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I am trying to work out the number of ways of choosing $a$ and $b$, $1 \le a, b \le 1000$, such that $ab = 2^6 \cdot 3^2 \cdot 5^4$.

Attempt 1) Times indices of the numbers together, since $6$ ways of choosing powers of two, $2$ for $3$, and $4$ for $5$, giving $6\cdot 2\cdot 4=48$

Attempt 2) writing out all possible ways that can factor $ab$.

Questions: 1) What are quick by hand methods for this type of problem? 2) How if any is this related to "$n$ choose $r$" formula for combinations?

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    what is ltealte? edit: ah makes sense now =) – 2017-01-12
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    Don't use _times_ as a verb. Use the grown-up version _multiply_. – 2017-01-12
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    One small note, there is also the option of choosing $0$ for each of the powers, so there are actually $(6+1)(2+1)(4+1)=105$ factors of $ab$ rather than $48$. – 2017-01-12
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    I would have said 4*2*3 distinct combinations – 2017-01-12

1 Answers 1

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It is enough to count the number of admissible choices for $$a=2^x\cdot 3^y\cdot 5^z\ ,\qquad x, \>y,\>z\in{\mathbb N}_{\geq0}\ .$$ The restrictions are $$x\leq6,\quad y\leq2,\quad z\leq4,\quad a\leq1000\ .$$

I propose a simple counting, according to the possible values of $z$:

Since $5^4=625$ the value $z=4$ enforces $x=y=0$. Gives ${\bf 1}$.

Since $5^3=125$ the value $z=3$ enforces $2^x\cdot 3^y\leq8$, hence $y=0$ and $0\leq x\leq3$, or $y=1$ and $0\leq x\leq1$. Gives ${\bf 6}$.

And so on: The value $z=2$ enforces $2^x\cdot 3^y\leq40$, $\ldots$