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Euler
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#1
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All the mathematicians seem to have gone sleeping... i need more challenging (but solvable) mathematic problems anyone? willing to volunteer?
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firebladez777.5
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1. A spider is sitting in the middle of one of the smallest walls in my living room and a fly is resting by the side of the window on the opposite wall, 1.5m above the ground and 0.5m from the adjacent wall.

What is the shortest distance the spider would have to crawl to catch the fly?

The room is 5m long, 4m wide and 2.5m high.


If the fly walks down the wall, is there a point at which the spider would be better changing its route.

2. Ask a friend to choose a number between 1 and 63 and make a mental note of it without telling you.

Show each of the six cards (you need to have those number cards, http://www.nrich.maths.org.uk/conten...able_cards.doc) in turn to your friend and ask them whether it contains your number.

Using this information and the cards it is possible to say what the number is.

For example your friend might choose the number 21.

This appears on three cards:
The one starting with a 4
The one starting with a 1
And the one starting with a 16.

That is interesting! 4+1+16 =21

Is this always the case?
Why?
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firebladez777.5
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Oh, here's another one:

Mr Smith and Mr Jones are two maths teachers, who meet up one day. Mr Smith lives in a house with a number between 13 and 1300. He informs Mr Jones of this fact, and challenges Mr Jones to work out the number by asking closed questions.

Mr Jones asks if the number is bigger than 500. Mr Smith answers, but he lies.

Mr Jones asks if the number is a perfect square. Mr Smith answers, but he lies.

Mr Jones asks if the number is a perfect cube. Mr Smith answers and (feeling a little guilty) tells the truth for once.

Mr Jones says he knows that the number is one of two possibilities, and if Mr Smith just tells him whether the second digit is 1, then he'll know the answer. Mr Smith tells him and Mr Jones says what he thinks the number is. He is, of course, wrong.

What is the number of Mr Smith's house?

Btw, dude.........get a life. It's the holidays!!!
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Euler
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(Original post by firebladez777.5)
Oh, here's another one:

Mr Smith and Mr Jones are two maths teachers, who meet up one day. Mr Smith lives in a house with a number between 13 and 1300. He informs Mr Jones of this fact, and challenges Mr Jones to work out the number by asking closed questions.

Mr Jones asks if the number is bigger than 500. Mr Smith answers, but he lies.

Mr Jones asks if the number is a perfect square. Mr Smith answers, but he lies.

Mr Jones asks if the number is a perfect cube. Mr Smith answers and (feeling a little guilty) tells the truth for once.

Mr Jones says he knows that the number is one of two possibilities, and if Mr Smith just tells him whether the second digit is 1, then he'll know the answer. Mr Smith tells him and Mr Jones says what he thinks the number is. He is, of course, wrong.

What is the number of Mr Smith's house?

Btw, dude.........get a life. It's the holidays!!!

Tell me the answers
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firebladez777.5
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I supply the questions. Never said anything about giving you the answers.
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firebladez777.5
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OK, here's the soltuion to the Card Numbers:

The solution to this problem is actually very simple. First consider all of the top left hand numbers: 1, 2, 4, 8, 16, 32. We shall call these the KEY numbers. It is these that are added to give the number you are thinking of.

Now, every number from 1 to 63 can be made by adding these numbers, the first 8 are given below:

1: - 1

2: - 2

3: - 2+1

4: - 4

5: - 4+1

6 :- 4+2

7: - 4+2+1

8: - 8

You can now add 1-7 to 8 to get all the numbers up to 16. At 16, you can add all the number 1-15 to 16, to get all the numbers to 32. And finally, you can add all the numbers 1-31 to 32, to get all the numbers up to 63 (made up of 1+2+4+8+16+32)

Each number has a unique way of being formed. Now, when making the cards, imagine that a person starts with blank cards except for the left hand corners which have the key numbers written on. All he has to do is go through all the numbers from 1-63, writing them on those cards whose key number makes it up.

For example, 39, would be written on those cards whose key numbers are 32, 4, 2 and 1. This way, the number 39 appears on just these cards, and the magician (or anyone) can quickly add the numbers (32+4+2+1).

Therefore, since every number is made up in a unique way from the key numbers, the key numbers on those cards selected will always add up to your number.
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firebladez777.5
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Hint to the spider/fly one: Sometimes problems like this are better "unwrapped". How about considering the net of the room?
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Euler
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(Original post by firebladez777.5)
Hint to the spider/fly one: Sometimes problems like this are better "unwrapped". How about considering the net of the room?

unwrapped? net of the room? :confused:
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firebladez777.5
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Net...like opening up a cube/cuboid so that it becomes two-dimensional...surely you know that?! :eek:
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Euler
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(Original post by firebladez777.5)
Net...like opening up a cube/cuboid so that it becomes two-dimensional...surely you know that?! :eek:

What about Mr jones/smith problem?


tell me the house number its driving me mad :mad:
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firebladez777.5
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(Original post by integral_neo)
What about Mr jones/smith problem?


tell me the house number its driving me mad :mad:
What do you think it is?
Tell me, then I promise I'll give you the answer.
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Euler
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#12
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(Original post by firebladez777.5)
What do you think it is?
Tell me, then I promise I'll give you the answer.

