Problem :
If $[x+0.19] +[x+0.20] +[x+0.21] +\cdots [x+0.91] =546$ find the value of $[100x]$ where [.] represents the greatest integer function less than equal to x.
My approach :
$x +1.19 = x + \frac{19}{100} = \frac{100x+19}{100}$
Similarly other terms
Not getting any clue further please suggest will be of great help.
Best Answer
First note that the number of terms is $73$. Also, if we look at $[x+a]$ where $a$ ranges from $0.19$ to $0.91$, then we see it can only reach two values; those are $n=[x+0.19]$ and possibly, but not necessarily, $n+1$. We know though that $$73n\leq [x+0.19]+[x+0.20]+\cdots+[x+0.91]<73(n+1)$$ and so we deduce $n=7$. We can also find exactly where the value of $[x+a]$ changes from $n$ to $n+1$ - since, $546-73\cdot 7=35$, so there are $35$ terms $[x+a]=n+1$, so the last $a$ such that $[x+a]=n$ is $a=0.56$. So, $[x+0.56]=7$ but $[x+0.57]=8$. This means that $7.43\leq x<7.44$. So, $743\leq 100x<744$, so $[100x]=743$.
Hope this helped!