# Thread: Cool problem of the day!

1. ## Re: Cool problem of the day!

i havent completed the question myself, but just posting some things i have noticed. z>2, x and y cannot both be even integers. LHS can never have 4 as a factor. xyz =/= 4k. only 1 term can be even. z is odd. The even integer must not be a factor of 4.

One solution is x=1,y=1,z=3

just updating my thoughts as i go.

2. ## Re: Cool problem of the day!

Originally Posted by barbernator
i havent completed the question myself, but just posting some things i have noticed. z>2, x and y cannot both be even integers.
Good start . This question is reasonably difficult.

3. ## Re: Cool problem of the day!

Originally Posted by barbernator
i havent completed the question myself, but just posting some things i have noticed. z>2, x and y cannot both be even integers. LHS can never have 4 as a factor. xyz =/= 4k. only 1 term can be even. z is odd. The even integer must not be a factor of 4.

One solution is x=1,y=1,z=3

just updating my thoughts as i go.
heres what i've got, but I actually have no idea to come to a "solution" as such other than manipulating the equation to find restrictions. Is there any way to manipulate the equation to bring it to 2 variables?

4. ## Re: Cool problem of the day!

Originally Posted by barbernator
heres what i've got, but I actually have no idea to come to a "solution" as such other than manipulating the equation to find restrictions. Is there any way to manipulate the equation to bring it to 2 variables?
Hint: There are no solutions if z>3. Once you have proven this, the problem is in two variables.

5. ## Re: Cool problem of the day!

I have an idea.

Each day or so, I'll post up a Number Theory problem (similar to seanieg89's question), a Geometry (including Analytic Geometry) problem, and an Algebra problem every day or so depending on if I have time.

This way you can answer which ever question you want to based on your strengths.

ie: My strength would be Series & Analysis-type questions whereas somebody else may have strengths in geometry.

Furthermore since many people here have little exposure to Elementary Number Theory (due to it not being in the HSC course), they may be more difficult to solve and I don't want this thread to die the moment a Number Theory question is added.

Whilst barbernator does seanieg89's question (I'm sure you'll get it, you are an intelligent student), here are my next set of problems, a little easier this time:

Number Theory:

Prove that the product of 4 consecutive integers is always one less than a perfect square.

Geometry:

The sides of a triangle (in the Euclidean Plane) are 6, 8 and x.

Find the range of x such that the triangle is acute.

Algebra:

A Triangle Number is defined by the following expression (for more reading, go here http://en.wikipedia.org/wiki/Triangular_number)

$T_n = \frac{n(n+1)}{2}$

Prove WITHOUT INDUCTION that:

$T_n = \sum_{k=0}^{n-1} (-1)^k (n-k)^2$

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Also, I think it's best that we let the HSC students try the questions. After all, this is for their benefit in terms of developing problem solving skills!

But if you are an Undergrad/Postgrad and you have a truly marvelous proof (something a HSC student wouldn't know) that is otherwise too large to fit in the margin of your copy of Arithmetica, then by all means post it here!

6. ## Re: Cool problem of the day!

Originally Posted by Carrotsticks
But if you are an Undergrad/Postgrad and you have a truly marvelous proof (something a HSC student wouldn't know) that is otherwise too large to fit in the margin of your copy of Arithmetica, then by all means post it here!
Haha, inb4nextfermat.

7. ## Re: Cool problem of the day!

Originally Posted by Carrotsticks
Number Theory:

Prove that the product of 4 consecutive integers is always one less than a perfect square.
$n(n+1)(n+2)(n+3) = n^4 + 6n^3 + 11n^2 + 6n$

We have to add a one on the end since it's one less than a perfect square though.

If this were a perfect square, there must be 3 terms all added together, which when squared equal the above expression.

The first term must be n^2 since $(n^2)^2 = n^4$

The third term must be 1 since $1^2=1$

Now we only have to find the middle term.

