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.

  • 2
    $\begingroup$ $x\in\mathbb{Z}$? $\endgroup$ – barak manos Jan 26 '16 at 12:48

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!


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.


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?


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 :


We get :


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 :


So finally :


An so $x=(8-1)+0.61=7.61$ and the final result for $\lfloor 100x\rfloor=761$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.