# Resolve into to factors

The number of ways the no. $10,800$ can be resolve as a product of $2$ factors, is

My Try :: $10,800 = 2^4 \times 3^3 \times 5^2$

So Total no. of factors $= (4+1)\times(3+1)\times(2+1) = 60$

But I did not understand how can i resolve into two factors

Thanks

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+1 for showing your effort –  Ross Millikan Feb 2 '13 at 4:29

Your expression for the number of factors is correct. So the number of ordered pairs $(u,v)$ of positive integers such that $uv=10800$ is indeed $60$.

The answer to "how many ways $\dots$" unfortunately depends on interpretation. If we consider the factoring $100\times 108$ to be the same as the factoring $108\times 100$, then the number $60$ should be divided by $2$, giving an answer of $30$. And then if we don't consider $1\times 10800$ as a factoring, we get $29$.

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Thanks André Nicolas I did not understand why we divide it by $2$ –  juantheron Feb 2 '13 at 4:16
@juantheron: You are welcome. As was pointed out in the answer, it is not absolutely clear how the question should be interpreted. Note that if our number is a perfect square, like $144=2^4\times 3^2$, there are $(5)(3)$ positive factors. The factoring $12\times 12$ does not change when we reverse the numbers, so the number of factorings (if order does not count) is $\frac{15-1}{2}+1$. –  André Nicolas Feb 2 '13 at 4:22
I don't understand shouldn't it be 60 sense $$\sum_{mk=n}=\sum_{d\mid n}1=d(n)$$ –  Ethan Feb 2 '13 at 4:57
@Ethan: If we consider $a\times b$ as the same factoring as $b\times a$, we get half of $60$. –  André Nicolas Feb 2 '13 at 5:06
I see now, for every $d$ there exists another divisor $\frac{n}{d}$, so we can pair them up together, such that the number of tuples $(n,\frac{n}{d})$ is $[\frac{d(n)}{2}]$, where [x] is the floor function –  Ethan Feb 2 '13 at 5:19

There are $4 \times 3 \times 2 = 24$ ways of picking 1 factor - the number of $2$s, $3$s and $4$s. Since each factor will have another complimentary number to multiply, the total number of pairs will be $$\frac {24} {2} = 12$$

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So, do you think $2^13^1$ has only $1\times1$ ways of picking $1$ factor? –  anon Feb 2 '13 at 4:52