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For school I have to solve some problems involving floor-functions, and I have no clue how to solve this one: for a given $n$, calculate the number of $k$'s, $k\lt n$ such that the number of multiples of 7 between and including $n$ and $n-k$ is bigger then the number of multiples of 7 lower than or equal to k.

I've managed to make it into a formula: $n$ and $k$ are right iff $$ \left \lfloor \frac{n}{7} \right \rfloor-\left\lfloor \frac{n-k}{7}\right \rfloor >\left \lfloor \frac{k}{7} \right \rfloor$$

For $n=17$, these $k$ work (I think): $4,5,6,11,12,13$

Could anyone give me a hint (not the full solution) on how to calculate the number of solutions for a certain n?

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1 Answer 1

up vote 1 down vote accepted

Hint: If $0 \leqslant k \leqslant n-7$, then

$$\left\lfloor \frac{n-(k+7)}{7}\right\rfloor + \left\lfloor \frac{k+7}{7}\right\rfloor = \left\lfloor \frac{n-k}{7}-1\right\rfloor + \left\lfloor \frac{k}{7}+1\right\rfloor = \left\lfloor \frac{n-k}{7}\right\rfloor + \left\lfloor \frac{k}{7}\right\rfloor.$$

Also, $\left\lfloor \frac{k}{7}\right\rfloor = 0$ for $0 \leqslant k < 7$.

For $0 \leqslant k < 7$, we have $\left\lfloor \frac{k}{7}\right\rfloor = 0$, and $\left\lfloor \frac{n}{7}\right\rfloor > \left\lfloor \frac{n-k}{7}\right\rfloor$ if and only if $k > r$, where $r$ is the least non-negative remainder of $n$ modulo $7$. So $\left\lfloor \frac{n}{7}\right\rfloor - \left\lfloor \frac{n-k}{7}\right\rfloor > \left\lfloor \frac{k}{7}\right\rfloor$ if and only if $k = q\cdot 7 + s$, with $r < s < 7$ and $0 \leqslant q < \left\lfloor \frac{n}{7}\right\rfloor$. The number of $k$s is therefore $\left\lfloor\frac{n}{7}\right\rfloor \cdot (6-r)$.

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