HSC 2015 MX2 Marathon (archive) (3 Viewers)

InteGrand · Jul 4, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
im assuming this works with n amount of factors aswell?

The factors need to be prime though. So like I said in my earlier post, first we need to write the integer under consideration as a product of powers of its prime factors.

Sy123 · Jul 4, 2015

Re: HSC 2015 4U Marathon

$\\ $Show that 1 less than the square of an odd number is divisible by 8$$

InteGrand · Jul 4, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
$\\ $Find the amount of positive integers that divide into$ \ 1^{a_1}2^{a_2}...n^{a_n} \ $for non-negative integers$ \$

$(a_1+1)(a_2+1)...(a_n+1)$ ?

No, since the factors must be prime factors. So we'd need to rewrite the given integer in terms of prime factors first, and THEN we could apply that formula (or figure it out).

Drsoccerball · Jul 4, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
No, since the factors must be prime factors. So we'd need to rewrite the given integer in terms of prime factors first, and THEN we could apply that formula (or figure it out).

But theres no common prime factor (even though impossible) wouldnt that be the formula

Drsoccerball · Jul 4, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Show that 1 less than the square of an odd number is divisible by 8$$

How do we prove that $n^{4m-2} -1$ is divisible by 8 D:
EDIT: i can only think of induction

InteGrand · Jul 4, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
But theres no common prime factor (even though impossible) wouldnt that be the formula

Not if they're not all primes.

E.g. in $N = 1^a 2^b 3^c 4^d$ , 4 is not a PRIME factor, so we would need to rewrite 4 in terms of its prime factors, which is just 4= 2², then write $N=2^b 3^c 2^{2d}= 2^{b+2d} 3^c$ (using index laws).

Now we have it in terms of PRIME factors, so we can apply the formula with the exponents in the primes in the last expression.

Sy123 · Jul 4, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
How do we prove that $n^{4m-2} -1$ is divisible by 8 D:
EDIT: i can only think of induction

Is that a typo?

The number would be of the form $(2n+1)^2 -1$

InteGrand · Jul 4, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
How do we prove that $n^{4m-2} -1$ is divisible by 8 D:
EDIT: i can only think of induction

$We can do the question by starting by letting our odd number be $2n-1$, then consider $(2n-1)^2 -1$.$

Sy123 · Jul 4, 2015

Re: HSC 2015 4U Marathon

$\\ $Show that if the polynomial$ \ x^3 + px + q = 0 \ $has exactly one real root then$ \ 4p^3 + 27q^2 > 0$

InteGrand · Jul 4, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
Is that a typo?

The number would be of the form $(2n+1)^2 -1$

I think he was thinking of $\left(n^{\text{odd number}}\right)^2$ , judging by what we wrote.

Sy123 · Jul 4, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
Is that a typo?

The number would be of the form $(2n+1)^2 -1$

Also try to do so without induction (though in an exam situation they probably wouldn't force you to think of a non-inductive solution)

Drsoccerball · Jul 4, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Show that if the polynomial$ \ x^3 + px + q = 0 \ $has exactly one real root then$ \ 4p^3 + 27q^2 > 0$

f'(x) 3x^2 +p

$x=\sqrt{\frac{-p}{3}}$

$Sub back into P(x)$

$P(p'(a)) <0$

$Get desired result$

I need to sleep D:

Drsoccerball · Jul 4, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
Is that a typo?

The number would be of the form $(2n+1)^2 -1$

You get 4n(n+1)
$\therefore $ prove $ : 2 divides n(n+1)$

Since all even numbers are divisible by 2 if any of the numbers are even then proof is complete
Since adding one changes it to odd from even vice versa and there is an n it will always be divided by 2
soz for dodge proof

Ekman · Jul 4, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
You get 4n(n+1)
$\therefore $ prove $ : 2 divides n(n+1)$

Since all even numbers are divisible by 2 if any of the numbers are even then proof is complete
Since adding one changes it to odd from even vice versa and there is an n it will always be divided by 2
soz for dodge proof

