Mathematics question (1 Viewer)

[SpiralFlex]

Hi all,

I have a few questions that I have been trying to work out during the last few days.

For Question 1, I have done all the proofs except for logbase3(5). It's proof by contradiction.



Thank you.

 

[Iruka]

Presumably you have been using the fact that an integer cannot be simultaneously odd and even to do Q1. You do can do the base 3 one by appealing of the Fundamental Theorem of Arithmetic:

http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic

(The odd/even stuff is just a special case of this.)

For Q2, I think you just write out the integers. For example, 1,2,4,5,7 and 8 are co prime with 9, 3 and 6 are not (they have a common factor of 3), hence phi(9) = 6.

There is a formula for Euler's Phi function which you can probably find on wikipedia if you are interested.

Q3 is just some slightly messy algebra. You can make it simpler by choosing the units of measurement so that the volume of the tank is 1.

 

[SpiralFlex]

Cheers Iruka.

 

[cutemouse]

Is the answer for the third question 39 and 37 minutes?

 

[SpiralFlex]

Is the answer for the third question 39 and 37 minutes?

2 hours and 4 hours. It's a basic quadratic question. I just do not know how to structure it.

 

[cutemouse]

Of course my answer's wrong. I should think before I post...

 

[SpiralFlex]

I'm not going to sleep until I solve this.

 

[hscishard]

Let the volume be T
T/80 = rate of both pipes

Let x be time in minutes
T/(x-120) + T/x = T/80 since it's both pipes working together

2x-120 = x(x-120)/80. Solve for x. x will be the larger solution as one of them will give a negative time value

 

[ohexploitable]

Let the volume be T
T/80 = rate of both pipes

Let x be time in minutes
T/(x-120) + T/x = T/80 since it's both pipes working together

2x-120 = x(x-120)/80. Solve for x. x will be the larger solution as one of them will give a negative time value

I remember asking you this question last year in the marathon thread. Or maybe you asked me?

 

[cutemouse]

For the other one...

Suppose that log_3 5 = p/q p,q relatively prime

5=3^(p/q)

5^q=3^p, which is impossible if p and q are relatively prime (5^q ends with 5 or 0 and 3^p ends with 3, 9, 7, 1 or 3)

 

[SpiralFlex]

There, solved my own question:

Let the small pipe be x hours and the large pipe will be x-2 hours.

In 1 hour, the small pipe alone will fill 1/x of the tank while the larger pipe will fill 1/(x-2) of the tank.

In 1 hour, the two pipes will fill 1/x + 1/(x-2) of the tank.

Given that the two pipes fill the whole tank in 80 minutes, in 1 hour, 3/4 tank is filled.
Therefore, 1/x + 1/(x-2) = 3/4
4(x-2) + 4x = 3x(x-2)
3x² − 14x + 8 = 0
x = 4
Therefore the small pipe takes 4 hours and the large pipe takes 2 hours.

1 question down 2 more to go!

 

[hscishard]

I remember asking you this question last year in the marathon thread. Or maybe you asked me?

I dunno. I remember seeing one that looked like that in cambridge yr11 3unit. I remember asking you about a circle question though...

 

[SpiralFlex]

For the other one...

Suppose that log_3 5 = p/q p,q relatively prime

5=3^(p/q)

5^q=3^p, which is impossible if p and q are relatively prime (5^q ends with 5 or 0 and 3^p ends with 3, 9, 7, 1 or 3)

I didn't considered that, I only tried the odd = even contradiction, thanks Jm.

 

[SpiralFlex]

I dunno. I remember seeing one that looked like that in cambridge yr11 3unit. I remember asking you about a circle question though...

It is in the Cambridge 3 Unit Year 11 book. =D