How many triplets (x,y,z) are there if xyz = 2012 and x ≤ y ≤ z?

Hi MWizzard :))

The questions use the current year!

I am just going to copy Heureka's method   :)

Given that xyz = 2016 and x, y and z are positive integers such that x < y < z, how many possible triples are there?

First you need to work out all the prime factors of 2016

\(2016=2^5*3^2*7\)

You can work out the third one of each trio yourself  :)

There is probably a much faster way.  But if there is no silly omissions that should be right    

I found 60 ordered trios

Are x,y and z integers?

I will assume that they are positive integers

2012=2*2*504

so

x=1, y=4, z=504

or

x=2 y=2  z=504

The divisors of 2012  are  1, 2, 4, 503, 1006, 2012

Assuming that x, y and z don't have to be unique [ all different], we have

x      y         z

1     1       2012

1     2       1006

1     4         503

2     2         503

4 triplets  ???

I think that's it   [ could you check this, heureka???.....your always way better at this than I am....!!!]

How many triplets (x,y,z) are there if xyz = 2012 and x ≤ y ≤ z?

\(\begin{array}{|c|c|c|c|c|} \hline & x & y & z & x \le y \le z\\ \hline 1 & 1 & 1 & 2012 & \surd \\ 2 & 1 & 2 & 1006 & \surd \\ 3 & 1 & 4 & 503 & \surd \\ 4 & 1 & 503 & 4 & \\ 5 & 1 & 1006 & 2 & \\ 6 & 1 & 2012 & 1 & \\ 7 & 2 & 1 & 1006 & \\ 8 & 2 & 2 & 503 & \surd \\ 9 & 2 & 503 & 2 & \\ 10 & 2 & 1006 & 1 & \\ 11 & 4 & 1 & 503 & \\ 12 & 4 & 503 & 1 & \\ 13 & 503 & 1 & 4 & \\ 14 & 503 & 2 & 2 & \\ 15 & 503 & 4 & 1 & \\ 16 & 1006 & 1 & 2 & \\ 17 & 1006 & 2 & 1 & \\ 18 & 2012 & 1 & 1 & \\ \hline \end{array}\)

It's okay CPhill !

Thanks, heureka!!!!!....BTW  .....I meant to type "you're" instead of "your"......doh  !!!!!!

Help me in Math Olympiad!!!

The questions use the current year!

Given that xyz = 2016 and x, y and z are positive integers such that x < y < z, how many possible triples are there?

Yes you are right MWizzard, 27 is not a factor of 2016  :)