Bounds on the size of rough numbers A positive integer $n$ can be described as $B$-rough if all of the prime factors of $n$ strictly exceed $B$.
The first five 2-rough numbers are 1, 3, 5, 7, 9. We always include 1 by convention.
It appears to be true that the $k$th $B$-rough number will never exceed $Bk$.
Answers to this question, for example, show why it is true for finite ranges of $k$ and $B$.  Is it possible to show that it is true in general without a major breakthrough in number theory?
 A: Warning: the argument extending j past B is flawed.  I will see what I can do to repair it.
Let $\phi(B,k)$ be the statement that there are more than $k$ many positive $B$-rough integers below $kB$, where both parameters are also positive integers. $\phi$ is false for $B \lt 4$, and is also false for $k=1$. Since there is only one prime between 5 and 10, $\phi$ is also false when $B=5$ and $k=2$.  However, there is a simple combinatorial argument which shows that $\phi$ is true for many pairs $(B,k)$.
$\phi$ is mildly stronger than what is asked in the posted question. When $B$ is prime, the question asks about a weak bound on totatives (numbers coprime) to $D$, $D$ being the product of the primes less than or equal to $B$. For prime $B$ greater than $3$ and $kB \geq D/2$, it is easy to compute that $\phi(k,B)$ holds. When $B=2$ the $k$th rough number is $2k-1$  and when $B=3$ the $k$th rough number is $3k-j$, where $j$ alternates between 2 and 1. We have $4k \gt 3k +3 -j$ when $k \gt 1$, so $\phi(4,k)$ is true for all $k \gt 1$. Similarly, for $B$ prime and $B+d$ less than the next prime after $B$ we have $\phi(B,k)$ implies $\phi(B+d,k)$, as the set of $B+d$-rough numbers is the same as the set of  $B$-rough numbers.
The combinatorial argument I give now won't take care of all the remaining cases, but it will bridge much of the gap between $k=2$ and $k \geq D/2B$.  Let $B$ and $q$ be consecutive primes with $q \lt B$. 
Let us use $P(a,x)$ to count the $a$-rough numbers in the interval $(0,x)$. We have
$$P(B,x)= P(q,x) - P(q,x/B).$$ This reflects the fact that a $B$-rough number is a $q$-rough number unless it is a multiple of $B$, or conversely $q$-rough numbers are just $B$-rough numbers with powers of $B$ times a $B$-rough number thrown in. This relation will allow us to transition between primes in proving $\phi$.
A totative counting argument gives us $\phi(5,k)$ for all $k \gt 5/2$. Suppose we have $\phi(q,k)$ for all $k \gt q/2$. Using the relation above and writing $j$ for the smallest integer above $q/2$, we have
$$P(B,jB) = P(q,jB) - P(q,j) \gt \lfloor jB/q \rfloor - 1.$$
We use the fact that $j \lt B$. Since $B$ and $q$ are consecutive odd primes, $B\geq q+2$ and so the right hand side is at least $\lfloor (q+3)/2 \rfloor - 1 = j$. This gives the implication 
$\phi(q,j)$ implies $\phi(B,j)$ when $j=\lceil q/2 \rceil$.
Now increase $j$ by one. $P(q,j)$ stays at 1 as long as $j \lt B$, while by assumption $P(q,jB)$ is still bigger than $j+1$ (because $j \gt q/2)$, so we get $\phi(B,j)$ for $B\gt j \gt q/2$, and this includes $j=(B+1)/2$ (remember $q \geq 5$). For even larger $j$, note that $P(q,j+2)-P(q,j)\leq 1$ while $P(q,B(j+2))$ has a lower bound of more than $j+2$ courtesy of the assumption $\phi(q,j+2)$. So we actually have $\phi(B,k)$ for all $k \gt q/2$.
Now we can repeat this argument to get $\phi(B,k)$ for all primes $B \gt 3$ and all $k \gt B/2$, using just combinatorial arguments.
With more care we can extend this argument to cover some $k$ below $q/2$. Let $r$ be the largest prime less than $q$ (so note that $r,q,B$ are consecutive primes) and replace $P(q,jB)$ in the equation above by $P(r,jB) - P(r,(jB)/q)$.  The point of this is that $B\geq r+6$ for $r\gt 3$, and we can for large enough $r$ start $j$ close to $r/3$, and go up to $j \lt q$. One then works with the quantity $P(r,jB)-2$ for $j\lt q$ to prove $\phi(B,k)$ for even smaller $k$.
It is not clear to me if one can use this to reach $\phi(B,2)$ for all prime $B$. However Jan-Christoph Schlage-Puchta has commented above that one can handle small $k$ for all but a small number of primes using Chebyshev-like bounds on prime numbers, so I am confident that $\phi$ holds in all the remaining cases.  The point of this post is to observe how far simple combinatorial arguments can go to establish a stronger statement.
Gerhard "Really Likes The Combinatorial Arguments" Paseman, 2019.02.27.
