Show that there is always one integer $t$ with a least prime factor $> 5$ where $x < t \le x+6$ Let $p_k$ be the $k$th prime.
Let $f_2(x) = \lfloor x\rfloor - \left\lfloor\dfrac{x}{2}\right\rfloor$ 
For $k > 1$, let:
$f_{p_k}(x) = f_{p_{k-1}}(\lfloor x\rfloor) - f_{p_{k-1}}\left(\left\lfloor\dfrac{x}{p_k}\right\rfloor\right)$ 
Let $x \ge 1$ be an integer.
Show that $f_5(x+6) - f_5(x) > 0$
I find this problem easy to solve but challenging to answer concisely.
Please let me know if I made a mistake or if my answer could be more concise.
$f_5(x+6) - f_5(x) = \sum\limits_{i|30}\left(\left\lfloor\dfrac{x+6}{i}\right\rfloor-\left\lfloor\dfrac{x}{i}\right\rfloor\right)\mu(i)$ 
where $\mu(i)$ is the möbius function.
Let $r(x,d)$ be the remainder of $x$ when divided by $d$.
Clearly:


*

*$r(x+6,2) = r(x,2)$

*$r(x+6,3) = r(x,3)$

*$r(x+6,6) = r(x,6)$
So that:
$f_5(x+6) - f_5(x) = (x+6 - x) - \left(\dfrac{x+6 - x}{2}\right) - \left(\dfrac{x+6 - x}{3}\right)- \left(\dfrac{x+6 - x - r(x+6,5) + r(x,5)}{5}\right)+ \left(\dfrac{x+6 - x}{6}\right)+ \left(\dfrac{x+6 - x - r(x+6,10) + r(x,10)}{10}\right) + \left(\dfrac{x+6 - x - r(x+6,15) + r(x,15)}{15}\right) - \left(\dfrac{x+6 - x - r(x+6,30) + r(x,30)}{30}\right)$
$= 2 - \left(\dfrac{6 - r(x+6,5) + r(x,5)}{5}\right)+ \left(\dfrac{6 - r(x+6,10) + r(x,10)}{10}\right) + \left(\dfrac{6 - r(x+6,15) + r(x,15)}{15}\right) - \left(\dfrac{6 - r(x+6,30) + r(x,30)}{30}\right)$ 
It follows that if $x \not\equiv 4 \pmod 5$:


*

*$\dfrac{6 - r(x+6,5) + r(x,5)}{5} = 1$
If $x \equiv 4 \pmod 5$:


*

*$\dfrac{6 - r(x+6,5) + r(x,5)}{5} - \dfrac{6 - r(x+6,10) + r(x,10)}{10} = 1$
if $r(x,30) < 24$:


*

*$\dfrac{6 - r(x+6,30) + r(x,30)}{30} = 0$
if $r(x,30) \ge 24$:


*

*$\dfrac{6 - r(x+6,15) + r(x,15)}{15} - \dfrac{6 - r(x+6,30) + r(x,30)}{30} = 0$
So that:
$f_5(x+6) - f_5(x) \ge 2 - 1 + 0 = 1$

Edit 1:
I forgot to mention that $x$ is an integer.  I have updated the question.

Edit 2:
Updated the recurrence relation in order to make it clearer.  One commenter said that the relation wasn't clear.

Edit 3:
Apologies I have no idea why I wrote "kth integer" instead of $k$th prime.  Thanks for your patience.  Now fixed.

You can actually solve this using (almost)  just very elementary set theory.
Let $X_5$ be the set of integers in $\{x+1,x+2,\ldots, x+6 \}$ that are a multiple of 5. Let $X_3$ be the set of integers in $\{x+1,x_2,\ldots, x+6 \}$ that are a multiple of 3, and let $X_2$ be the set of integers $\{x+1,x+2,x+3,\ldots, x+6 \}$ that are a multiple of 2.
Then $|X_5|$ is either 1 or 2, and if $|X_5|$ is 2 then $X|_5 \cap X_2|$ is at least 1.
Then $|X_3|$ is 2 and $|X_2|$ is 3. And $|X_2 \cap X_3|$ is 1
Case 1: $|X_5| = 2$. Then if $|X_3 \cap X_5 \cap X_2|$ is 1,
then $|X_2 \cup X_3 \cup X_5| = |X_2| + |X_3| + |X_5| - |X_2 \cap X_3| - |X_2 \cap X_5| -|X_3 \cap X_5| + |X_2+X_3+x_5|$ $=3+2+2 -1 -1 -1 + 1 < 6$ so there is an element in $\{x+1,\ldots, x+6 \}$ that is not a multiple of 2, 3, or 5.
If $|X_3 \cap X_5 \cap X_2|$ is 0,
then $|X_2 \cup X_3 \cup X_5| = |X_2| + |X_3| + |X_5| - |X_2 \cap X_3| - |X_2 \cap X_5| + |X_2+X_3+x_5|$ $\geq$ $3+2+2 -1 -1  < 6$ so there is an element in 
$\{x+1,\ldots, x+6 \}$ that is not a multiple of 2, 3, or 5.
Can you work through Case 2: $|X_5| =1$?
