Find the greatest positive integer value of $n$ such that $$\left[\frac{n}{3}\right]+\left[\frac{n}{35}\right]=\left[\frac{n}{5}\right]+\left[\frac{n}{7}\right]$$
where $[.]$ denotes the greatest integer function.
We can rewrite the equation as $$\frac{n}{3}-\left\{\frac{n}{3}\right\}+\frac{n}{35}-\left\{\frac{n}{35}\right\}=\frac{n}{5}-\left\{\frac{n}{5}\right\}+\frac{n}{7}-\left\{\frac{n}{7}\right\}$$
where $\{x\}$ denotes fractional part of $x$.
This can rephrased as $$\frac{2n}{105}=\left\{\frac{n}{3}\right\}+\left\{\frac{n}{35}\right\}-\left\{\frac{n}{5}\right\}-\left\{\frac{n}{7}\right\}$$
$$\implies \frac{2n}{105}<2$$
$$\implies n<105$$
I'm stuck after this. How to proceed from here$?$ The solution that I have to this problem uses trial and error as the method. In general how to solve equations involving greatest integer function$?$
Any help is greatly appreciated.
Best Answer
Note that as long as $n$ is not a multiple of $5$ or $7$, then decreasing $n$ by $1$ increases $\{n/35\}-\{n/5\}-\{n/7\}$ by $11/35$. Therefore, we know that whatever $n$ maximizes this, it is divisible by $5$ or $7$, and so $\{n/5\}+\{n/7\}$ is a multiple of $1/7$ or $1/5$. The first candidate is $n\equiv 30 \mod 35$, which gives $\{n/5\}+\{n/7\} = 2/7$. There are only two remainders that give a smaller value, $21$ and $15$, and both have a smaller value of $\{n/35\}-\{n/5\}-\{n/7\}$. Therefore, we have that $$ \left\{\frac{n}{35} \right\}-\left\{\frac{n}{5}\right\}-\left\{\frac{n}{7}\right\}\le \frac{30}{35}-\frac{0}{5}-\frac{2}{7} = \frac{4}{7} $$ Combining this with the bound $\{n/3\} \le 2/3$ gives $$ n\le \frac{105}{2}\left(\frac{2}{3} + \frac{4}{7}\right) = 65. $$ And since $n= 65$ satisfies the equation, it is the largest such value.