HSC 2013 MX2 Marathon (archive) (2 Viewers)

RealiseNothing · Dec 6, 2013

Re: HSC 2014 4U Marathon

Sy123 said:
Could you clarify what you mean by $a_m = a_{n+m}$ , there is no term in the sum beyond $a_{n+1}$ , so I was wondering if it were perhaps something like $a_m = a_{n-m}$

Basically it's just a relation to account for the $a_{n+1}$ on the end of the summation, i.e. let $a_{n+1} = a_1$ so you obtain $\sqrt[n]{\frac{a_n}{a_1}}$

Sy123 · Dec 6, 2013

Re: HSC 2014 4U Marathon

Hopefully this works:

$Without loss of generality, we assume the sequence$ \{a_k \} \ $from$ \ k=1, 2, \dots , n \ $is increasing$

$$Therefore$ \ \ \frac{a_k}{a_{k+1}} < 1$

$$Therefore$ \ \ \frac{a_{k}}{a_{k+1}} < \sqrt[k]{\frac{a_{k}}{a_{k+1}}}$

$\Rightarrow \ \ \sum_{k=1}^n \sqrt[k]{\frac{a_k}{a_{k+1}}} > \sum_{k=1}^n \frac{a_k}{a_{k+1}}$

$$By the AM-GM inequality, we find that$ \ \ \sum_{k=1}^n \frac{a_k}{a_{k+1}} \geq n \sqrt[n]{\frac{a_1}{a_{n+1}}} = n \ \ \ \ \ \ $Since$ \ a_1 = a_{n+1} \ \dots (*)$

Now,

$\frac{n(n+1)}{2}\left(\frac{(\prod_{k=1}^{n-1} (k!)^2 )^{1/n}}{(n!)^2} \right)^{1/(n+1)} = \frac{n(n+1)}{2} \left(\prod_{m=1}^{n} \frac{(m-1)!}{n!} \right)^{\frac{2}{n(n+1)}}$

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

That product in the middle, we split it up individually, we are left with $1 + 2 + \dots + = \frac{n(n+1)}{2}$ number of terms. Therefore by the AM-GM inequality again.

$\left(\frac{1}{n} + \frac{1}{n} + \dots + \frac{1}{n} \right) + \dots + \left( \frac{1}{2} + \frac{1}{2} \right) + 1 \geq \frac{n(n+1)}{2} \left( \frac{1}{n^n} \frac{1}{(n-1)^{n-1}} \dots \frac{1}{2^2} \frac{1}{1} \right)^{\frac{2}{n(n+1)}$

The LHS = n, therefore:

$n \geq \frac{n(n+1)}{2}\left(\frac{(\prod_{k=1}^{n-1} (k!)^2 )^{1/n}}{(n!)^2} \right)^{1/(n+1)}$

And by, (*)

$\sum_{k=1}^n \frac{a_k}{a_{k+1}} \geq \frac{n(n+1)}{2}\left(\frac{(\prod_{k=1}^{n-1} (k!)^2 )^{1/n}}{(n!)^2} \right)^{1/(n+1)}$

And due to our initial condition

$\sum_{k=1}^n \sqrt[k]{\frac{a_k}{a_{k+1}}} \geq \frac{n(n+1)}{2}\left(\frac{(\prod_{k=1}^{n-1} (k!)^2 )^{1/n}}{(n!)^2} \right)^{1/(n+1)}$

Proof is complete?

RealiseNothing · Dec 7, 2013

Re: HSC 2014 4U Marathon

I'm not sure if you can assume $a_k < a_{k+1}$ since we have different powers for each term.

Sy123 · Dec 7, 2013

Re: HSC 2014 4U Marathon

RealiseNothing said:
I'm not sure if you can assume $a_k < a_{k+1}$ since we have different powers for each term.

What do you mean?

RealiseNothing · Dec 7, 2013

Re: HSC 2014 4U Marathon

Sy123 said:
What do you mean?

As in, if you assume the sequence is increasing, is there really no loss of generality?

I'm pretty sure it loses generality. It only really works when all terms are essentially arbitrary with respect to eachother (for example if you could replace $a_1$ with $a_5$ and it wouldn't change anything). However since the k-th term depends on whatever term is in the numerator (as the k-th term would be the k-th root), then we can not just replace terms with each other as this would change the actual summation.

You assumed that the sequence was increasing. Now suppose otherwise and the sequence is decreasing, we get:

$1 < \frac{a_k}{a_{k+1}}$

$\sqrt[k]{\frac{a_k}{a_{k+1}}} < \frac{a_k}{a_{k+1}}$

And this would lead to your last step, where you use the initial condition, becoming invalid. This is because it is definitely possible to have a decreasing sequence.

