2^(3n+1) + 2k - 1 = 2(7 +1)^n + 2k - 1.
Remainder of f(1) / 7 = 2 + 2k - 1 = 2k +1.
k = 3, 10, 17, 24, …, 94.
Total 14 number of positive integers k.

[CoffeeCat]

Ok i understood your working. Somehow you managed to filter out the terms that will be divisible by 7. However just curious, were you taught to use this function method by anyone?

I know conceptually you are trying to say that when you say that expression leave the same remainder when the 8 (or x) is replaced by a 1 (well done for seeing that), but unless your teacher approve that method, I think the way you present your working is mathematically weird. have you learn modular arithmetic or congruences? that’s the tool you might need to express that conceptual idea.