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.
On
There are 91-18=73 terms. And 73*7=511 < 546 < 73*8. So maybe x should be between 7 and 8.
How do you manage the number of 7s and 8s?
On
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!
On
There's a way to solve this ( for positive x ) by first observing that your sum is the difference of two sums of the same form :
$$S_n=\sum_{k=0}^{n}\lfloor x+ak\rfloor$$
Let's start by defining $x_0=\lfloor x\rfloor$ and $f!=x-x_0$ so we have the integer and fractional part of $x$ handy.
Now we can see that :
$$|x+ak|=x_0+\lfloor f+ak \rfloor$$
Let's assume $an < 1$ for convenience ( as it suffices in this problem ).
There is some $k_0$ where $f+ak >= 1$ that contributes $+1$ to the sum and that other values contribute nothing. This $k_0$ is given by :
$$k_0=\left\lceil \frac{1-f}{a} \right\rceil$$
And we can write the sum $S_n$ easily as :
$$S_n = nx_0 + ( n+1-k_0)$$
when $n >= k_0$ and
$$S_n = nx_0$$
when $n < k_0$
And now we have a way to solve the problem at hand.
We can see that, for $a=0.01$ either $S_{91}-S_{18}=(91-18)(x_0+1)=546$ which is not the case as $73$ does is not a factor of $546$ or :
$$S_{91}-S_{18}=x_0(91-18)+(91-18)-k_0+1=546$$
We get :
$$545=73(x_0+1)-k_0$$
Taking the modulus 73 we get :
$$k_0\,mod\,73 =73-( 545\, mod\, 73 )=39$$
And as this is the only possible value in the range $18$ to $91$ we can say $k_0=39$.
Using this we get for $f=1-ak_0=1-(0.01)(39)=0.61$ and using this in our formula for the sum we can get :
$$546=73(x_0+1)-k_0+1=73(x_0+1)-38$$
So finally :
$$x_0=\frac{546+38}{73}=8$$
An so $x=(8-1)+0.61=7.61$ and the final result for $\lfloor 100x\rfloor=761$
For a clue note that the original equation is the sum of 73 terms. The first and last term differ by at most $1$, so the sum is the total of a number of terms at the lower value and a number of items at the higher value. You should be able to work out how many of each, and this will tell you where the value steps up by $1$. This will in turn give you information to bound $x$ sufficiently to answer the question.