RealiseNothing · Dec 7, 2013

Re: HSC 2014 4U Marathon

Two nice questions:

1) The number $12^{a} \times 35^{b} \times 40^{c}$ ends in a sequence of $n$ zeroes. Find $n$ in terms of $a, b, c$ if $a>b>c$

(i.e. if the number was 5000, then n=3 as it ends with 3 zeroes)

2) Consider the product/number $1 \times 3 \times 7 \times 9 \times 11 \times 13 \times 17 \times 19 \times 21 \times ... \times (10^n-1)$

Find the last digit of this number (i.e. the units digit).

RealiseNothing · Dec 7, 2013

Re: HSC 2014 4U Marathon

Consider an arbitrary chord in a circle.

Let the midpoint of this chord be the point P.

Prove whether or not we can construct another chord in the same circle such that it's midpoint coincides with P.

braintic · Dec 7, 2013

Re: HSC 2014 4U Marathon

RealiseNothing said:
Consider an arbitrary chord in a circle.

Let the midpoint of this chord be the point P.

Prove whether or not we can construct another chord in the same circle such that it's midpoint coincides with P.

Doesn't this depend on whether the arbitrary chord is in fact a diameter?

Drongoski · Dec 7, 2013

Re: HSC 2014 4U Marathon

I think this is equivalent to showing that this chord, with P as its mid-point, is the shortest chord passing thru P.

Edit: It is sufficient to show any chord thru P has a length different from that of this chord.

Sy123 · Dec 7, 2013

Re: HSC 2014 4U Marathon

RealiseNothing said:
As in, if you assume the sequence is increasing, is there really no loss of generality?

I'm pretty sure it loses generality. It only really works when all terms are essentially arbitrary with respect to eachother (for example if you could replace $a_1$ with $a_5$ and it wouldn't change anything). However since the k-th term depends on whatever term is in the numerator (as the k-th term would be the k-th root), then we can not just replace terms with each other as this would change the actual summation.

You assumed that the sequence was increasing. Now suppose otherwise and the sequence is decreasing, we get:

$1 < \frac{a_k}{a_{k+1}}$

$\sqrt[k]{\frac{a_k}{a_{k+1}}} < \frac{a_k}{a_{k+1}}$

And this would lead to your last step, where you use the initial condition, becoming invalid. This is because it is definitely possible to have a decreasing sequence.

Hmm I see.....

In case of montone decreasing however, we simply need to see that:

$\sum_{k=1}^n \sqrt[k]{\frac{a_k}{a_{k+1}}} > \sum_{k=1}^n \sqrt[n]{\frac{a_k}{a_{k+1}}}$

Which would make a process identical to the montone increasing one

Though for non-monotonic sequences, I think we would need to:

$\\ $Prove that there exists an integer$ \ m \ $such that for some sequence$ \ \ {a_k} \\ \\ \sum_{k=1}^n \sqrt[k]{\frac{a_k}{a_{k+1}}} > \sum_{k=1}^n \sqrt[m]{\frac{a_k}{a_{k+1}}}$

Whether this is true or not I don't know, but if we can prove that then we have proven it for the general case of any sequence.

I'll see if I can think of another way though

Sy123 · Dec 7, 2013

Re: HSC 2014 4U Marathon

RealiseNothing said:
Two nice questions:

1) The number $12^{a} \times 35^{b} \times 40^{c}$ ends in a sequence of $n$ zeroes. Find $n$ in terms of $a, b, c$ if $a>b>c$

(i.e. if the number was 5000, then n=3 as it ends with 3 zeroes)

2) Consider the product/number $1 \times 3 \times 7 \times 9 \times 11 \times 13 \times 17 \times 19 \times 21 \times ... \times (10^n-1)$

Find the last digit of this number (i.e. the units digit).

$$1)$ \ 12^a \times 35^b \times 40^c = 2^{2a+3c} 5^{b+c} 3^a 7^b$

$Hence we must find the largest$ \ n \ $such that$

$\frac{2^{2a+3c} 5^{b+c} 3^a 7^b}{2^{n} 5^n } = K \ $for integer$ \ K$

$\therefore \ \ 2a+3c = n \ $and$ \ b+c \geq n$

$$OR$ \ b+c = n \ $and$ \ 2a+3c \geq n$

$$The first option implies$ \ \ b \geq 2a + 2c \ $which is not true as of$ \ a>b>c$

$$Therefore$ \ \ n = b+c$

Sy123 · Dec 7, 2013

Re: HSC 2014 4U Marathon

RealiseNothing said:
Consider an arbitrary chord in a circle.

