Problem - 1278
Prove: there exists a rational number $\frac{c}{d}$, where $d<1000$, such that $$\Big[k\cdot\frac{c}{d}\Big]=\Big[k\cdot\frac{73}{100}\Big]$$ holds for every positive integer $k$ that is less than 1000. Here $\Big[x\Big]$ denotes the largest integer that is not exceeding $x$.
Note that 73 and 100 are relatively prime, therefore there exist integers $c$ and $d$ such that $$73d - 100c=1$$
We now prove such $c$ and $d$ satisfy the given requirement.
Let $\Big[k\cdot\frac{c}{d}\Big] = n$. Because $k<1000$, we have $$\frac{73}{100}k-\frac{c}{d}k=\frac{k(73d-100c)}{100d}=\frac{k}{100d}$$
This implies: $0< \frac{73}{100}k-\frac{c}{d}k < \frac{1}{d}$
Because $\Big[\frac{kc}{d}\Big]=n$, we have $\frac{kc}{d} < n+1 = \frac{n+1}{d}{d}$. Therefore $$\frac{73k}{100} < \frac{kc+1}{d} \le \frac{(n+1)d}{d}=n+1$$. This means: $$\Big[\frac{73k}{100}\Big]=n=\Big[\frac{kc}{d}\Big]$$