MHB Number of Terms in Sequence $\left\lfloor \dfrac{x^2}{1998} \right\rfloor$

[anemone]

Find the number of different terms of the finite sequence $\left\lfloor \dfrac{x^2}{1998} \right\rfloor$ where $x=1,\,2,\,3,\,\cdots,\,1997$.

 

[chisigma]

Find the number of different terms of the finite sequence $\left\lfloor \dfrac{x^2}{1998} \right\rfloor$ where $x=1,\,2,\,3,\,\cdots,\,1997$.

[sp]According to the identity $\displaystyle \frac{x^{2}}{x+1} = \frac{x^{2} - 1 + 1}{x+ 1} = x-1 + \frac{1}{x+1}$ the number of different terms are $\displaystyle \left\lfloor \dfrac{1997^2}{1998} \right\rfloor = 1997-1 = 1996$... [/sp]Kind regards $\chi$ $\sigma$

 

[anemone]

[sp]According to the identity $\displaystyle \frac{x^{2}}{x+1} = \frac{x^{2} - 1 + 1}{x+ 1} = x-1 + \frac{1}{x+1}$ the number of different terms are $\displaystyle \left\lfloor \dfrac{1997^2}{1998} \right\rfloor = 1997-1 = 1996$... [/sp]Kind regards $\chi$ $\sigma$

Hi chisigma,:) thanks for participating but I am sorry, your answer isn't correct...:(

 

[magneto1]

Spoiler

We have:
\[
\frac{(k+1)^2}{1998} - \frac{k^2}{1998} = \frac{2k+1}{1998} =: d(k).
\]

$d(k) < 1$ when $1 \leq k \leq 998$. So we have:
\[
\left\lfloor \frac{1^2}{1998} \right\rfloor = 0; \left\lfloor \frac{998^2}{1998} \right\rfloor = 498.
\]
So there are $499$ different values in this range of $k$ (going from $0$ to $498$).

$d(k) > 1$ when $999 \leq k \leq 1997$. For these values of $k$, we know the floor function takes on
all distinct values, so there are $1997 - 999 + 1 = 999$ distinct values for this range of $k$.

The total is therefore $999 + 499 = 1498$.

 

[kaliprasad]

Find the number of different terms of the finite sequence $\left\lfloor \dfrac{x^2}{1998} \right\rfloor$ where $x=1,\,2,\,3,\,\cdots,\,1997$.

Spoiler

x = 1 gives the value 0
x = 1997 gives the value 1996 as 1997 * 1997 - 1 = 1996 * 10998

so 1997 values are taken all values from 0 to 1996

 

[anemone]

Spoiler

We have:
\[
\frac{(k+1)^2}{1998} - \frac{k^2}{1998} = \frac{2k+1}{1998} =: d(k).
\]

$d(k) < 1$ when $1 \leq k \leq 998$. So we have:
\[
\left\lfloor \frac{1^2}{1998} \right\rfloor = 0; \left\lfloor \frac{998^2}{1998} \right\rfloor = 498.
\]
So there are $499$ different values in this range of $k$ (going from $0$ to $498$).

$d(k) > 1$ when $999 \leq k \leq 1997$. For these values of $k$, we know the floor function takes on
all distinct values, so there are $1997 - 999 + 1 = 999$ distinct values for this range of $k$.

The total is therefore $999 + 499 = 1498$.

Well done, magneto and thanks for participating! But having said this, I wanted to give some credit to chisigma as well!:)

 

[anemone]

Spoiler

x = 1 gives the value 0
x = 1997 gives the value 1996 as 1997 * 1997 - 1 = 1996 * 10998

so 1997 values are taken all values from 0 to 1996

Hi kaliprasad, somehow I missed to read your post. Sorry about that!

Hmm, what do you think of the following

$\left\lfloor \dfrac{1^2}{1998} \right\rfloor =0,\,\left\lfloor \dfrac{2^2}{1998} \right\rfloor =0,\cdots,\,\left\lfloor \dfrac{44^2}{1998} \right\rfloor =0$

That surely gives us less than 1997 distinct terms in that sequence when $x$ ranges from 1 to 1997, doesn't it?

 

[kaliprasad]

Hi kaliprasad, somehow I missed to read your post. Sorry about that!

Hmm, what do you think of the following

$\left\lfloor \dfrac{1^2}{1998} \right\rfloor =0,\,\left\lfloor \dfrac{2^2}{1998} \right\rfloor =0,\cdots,\,\left\lfloor \dfrac{44^2}{1998} \right\rfloor =0$

That surely gives us less than 1997 distinct terms in that sequence when $x$ ranges from 1 to 1997, doesn't it?

You are right. Wrongly I presumed that all values are taken. I am wrong fully

 

[anemone]

You are right. Wrongly I presumed that all values are taken. I am wrong fully

Don't worry about it, kali!:)