Let the midpoint of this chord be the point P.

Prove whether or not we can construct another chord in the same circle such that it's midpoint coincides with P.

$$Draw a circle with chord$ \ AB \ $midpoint$ \ P \ $draw another chord$ \ CD \ $through$ \ P$

$$Let$ \ \angle CAB = \angle CDB = \beta$

$\angle ACD = \angle ABD = \alpha$

$$Let$ \ AP = BP = a$

$By the sine rule$

$\frac{CP}{\sin \beta} = \frac{a}{\sin \alpha}$

$\frac{DP}{\sin \alpha} = \frac{a}{\sin \beta }$

$$We assume$ \ \alpha, \beta \neq 0 \ $otherwise$ \ CD \ $co-incides with$ \ AB$

$$Therefore if$ \ CP = DP$

$\sin \alpha = \sin \beta$

$\rightarrow \ \alpha = \beta$

$$Hence$ \ \triangle CPA = \triangle BPD \ $and$ \ CA || BD \ $and$ \ CB || AD$

$Therefore$ \ ABCD \ $is a parralellogram$

$But since it is cyclic, therefore each angle must be a right angle$

$Therefore$ \ ABCD \ $is a rectangle$

$\\ $But a chord only subtends a right angle if it is the diameter, therefore the only way$ \ CP = PD \ $is iff$ \ P \ $is the center$$

Sy123 · Dec 7, 2013

Re: HSC 2014 4U Marathon

RealiseNothing said:
Two nice questions:

1) The number $12^{a} \times 35^{b} \times 40^{c}$ ends in a sequence of $n$ zeroes. Find $n$ in terms of $a, b, c$ if $a>b>c$

(i.e. if the number was 5000, then n=3 as it ends with 3 zeroes)

2) Consider the product/number $1 \times 3 \times 7 \times 9 \times 11 \times 13 \times 17 \times 19 \times 21 \times ... \times (10^n-1)$

Find the last digit of this number (i.e. the units digit).

I hope this is rigorous enough:

$2) Inside the product, we will write each number in the form$

$a= a_k 10^k + a_{k-1}10^{k-1} + \dots + a_1 10 + a_0$

Hence our product is:

$1\cdot 3 \cdot 7 \cdot 7 \cdot 9 \cdot (10 + 1) \cdot (10 + 3) \cdot (10+7) \cdot (10+9) \cdot \dots \cdot (9 \cdot 10^{n-1} + \dots + 9)$

The last digit in this product is the last digit of the number obtained by multiplying all the units terms.

Therefore the last digit of the product is the last digit of

$1 \cdot 3 \cdot 7 \cdot 9 \cdot 1 \cdot 3 \cdot 7 \cdot 9 \dots \cdot 9$

$= (3 \times 7 \times 9)^{10^{n-1}} = 3^{10^{n-1}} \times 7^{10^{n-1}} \times 9^{10^{n-1}}$

So now we find the last digits of each of these terms, and multiply them together.

We first notice that:

3^1 = 3

3^2 = 9

3^3 = 2 7

3^4 = 8 1

3^5 = ....3

And so on, we see that the last digit is periodic every 4 powers, alternating through 3,7,9,1. We can prove this if necessary through modular arithmetic

Meaning, we simply need to determine whether the number 10^{n-1} is in the form:

4k + 1, 4k+2 , 4k+3, 4k+4.
It cannot be 4k+1, 4k+3 since it is even, and since 10^{n-1} = 2^{n-1} 5^{n-1}, if n > 2 then it is divisible by 4, hence in the form 4k+4, thus the last digit is 1 if n > 2

Similarly we do the same for 7 and 9, and notice that 7 is periodic by 4 through:

7,9,3,1

Thus for n > 2, 7^{10^{n-1}} has last digit of 1.

For 9, it is periodic of 2 through, 9,1.

If 10^{n-1} is even, last digit is 1, thus it is 1.

---

Now if n=2, then the power of 3^(10^{n-1}} is in the form 4k+2, meaning the last digit is 9
Similar for 7, last digit is 9, m,ultiply together to get 81, times 1 from the 9, thus we take last digit from that which is 1.