We know the middle term must be a factor of $n$, so let's express it as $n(x)$

So $(n^2 + n(x) + 1)^2 = n^4 + 6n^3 + 11n^2 + 6n + 1$

$n^4 + n^3(x) + n^2 + n^3(x) + n^2(x)^2 + n(x) + n^2 + n(x) + 1 = n^4 + 6n^3 + 11n^2 + 6n + 1$

$n^4 + 2xn^3 + (x^2+2)n^2 + 2xn + 1 = n^4 + 6n^3 + 11n^2 + 6n + 1$

Hence $x=3$ by equating co-efficients.

Therefore $(n^2 + 3n + 1)^2 = n^4 +6n^3 + 11n^2 + 6n + 1$

Hence the product of 4 consecutive integers is always one less than a perfect square.

8. ## Re: Cool problem of the day!

An even easier way!

Let the four consecutive integers be $n~(n+1)~(n+2)~(n+3)$

Multiply the two inner terms, and the two outer terms. We get:

$(n+1)(n+2)=n^2 + 3n + 2$

$n(n+3)=n^2 + 3n$

We can rewrite these as:

$(n+1)(n+2)= (n^2 + 3n + 1) + 1$

$n(n+3)= (n^2 + 3n + 1) - 1$

To get the product of all 4 terms, we just multiply these two expressions together.

$[(n^2 + 3n + 1) + 1][(n^2 + 3n + 1) - 1]$

But! They are actually a difference of two squares!

Hence the product of the 4 integers is:

$(n^2 + 3n +1)^2 - 1$

Which is one less than a perfect square, as required.

9. ## Re: Cool problem of the day!

Originally Posted by Carrotsticks

Geometry:

The sides of a triangle (in the Euclidean Plane) are 6, 8 and x.

Find the range of x such that the triangle is acute.

$6 + 8 > x ~ (Triangle inequality)$

$x < 14$

$Similarly,~ x + 6 > 8$

$x > 2$

$\therefore 2 < x < 14$

$By the cosine rule, cos\alpha = \frac{100 - x^2}{96}$

$If cos\alpha > 0, this means that 0 < \alpha < \frac{\pi}{2}$

$\frac{100 - x^2}{96} > 0$

$x^2 - 100 < 0$

$\therefore 0 < x < 10, x \nleqslant 0$

$Similarly,~ cos\beta = \frac{x^2 - 32}{12x}$

$x^2 - 32 > 0$

$x > 2\sqrt{7}$

$0 < x < 10 \cap 2 < x < 14 \cap x > 2\sqrt{7}$

$\therefore 2\sqrt{7} < x < 10$

10. ## Re: Cool problem of the day!

Very nice Realise!

nightweaver, you are very close, but not quite.

A few things:

1. It's called the Triangle Inequality, not the 'Axiom of Triangle'. It is something prove-able.

2. 2 is indeed a lower bound for x, but it isn't necessarily the LARGEST lower bound. Consider the limiting cases when the triangle is right angled.

11. ## Re: Cool problem of the day!

Originally Posted by Carrotsticks
Very nice Realise!

nightweaver, you are very close, but not quite.

A few things:

1. It's called the Triangle Inequality, not the 'Axiom of Triangle'. It is something prove-able.

2. 2 is indeed a lower bound for x, but it isn't necessarily the LARGEST lower bound. Consider the limiting cases when the triangle is right angled.
Updated.

12. ## Re: Cool problem of the day!

Originally Posted by nightweaver066
Updated.
Haha you got 'axiom of triangle' from Harry didn't you?

13. ## Re: Cool problem of the day!

Originally Posted by Bored_of_HSC
Haha you got 'axiom of triangle' from Harry didn't you?
Who else.. haha..

Originally Posted by barbernator
2 root 7 instead of 4 root 2.
I can't do simple arithmetic. Thank you.

14. ## Re: Cool problem of the day!

Originally Posted by nightweaver066
Who else.. haha..
I think best to use 'Triangle Inequality' from now on.

'Axiom of triangle' just sounds silly imo as it is not an axiom.

15. ## Re: Cool problem of the day!

sorry I phrased the inequality completely wrongly. I don't think it's really an inequality I have to solve as an upper bound of seaniegs question now i think of it. The lower bound was just x^2 + y^2 +1 > 2xy, but i've got no idea for how to prove that z<4. It's probably something to do with the discriminant being a perfect square when i solve the quadratic for x possibly, but I don't really know how to prove that anyway. Oh well.

