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In how many ways can the number 1;000;000 (one million) be written as the product of three positive integers $a, b, c,$ where $a \le b \le c$?

(A) 139

(B) 196

(C) 219

(D) 784

(E) None of the above

This is my working out so far:

$1000000 = 10^{6} = 2^{6} \cdot 5^{6}$ and then to consider this as the product of three factors i.e.

$10^6 = 2^6 \cdot 5^6$

$= 2^a 5^p \cdot 2^b5^q \cdot 2^c 5^r$ (where $a+b+c = 6$ and $p+q+r = 6$).

However there are repetitions here because $2^3\,5^3 \cdot 2^2\,5^2 \cdot 2^1\,5^1$ is the product of the same three factors as $2^2\,5^2 \cdot 2^3\,5^3 \cdot 2^1\,5^1$.

I think there are 139 such factors.

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I think some working or demonstration of effort should be given. As a very general comment, this kind of problem calls for some careful organisation of cases - so some clarity of thought. It is worth trying this out for yourself before asking for help, because it always looks so blindingly obvious when you are told a good way of doing it - and that doesn't do anything much to help you solve the next problem. – Mark Bennet Jan 8 '13 at 23:17
Sorry didn't have time to write my working out beforehand, just came back to write my working out now. – DeeDee Jan 8 '13 at 23:35
up vote 4 down vote accepted

Hint: Solve the problem for $abc = 10^6$. This has $ {8 \choose 2}^2=784$ solutions.

Count the number of solutions where $a=b=c$.

Count the number of solutions where $a=b$ or $b=c$ or $c=a$.

Count the number of solutions where $a, b, c$ are pairwise distinct.

Account for your repeated counting above, to find the cases where $a \leq b \leq c$.

Motivation: Bars and stripes, Orbit Stabilizier Theorem.

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My friend tried using combinations of factors. 1 x 1 x 1000000, 1 x 2 x 500000, 1 x 4 x 250000, ... ... 80 x 100 x 125, 100 x 100 x 100. – DeeDee Jan 8 '13 at 23:55
@user55613 I designed this problem for, though it's basic enough that you could conceivably find it elsewhere, which is why I withdrew my objection. The above is a straightforward approach to work on this very easily, having understood the effects of double counting. Of course, listing out all the factors is one brute force way to do it, but it doesn't add to your understanding of the issues at hand. – Calvin Lin Jan 8 '13 at 23:57
@CalvinLin: furthermore, this question counts $1\cdot2^6\cdot5^6$ and $2^6\cdot1\cdot5^6$ as the same, whereas the question counts them as different (as asked). – robjohn Jan 9 '13 at 0:17
@robjohn I had a version that was the same as this, but responded before I checked that I sent the correct link. OP mentioned that this is MCQ, which will be very different from a short answer qn, which is why I withdrew my objection (deleted comments, but couldn't pull chat or flag). I gave an explanation of an approach that is logical, and not simply brute force / coded, so that even if someone sees this, they do not get the answer directly, but have to work through it and understand what is happening. – Calvin Lin Jan 9 '13 at 0:23

As we can factorise one million as $5^6 \times 2^6 $, it becomes a matter of how many ways we can split up a set of size 12 into three disjoint subsets. This is more of a combinatronics problem, really.

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I don't understand the downvote. This answer is exactly headed in the right directly. Of course, it misses the detail of keeping track of repetitions. – Code-Guru Jan 8 '13 at 22:59
This is not true. Perhaps if it's 2 sets of size 6, split into 3 disjoint subsets, AND then keeping track of repetitions – Calvin Lin Jan 8 '13 at 23:00
Yeah, I forgot about the fact we'd be counting extra numbers of numbers (we'd be counting three for every one we need, I believe, although this is a quick guess) - as for the 2 sets of size 6, I fail to see why this is any different to one set of size 12 (as the 2 sets of size 6 will be disjoint). – Andrew D Jan 8 '13 at 23:03
Actually, now I've had a proper look at it, I can see why having 2 sets of size 6 is useful - it cuts down on a lot of repetitions doing it with a set of size 12. – Andrew D Jan 8 '13 at 23:20

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