Floor problems

Given that

$\hspace{10mm}$ $\left\lfloor 2n \right\rfloor + \left\lfloor 3n \right\rfloor$

where n is any real integer.

We shall consider the following four cases:-

i) $\hspace{3mm} 0\leq\; n \; <\; 0.33$

$\hspace{5mm} \therefore\; 0\leq 2n <\;0.66 \hspace{4mm}$ and $\hspace{5mm} 0\leq 3n <\;0.99 \hspace{4mm}$

$\hspace{5mm} \therefore\;\left\lfloor 2n \right\rfloor = 2n \hspace{4mm}$ and $\hspace{5mm} \left\lfloor 3n \right\rfloor = 3n$

$\hspace{5mm}$ So the number is of form:-

$\hspace{7mm} \left\lfloor 2n \right\rfloor + \left\lfloor 3n \right\rfloor = 2n + 3n = 5n$

$\hspace{7mm}$ i.e. $\hspace{3mm} 5,10,\dots,1000$

$\hspace{5mm}$ We can easily find no of terms in this A.P. which are equal to $\hspace{3mm}\boxed{200}$

ii) $\hspace{3mm} 0.33<\;n < 0.5$

$\hspace{5mm} \therefore\; 0.66 < 2n <1 \hspace{4mm}$ and $\hspace{5mm} 0.99< 3n <1.5 \hspace{4mm}$

$\hspace{5mm} \therefore\;\left\lfloor 2n \right\rfloor = 2n \hspace{4mm}$ and $\hspace{5mm} \left\lfloor 3n \right\rfloor = 3n + 1$

$\hspace{5mm}$ So the number is of form:-

$\hspace{7mm} \left\lfloor 2n \right\rfloor + \left\lfloor 3n \right\rfloor = 2n + 3n + 1 = 5n + 1$

$\hspace{7mm}$ i.e. $\hspace{3mm} 1,6,\dots,996$

$\hspace{5mm}$ This is also an A.P. which is having $\hspace{3mm}\boxed{200}\hspace{3mm}$ terms.

iii) $\hspace{3mm} 0.5\leq\;n <0.66$

$\hspace{5mm} \therefore\; 1 \leq 2n <1.32 \hspace{4mm}$ and $\hspace{5mm} 1.5\leq 3n <1.98 \hspace{4mm}$

$\hspace{5mm} \therefore\;\left\lfloor 2n \right\rfloor = 2n + 1\hspace{4mm}$ and $\hspace{5mm} \left\lfloor 3n \right\rfloor = 3n + 1$

$\hspace{5mm}$ So the number is of form:-

$\hspace{7mm} \left\lfloor 2n \right\rfloor + \left\lfloor 3n \right\rfloor = 2n + 1 + 3n + 1 = 5n + 2$

$\hspace{7mm}$ i.e. $\hspace{3mm} 2,7,\dots,997$

$\hspace{5mm}$ This A.P. is having $\hspace{3mm}\boxed{200}\hspace{3mm}$ terms.

iv) $\hspace{3mm} 0.66< n <1$

$\hspace{5mm} \therefore\; 1.32 < 2n <2 \hspace{4mm}$ and $\hspace{5mm} 1.99< 3n <3 \hspace{4mm}$

$\hspace{5mm} \therefore\;\left\lfloor 2n \right\rfloor = 2n + 1\hspace{4mm}$ and $\hspace{5mm} \left\lfloor 3n \right\rfloor = 3n + 2$

$\hspace{5mm}$ So the number is of form:-

$\hspace{7mm} \left\lfloor 2n \right\rfloor + \left\lfloor 3n \right\rfloor = 2n + 1 + 3n + 2 = 5n + 3$

$\hspace{7mm}$ i.e. $\hspace{3mm} 3,8,\dots,998$

$\hspace{5mm}$ This A.P. is having $\hspace{3mm}\boxed{200}\hspace{3mm}$ terms.

$\hspace{4mm}\Rightarrow$ Answer is $\hspace{1mm}\boxed{800}$

First, I defined the function f(x)=⌊2x⌋+⌊3x⌋ for all real numbers x between 1 and 1000 inclusive. The problem can then be restated as: "Find the range R of f(x), and count the elements in R."

Then I assumed that the range of f(x) consisted of all the integers in the given interval, and tried to find a real number x to satisfy each value for f(x). Experimenting with the first few integers allowed me to deduce that f(x) = 1 when x = 1/3 (say), f(x) = 2 when x = 1/2, and f(x) = 3 when x = 2/3. All efforts to find a value for x so that f(x) = 4, however, were fruitless. I quickly convinced myself that f(x) could never equal 4 by the following (non-rigorous) lines of thought:

If x=1, f(x) = ⌊2(1)⌋+⌊3(1)⌋ = 2+3 = 5.

If x<1 by a very small amount, then ⌊2x⌋=1 by definition of the floor function, and⌊3x⌋ = 2 similarly. Hence, f(x) would equal 1+2, or 3.

I then observed that the same behavior would continue for all integral values of f(x) of the form 5n+4, where n = 0,1,2,... since f(x) is montonically increasing and the range (again, just considering integer values for f(x)) would consist of multiples of 5 if the floor brackets were absent from the 2x and 3x terms. There were 200 such integers in the interval [1,1000]; hence, there must be 1000 - 200 = 800 calculable values for the range of f(x).

Thus, the answer to be sought in the original problem is 800.

Q.E.D.

Let n=i+f, where i is the integer part & f is the fractional part

[2n]=[2i+2f]=2i+[2f]

[3n]=[3i+3f]=3i+[3f]

[2n]+[3n]= 5i+[2f]+[3f]

Case I: $0≤f<0.333$

[2f]=0 & [3f]=0

Case 2: $0.333≤f<0.5$

[2f]=0 & [3f]=1

Case 3: $0.5≤f<0.667$

[2f]=1 & [3f]=1

Case 4: $0.667≤f<1$

[2f]=1 & [3f]=2

So [2f]+[3f]=0,1,2,3

Hence we have to count numbers of form 5i, 5i+1, 5i+2, 5i+3

Or we can subtract 4, 9, 14, 19,....

Which will count to 200

Hence answer is 800

Define $f(m) = \left\lfloor \frac{m}{3} \right\rfloor + \left\lfloor \frac{m}{2} \right\rfloor$ . Obviously this has the same range as the given expression as a function of $n$ .

Clearly, $f(m) = f(\lfloor m \rfloor)$ , so we need only consider integer values of $m$ .

Observe that $f(0) = 0 \\ f(1) = 0 \\ f(2) = 1 \\ f(3) = 2 \\ f(4) = 3 \\ f(5) = 3 \\ f(m + 6) = f(m) +\ 5$ so that for integer $k$ , $\exists m\ f(m) = k$ if and only if $k \not\equiv 4\ (\mathrm{mod}\ 5)$ .

Obviously in $[1, 1000]$ , there are $\boxed{800}$ such integers.

Hi friends, after solving this question just for the fun of it I tried a different approach .

I drew the graph of y =floor(2x) + floor(3x) .

I found that for the first 63 natural numbers, there were 50 solutions , therefore for 1000, how much ??? (using Proportionality relations) 793.65 ; from the first 25 solutions I got 781.25 solutions for 1000 natural numbers.

Now for 101 natural numbers I found 80 solutions implying 792.08 solutions for 1000 natural numbers .

Similarly , for 126 natural numbers I got 100 solutions implying 793.65 solutions for 1000 natural numbers .

See, the value progresses steadily towards 800 , isn't it interesting ?