O-Level Additional Math

[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.

[OK Lor]

Hi CoffeeCat,


Thanks. The method was suggested by my dad . From the Remainder Theorem, when f(x) is divided by x-1 (or 7), the remainder will be f(1).
Modular arithmetic -> lx+4l >5? ok lor.
Congruences -> plane geometry? (needs to use a lot of right brain :lol: )

[CoffeeCat]

OK Lor:

Hi CoffeeCat,


Thanks. The method was suggested by my dad . From the Remainder Theorem, when f(x) is divided by x-1 (or 7), the remainder will be f(1).
Modular arithmetic -> lx+4l >5? ok lor.
Congruences -> plane geometry? (needs to use a lot of right brain :lol: )

no la modular arithmetic is merely a branch of mathematics with notation to deal with remainder, just like how logarithms is invented just to deal with exponents.
The cool thing about modular arithmetic and congruences is it make our life easier when expressing our ideas.

[OK Lor]

Hi CoffeeCat,


Cool! It's useful for SMO, will put in some efforts to understand the concept. Thanks.
a = b (mod m).

[OK Lor]

Hi Sir,


Please help (SMO 2010):

1. Find the value of (14^3 + 15^3 + 16^3 + … + 24^3 + 25^3)^(1/2).
2. Find the least possible value of f(x) = 9/(1+cos2x) + 25/(1-cos2x), where x ranges over all real numbers for which f(x) is defined.

Thanks.