Therefore we see as our final answer:

$$Last digit$ \ = \begin{cases} 9, \ \ \ \ n=1 \\ \\ 1, \ \ \ \ n\geq 2 \end{cases}$

I hope that's right lol

Sy123 · Dec 7, 2013

Re: HSC 2014 4U Marathon

Strong quintuple post:

$Sum the series$

$1\cdot 2 \cdot 3 \cdot 4 + 2\cdot 3 \cdot 4 \cdot 5 + 3 \cdot 4 \cdot 5 \cdot 6 + \dots + n \cdot (n+1) \cdot (n+2) \cdot (n+3)$

RealiseNothing · Dec 7, 2013

Re: HSC 2014 4U Marathon

Sy123 said:
I hope this is rigorous enough:

$2) Inside the product, we will write each number in the form$

$a= a_k 10^k + a_{k-1}10^{k-1} + \dots + a_1 10 + a_0$

Hence our product is:

$1\cdot 3 \cdot 7 \cdot 7 \cdot 9 \cdot (10 + 1) \cdot (10 + 3) \cdot (10+7) \cdot (10+9) \cdot \dots \cdot (9 \cdot 10^{n-1} + \dots + 9)$

The last digit in this product is the last digit of the number obtained by multiplying all the units terms.

Therefore the last digit of the product is the last digit of

$1 \cdot 3 \cdot 7 \cdot 9 \cdot 1 \cdot 3 \cdot 7 \cdot 9 \dots \cdot 9$

$= (3 \times 7 \times 9)^{10^{n-1}} = 3^{10^{n-1}} \times 7^{10^{n-1}} \times 9^{10^{n-1}}$

So now we find the last digits of each of these terms, and multiply them together.

We first notice that:

3^1 = 3

3^2 = 9

3^3 = 2 7

3^4 = 8 1

3^5 = ....3

And so on, we see that the last digit is periodic every 4 powers, alternating through 3,7,9,1. We can prove this if necessary through modular arithmetic

Meaning, we simply need to determine whether the number 10^{n-1} is in the form:

4k + 1, 4k+2 , 4k+3, 4k+4.
It cannot be 4k+1, 4k+3 since it is even, and since 10^{n-1} = 2^{n-1} 5^{n-1}, if n > 2 then it is divisible by 4, hence in the form 4k+4, thus the last digit is 1 if n > 2

Similarly we do the same for 7 and 9, and notice that 7 is periodic by 4 through:

7,9,3,1

Thus for n > 2, 7^{10^{n-1}} has last digit of 1.

For 9, it is periodic of 2 through, 9,1.

If 10^{n-1} is even, last digit is 1, thus it is 1.

---

Now if n=2, then the power of 3^(10^{n-1}} is in the form 4k+2, meaning the last digit is 9
Similar for 7, last digit is 9, m,ultiply together to get 81, times 1 from the 9, thus we take last digit from that which is 1.

Therefore we see as our final answer:

$$Last digit$ \ = \begin{cases} 9, \ \ \ \ n=1 \\ \\ 1, \ \ \ \ n\geq 2 \end{cases}$

I hope that's right lol

Yep that's right, though it's easier to do this instead:

$3^{10^{n-1}} \times 7^{10^{n-1}} \times 9^{10^{n-1}} = (3\times7\times9)^{10^{n-1}} = 189^{10^{n-1}}$

So we only need to find the last digit of the powers of 9.

asianese · Dec 7, 2013

Re: HSC 2014 4U Marathon

Love this post HSC loving you two have of this thread.

RealiseNothing · Dec 9, 2013

Re: HSC 2014 4U Marathon

Sy123 said:
Strong quintuple post:

$Sum the series$

$1\cdot 2 \cdot 3 \cdot 4 + 2\cdot 3 \cdot 4 \cdot 5 + 3 \cdot 4 \cdot 5 \cdot 6 + \dots + n \cdot (n+1) \cdot (n+2) \cdot (n+3)$

The pattern isn't too hard to see, and from there it's a straight forward induction.

However I'm trying to find a non-induction method now.

Ikki · Dec 9, 2013

Re: HSC 2014 4U Marathon

Is series a topic in 4u?

Sy123 · Dec 9, 2013

Re: HSC 2014 4U Marathon

Ikki said:
Is series a topic in 4u?

Well, lets just say its Harder 3U, where the 3U topic is Harder 2U

seanieg89 · Dec 9, 2013

Re: HSC 2014 4U Marathon

RealiseNothing said:
The pattern isn't too hard to see, and from there it's a straight forward induction.

However I'm trying to find a non-induction method now.

$\sum_{k=1}^n k(k+1)(k+2)(k+3)=\\ \frac{1}{5}\sum_{k=1}^n \left[k(k+1)(k+2)(k+3)(k+4)-(k-1)(k)(k+1)(k+2)(k+3)\right]\\ = \frac{1}{5}n(n+1)(n+2)(n+3)(n+4).$

HSC 2013 MX2 Marathon (archive) (2 Viewers)

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