Cool problem of the day! (1 Viewer)

Carrotsticks · May 14, 2012

seanieg89 said:
Only nontrivial solns are the pairs (4,2), (2,4). Proof:

$$For $x\geq 3>e$, and $k$ any positive integer we have:\\$(x+k)^x=x^x(1+k/x)^x<x^xe^k<x^{x+k}$

so we cannot have equality if x and y are not equal. Here I have assumed a commonly known limit expression for e (and the fact that this expression is increasing...an easy thing to check).

Nice! The same question can also be posed, but in Q instead of Z.

Define the set of Prime Numbers: {p_1, p_2, p_3,...,p_n}

They are not necessarily consecutive.

Show that:

$\sum_{i=1}^{n} \frac{1}{p_i} \notin \mathbb{Z} \qquad \forall n \in \mathbb{N}$

barbernator · May 14, 2012

Carrotsticks said:
Nice! The same question can also be posed, but in Q instead of Z.

Define the set of Prime Numbers: {p_1, p_2, p_3,...,p_n}

They are not necessarily consecutive.

Show that:

$\sum_{i=1}^{n} \frac{1}{p_i} \notin \mathbb{Z} \qquad \forall n \in \mathbb{N}$

i need to learn some proper notation.

kazemagic · May 14, 2012

barbernator said:
did it again and got this. I will post up a solution if people want a bit later.

Yes pl0xxx

AAEldar · May 14, 2012

barbernator said:
i need to learn some proper notation.

It makes things easier to write out!

4025808 · May 14, 2012

Just read the first question up on the top. Managed to get it using circle geo techniques normally used in 4U; making and implying that there's a circle over a triangle that is 90*s. Used it twice though.

RealiseNothing · May 14, 2012

Carrotsticks said:
Nice! The same question can also be posed, but in Q instead of Z.

Define the set of Prime Numbers: {p_1, p_2, p_3,...,p_n}

They are not necessarily consecutive.

Show that:

$\sum_{i=1}^{n} \frac{1}{p_i} \notin \mathbb{Z} \qquad \forall n \in \mathbb{N}$

Is it a proof by contradiction?

Carrotsticks · May 14, 2012

RealiseNothing said:
Is it a proof by contradiction?

As long as your proof is valid and makes sense, then it is acceptable, whether it be by contradiction or not.

barbernator · May 14, 2012

here is my solution, it skips a few working steps to stay short and on one page.

seanieg89 · May 14, 2012

Suppose the sum is equal to some integer N. Multiply both sides by p1p2...pn and you will get an equation, the right side of which is divisible by every prime in the set, the left side of which isn't. Contradiction.

seanieg89 · May 14, 2012

A harder question along those lines, prove that the sum: 1+1/2+1/3+...+1/n is not an integer for any n>1. Also, I posted a question on the previous page.

barbernator · May 14, 2012

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.

seanieg89 · May 15, 2012

barbernator said:
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.

barbernator · May 15, 2012

barbernator said:
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?

seanieg89 · May 15, 2012

barbernator said:
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.

Carrotsticks · May 16, 2012

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:

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

-----------------------------------------------------------------------------

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!

RealiseNothing · May 16, 2012

Carrotsticks said:
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.

RealiseNothing · May 16, 2012

Carrotsticks said:
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.

RealiseNothing · May 16, 2012

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.

nightweaver066 · May 16, 2012

Carrotsticks said:
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}$

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

$x^2 - 100 < 0$

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

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

$x^2 - 32 > 0$

$x > 2\sqrt{7}$

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

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

Carrotsticks · May 16, 2012

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.

Cool problem of the day! (1 Viewer)

Retired

Active Member

Member

Premium Member

Well-Known Member

what is that?It is Cowpea

Retired

Active Member

Well-Known Member

Well-Known Member

Active Member

Well-Known Member

Active Member

Well-Known Member

Retired

what is that?It is Cowpea

what is that?It is Cowpea

what is that?It is Cowpea

Well-Known Member

Retired

Users Who Are Viewing This Thread (Users: 0, Guests: 1)