q from UNSW maths assessment (from back in June I think?) (1 Viewer)

[Dumsum]

It's the second last one:

The letters m and n are positive integers, and n³ = m⁵. If n < 3 000 000 how many values can m take?

I cannot for the life of me figure out a logical way of approaching this question, but I'm comforted by the fact that only 3% of the state got it correct.

 

[turtle_2468]

If you split n up using prime decomposition you will see that n has to be a power of 5. Therefore take 5th root of 3000000 - floor of that number should be possible values of (5th root of n) and hence the possible number of values n can take. This is the same as the number of values m can take.

In a rush... msg again if you don't get this, I'll explain in more detail.

 

[Dumsum]

Hmm, 5th root of 3,000,000 = 19.74... and the answer is 19... wow.

But yeah, it kinda went over my head a bit... I understand what prime decomposition is, but how do I apply that in this situation?

 

[turtle_2468]

Hmm, 5th root of 3,000,000 = 19.74... and the answer is 19... wow.

But yeah, it kinda went over my head a bit... I understand what prime decomposition is, but how do I apply that in this situation?

Umm... ok.
So n^3=m^5.

Let n=p1^a1*p2^a2*p3^a3.....*pn^an. (too lazy... you know a1 is a subscript 1 etc)
Obviously m has to have the same factors...
so m=p1^b1*p2^b2*...*pn^bn.

n^3=p1^3a1*p2^3a2*p3^3a3.....*pn^3an.
m^5=p1^5b1*p2^5b2*...*pn^5bn.

So because the two numbers are the same.. 3a1=5b1 etc.
So a1, a2, ..., an are all divisible by 5.

In addition, if we chuck any set of numbers a1, a2, ... which are all divisible by 5 in, there are valid values for bn, so we have a solution for (n,m). Therefore we have all the solutions of (n,m).

Now we need to figure out how many n's we can have... well we know that 1) it is a 5th power and 2) all 5th powers work. ie if we let n=x^5, we have a "good" solution if x is an integer. So we just need to solve x^5<3000000 which we did..

 

[Stefano]

For those of us that do not understand what prime decomposition is or how it works could you briefly explain or chuck us a url ?

Thanks.

 

[turtle_2468]

basically, just writing a number as the product of prime numbers.
So for example, 24 is not a prime number. but we can divide out by 2 to get 12, again by 2 twice to get 3 and by 3 to get 1
so 24 = 2^3*3.
Similarly 120=2^2*3*5. Note that all the things which are raised to powers (here 2,3 and 5) are prime.
Strangely enough this requires a fundamental theorem of (algebra?) but it's quite easy to prove...
anyway. To do a prime decomposition on a number, just find a prime it's divisible by and repeat.

eg take 920. You know it's divisible by 2, so 920=2*460
2*460=2*2*230
=2*2*2*115
=2*2*2*5*23 which are all prime
hence 2^3*5*23=920.
The form I put above (p1^a1) etc is just putting this forumla into unknowns...

 

[Templar]

Strangely enough this requires a fundamental theorem of (algebra?) but it's quite easy to prove...

Each number has an unique prime factorisation (as stated by the fundamental theorem of arithmetic).

 

[Stefano]

Ahh, i see now. Very interesting! Thanks turtle_2468.

My next question is: Why is Dumsum ( A year 12 student ) doing University math questions? If you're that good mate and you got spare time, I'd love some help

 

[Dumsum]

Ahh, i see now. Very interesting! Thanks turtle_2468.

My next question is: Why is Dumsum ( A year 12 student ) doing University math questions? If you're that good mate and you got spare time, I'd love some help

It was a question in the Mathematics Assessment... http://www.etc.unsw.edu.au/etc/schools_competitions/australasian_competitions/mathematics Not my fault if it's uni level lol. Remember only 3% of entrants got it out.