What is $c$ in $\left\lfloor\frac{a}{bc}\right\rfloor=d$

Question

As part of my attempts to solve integer factorization problem. I came to this equation:

$$\left\lfloor\frac{a}{bc}\right\rfloor=d$$

• $a,b,c,d$ are positive integer values
• $\frac{a}{b}$ is an integer value

I know the values of $a,b,d$; how can I find $c$?

Answer 1

Having $\lfloor\frac{a}{bc}\rfloor=d$ is the same thing as saying:

$$d\leq\frac{a}{bc}<d+1$$

So assuming both $a$ and $b$ are positive, you obtain the following set of solutions for $c$:

$$\frac{a}{b(d+1)}<c\leq\frac{a}{bd}$$

Answer 2

Note that the question can be simplified: you have positive integers $a,b$, and $n$, you know $a$ and $b$, and you know that $\left\lfloor\frac{a}n\right\rfloor=b$, and you want to find $n$. (My $a$ is your $a/b$, my $n$ is your $c$, and my $b$ is your $d$.)

We know that $b\le\frac{a}n<b+1$, so $bn\le a<(b+1)n$, and therefore

$$\frac{a}{b+1}<n\le\frac{a}b\;;\tag{1}$$

any integer $n$ satisfying $(1)$ will serve.

Answer 3

Expanding on ruadan's answer:

If $c$ is to be uniquely determined, we must have the two bounds within $1$, or $1 \ge \frac{a}{bd}-\frac{a}{b(d+1)} = \frac{a(d+1-d)}{bd(d+1)} = \frac{a}{bd(d+1)} $ or $a \le bd(d+1)$.