16. ## Re: Cool problem of the day!

If only I was capable enough to do the questions which Carrotsticks posts up daily. I must be missing out.

17. ## Re: Cool problem of the day!

Originally Posted by Demento1
If only I was capable enough to do the questions which Carrotsticks posts up daily. I must be missing out.
in time you will btw, i have a good one which i will suggest to carrot for the next round of q's

18. ## Re: Cool problem of the day!

Originally Posted by barbernator
sorry I phrased the inequality completely wrongly. I don't think it's really an inequality I have to solve as an upper bound of seaniegs question now i think of it. The lower bound was just x^2 + y^2 +1 > 2xy, but i've got no idea for how to prove that z<4. It's probably something to do with the discriminant being a perfect square when i solve the quadratic for x possibly, but I don't really know how to prove that anyway. Oh well.
Proving that z>2 is trivial, but proving z<4 is not...Assume solutions with z>3 exists and look for a contradiction. I will post a solution tomorrow.

19. ## Re: Cool problem of the day!

Hint for the Triangular Number question:

First prove that the sum of two consecutive triangular numbers is a perfect square....

$T_n + T_{n-1} = ....$

20. ## Re: Cool problem of the day!

$T_n+T_{n-1}=n^2\\ \Rightarrow T_n=(-1)^nT_0+\sum_{k=0}^{n-1} (-1)^k(T_{n-k}+T_{n-k-1})=\sum_{k=0}^{n-1}(-1)^k(n-k)^2$

21. ## Re: Cool problem of the day!

Sweet!

Here is the next set of problems:

Geometry (courtesy of barbernator):

Consider a circular paddock with radius R such that R > 5 metres.

Along the circumference of the paddock, a post hammered into the ground and a sheep (poor sheep) is tied to the post by a 10 metre long rope.

In terms of R, what percentage of the paddock can the sheep roam?

Note: To check if your answer is correct or not, you should be able to show that when R --> Infinity, the percentage approaches 0 and when R --> 5, the percentage approaches 100.

Algebra:

Prove that the sum of the reciprocals of INFINITE Triangular Numbers is 2.

Number Theory:

seanieg89 will be providing the Number Theory questions from now on. Much appreciated =)

22. ## Re: Cool problem of the day!

Originally Posted by Carrotsticks
Algebra:

Prove that the sum of the reciprocals of INFINITE Triangular Numbers is 2.
I've noticed that the triangular numbers are either multiples of 3, or you can make a series of the non-multiples of 3 which differ by increasing factors of 9.

This have anything to do with it?

23. ## Re: Cool problem of the day!

$\sum_{n=1}^N T_n^{-1}=2\sum_{n=1}^N \frac{1}{n(n+1)}=2\sum_{n=1}^N \frac{1}{n}-\frac{1}{n+1}=2-\frac{2}{N+1}\rightarrow 2\\ \\as N\rightarrow \infty$

New number theory questions, ordered roughly by increasing difficulty:

1. Prove that the product of 5 consecutive positive integers cannot be the square of a positive integer.

2. The positive integers m and n satisfy: $2001m^2+m=2002n^2+n$. Prove that m-n is a perfect square.

3. An odd positive integer n is said to be 'sexy' if n divides: $1\cdot 3\cdot 5\cdot \ldots \cdot (n-2)+2\cdot 4\cdot 6 \cdot \ldots\cdot (n-1).$ Prove that for any twin prime pair (p,q), EXACTLY one of p,q is sexy.

4. Let a be an arbitrary irrational number. Prove that there are irrational numbers b and b' such that:

-a+b, ab' are both rational,

-a+b', ab are both irrational.

24. ## Re: Cool problem of the day!

I never saw the day coming where you would prove something is sexy...

25. ## Re: Cool problem of the day!

And a quick sketch of my solution to the question I posted earlier, I might tidy it up a bit later.amstemplate.pdf

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