627
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Cool Lad
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The function f is defined by

f:x - In(5x - 2), x>3/2

a. Find an expression for f^-1(x)
b. Write down the domain of f^-1
c. Solve, giving your answer to 3dp,

In(5x - 2) = 2
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firebladez777.5
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(Original post by integral_neo)
627
Ha ha ha.... (ok, I've had my daily hysterical laughter fit...good try though).

Here's the solution: The first clue we have is that the number is between 13 -1300. The second clue we have is that it may be above 500. The third clue is that the answer may be a square number and the fourth that the number may be a cube.

So, to follow the third and fourth clues, I listed all the square numbers and all the cube numbers between 13 -1300. You could generate these very quickly on a spreadsheet.

The fifth clue was that whatever Mr Smith said, Jones believed. So Jones followed what he thought was true and came up with two answers. One of which had "1" as its second digit.


If Mr Jones was told that the number was more than 500, and that the number wasn't a square number or a perfect cube, then we would end up with over 500 answers, which doesn't work. The same goes if Mr Jones was told that the number was less than 500, and that the number wasn't a square number or a perfect cube. It doesn't work either.


So let's presume that Mr Jones was told that the number was more than 500 and that the number was a square but not a cube. It's impossible because no square number bigger than 500 has 1 as a second digit.


If Mr Jones was told that the number was smaller than 500, and that the number was a square number but not a perfect cube, then we have met the criteria that one of the solutions has 1 as the second digit, but there are a lot more than just 2 solutions, which is what we were looking for.


Let's presume that Mr Smith said that the number was more than 500, and that the number wasn't a square but a perfect cube. 27,64,125, 216,343 are out because they are smaller than 500. Looking at the list of square numbers, 729 is out because it appears there. Leaving 512, 1000 left. It meets the criteria that there are two solutions left. And 512 has a 1 as a second digit.


This means that Mr Smith told Jones that the number was more than 500, not a square number but a perfect cube.


We know that Mr Smith lied to Jones about whether the number was 500 or not, and whether the number was a square or not. Hence, we can come up with the conclusion that:

Mr Smith's house is less than 500

His house is a square number and a perfect cube


Looking through the solutions we can get, there is only one solution, and that Mr Smith's house number is 64 (less than 500, square number, and a perfect cube).


So the answer is 64.
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Euler
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(Original post by Cool Lad)
The function f is defined by

f:x - In(5x - 2), x>3/2

a. Find an expression for f^-1(x)
b. Write down the domain of f^-1
c. Solve, giving your answer to 3dp,

In(5x - 2) = 2
In??? whats that
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firebladez777.5
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(Original post by integral_neo)
In??? whats that
I think he means ln (you know...natural logarithm?). i.e. taking logs to base e
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Euler
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#17
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(Original post by firebladez777.5)
I think he means ln (you know...natural logarithm?). i.e. taking logs to base e
i thought so
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Euler
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#18
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(Original post by Cool Lad)
The function f is defined by

f:x - ln(5x - 2), x>3/2

a. Find an expression for f^-1(x)
b. Write down the domain of f^-1
c. Solve, giving your answer to 3dp,

ln(5x - 2) = 2
a. f^-1(x) = (e^x + 2)/5

b. x>3/2

c. solve what?!
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Fermat
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(Original post by integral_neo)
a. f^-1(x) = (e^x + 2)/5

b. x>3/2

.....
f(x) = ln(5x - 2) , x > 3/2

if x>3/2 then f(x) > ln(15/2 - 2) = ln(11/2)

i.e codomain of f(x) is ln(11/2) which becomes the domain of its inverse, f^(-1).

b) x > ln(11/2)
==========

(Original post by integral_neo)
......

c. solve what?!
Solve this,

ln(5x - 2) = 2

therefore,

5x - 2 = e^2 - ( from definition of a logarithm)
x = (2 + e^2)/5
x = 2.956
======

Alternatively,

f(x) = ln(5x - 2)
g(x) = (2 + e^x)/5, where g = f^(-1)

and f(g(x)) = g(f(x)) = id(x) = x where id() is the identity function.

We have f(x) = ln(5x - 2) = 2

So,

g(f(x)) = g(2)
x = g(2)
x = (2 + e^2)/5
x = 2.956
======
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yellowtaxi
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#20
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(Original post by integral_neo)
All the mathematicians seem to have gone sleeping... i need more challenging (but solvable) mathematic problems anyone? willing to volunteer?
OH MY GOD

how sad are all yous...."i need more work"....gawd :rolleyes:
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