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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.

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    $\begingroup$ $x\in\mathbb{Z}$? $\endgroup$ – barak manos Jan 26 '16 at 12:48
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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!

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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.

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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?

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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 $\lfloor 100x\rfloor=761$

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  • $\begingroup$ Unfortunately, there must be some mistake somewhere in your calculations, because $[x + 0.19] + \dots + [x + 0.91] = 564$ when $x = 7.61$, while the sum should have been $546$. The Mathematica code does the sum for you: f[x_] := Sum[Floor[x + k], {k, 0.19, 0.91, 0.01}]; f[7.61]. For the correct value $x = 7.43$ see vrugtehagel's answer. $\endgroup$ – Alex M. Jan 26 '16 at 18:11

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