I would of mentioned that 2 is a prime number and that is able to divide all even numbers.
Hence: n(n+1) is always even for all real values of n. (As you get even number x odd number=even number)
As n(n+1) is always even, it is divisible by 2 as 2 is the base prime number for all even numbers

Sy123 · Jul 4, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
You get 4n(n+1)
$\therefore $ prove $ : 2 divides n(n+1)$

Since all even numbers are divisible by 2 if any of the numbers are even then proof is complete
Since adding one changes it to odd from even vice versa and there is an n it will always be divided by 2
soz for dodge proof

Ekman said:
I would of mentioned that 2 is a prime number and that is able to divide all even numbers.
Hence: n(n+1) is always even for all real values of n. (As you get even number x odd number=even number)
As n(n+1) is always even, it is divisible by 2 as 2 is the base prime number for all even numbers

These are very similar proofs just worded differently. For future reference a more rigorous way of writing it would be (I'm not sure how nitpicky they would be in the HSC, its good to be safe though):

$(2n+1)^2 - 1 = 4n(n+1) = 8 \left(\frac{n(n+1)}{2} \right)$

$\\ $If$ \ n \ $is even, then$ \ \frac{n}{2} \ $is an integer (by def. of even number), so$ \ \frac{n}{2}(n+1) \ $is an integer, so$ \ 8 \frac{n}{2}(n+1) \ $is divisible by$ \ 8$

$\\ $If$ \ n \ $is odd, then$ \ n+1 \ $is even, so$ \ \frac{n+1}{2} \ $is an integer (by def. of even number), so$ \ n \frac{n+1}{2} \ $is an integer, so$ \ 8n \frac{n+1}{2} \ $is divisible by$ \ 8$

$\\ $Thus, since$ \ n \ $is either even or odd, and both result in the expression being divisible by$ \ 8 \ $thus the expression is divisible by$ \ 8$

--------

Alternatively to avoid this lengthy wordy proof:

$\\(2n+1)^2 - 1 = 4n(n+1) = 8 \left(\frac{n}{2}(n+1) \right) = 8 (1+2+3+\dots + n) = 8P \ $where$ \ P \ $integer$ \\ \\ \therefore \ (2n+1)^2 - 1 \ $is divisible by$ \ 8$

Sy123 · Jul 4, 2015

Re: HSC 2015 4U Marathon

$\\ $Prove that$ \ \frac{1}{6} n(n+1)(n+2) \ $is an integer for integers$ \ n \geq 1$

Try to prove it without induction (try to generalize the above method of using evens/odds, or go for a purely algebraic approach)

Ekman · Jul 4, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Prove that$ \ \frac{1}{6} n(n+1)(n+2) \ $is an integer for integers$ \ n \geq 1$

Try to prove it without induction (try to generalize the above method of using evens/odds, or go for a purely algebraic approach)

$\frac{1}{6} n(n+1)(n+2) = \frac{1}{6}\times\frac{(n+2)!}{(n-1)!} = \binom{n+2}{n-1}$

$\therefore $ Since $ \binom{n+2}{n-1} $ is an integer for $ n\geq1 $ hence $ \frac{1}{6} n(n+1)(n+2) $ is also an integer $$

InteGrand · Jul 4, 2015

Re: HSC 2015 4U Marathon

Another way:

$Alternatively, use the fact that $n,n+1,n+2$ are three consecutive integers, so they contain both a multiple of 2 and a multiple of 3.$

Speed6 · Jul 4, 2015

Re: HSC 2015 4U Marathon

How many marks will that question generally be?

Sy123 · Jul 4, 2015

Re: HSC 2015 4U Marathon

Speed6 said:
How many marks will that question generally be?

Probably 2

HSC 2015 MX2 Marathon (archive) (3 Viewers)

Well-Known Member

This too shall pass

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

This too shall pass

Well-Known Member

This too shall pass

Well-Known Member

This too shall pass

Well-Known Member

Well-Known Member

Well-Known Member

This too shall pass

This too shall pass

Well-Known Member

Well-Known Member

Retired '16

This too shall pass

Users Who Are Viewing This Thread (Users: 0, Guests: 3)