Question

How many non perfect square numbers are there between \[{17^2}\] and ${18^2}$.

Answer 1

Hint:There are $2n$ natural numbers lying between two consecutive perfect square numbers ${n^2}$ and ${\left( {n + 1} \right)^2}$. We have to count only natural numbers which are not perfect squares.

Complete step-by-step answer:
Given numbers are \[{17^2}\] and ${18^2}$
So we can write ${18^2}$ as ${\left( {17 + 1} \right)^2}$
We have to find non perfect square number are there between \[{17^2}\] and ${\left( {17 + 1} \right)^2}$
By comparing with hint we say that ${n^2} = {17^2}$ and ${\left( {n + 1} \right)^2} = {\left( {17 + 1} \right)^2}$
Now from this we find that the value of $n = 17$
So now
 There are $2n$ natural numbers lying between two consecutive perfect square numbers ${n^2}$ and ${\left( {n + 1} \right)^2}$.
From this our required answer is $2n$
We know $n = 17$
So $2 \times 17$
$ \Rightarrow 34$
So there is $34$ non perfect square number between \[{17^2}\] and ${18^2}$.

Note:Alternative method:
For any two given natural numbers $n$ and $m$ where $n > m$. There are $\left( {n - m - 1} \right)$ natural numbers between $n$ and $m$. So for this method we have to find the square of the given number.
Given square number is \[{17^2}\] and ${18^2}$
Now solve square
${17^2} = 17 \times 17$ $ \Rightarrow 289$
And similarly
${18^2} = 18 \times 18$ $ \Rightarrow 324$
Now as given in hint two given natural numbers $n$ and $m$ where $n > m$
So $m = 289$ and $n = 324$
Now There are $\left( {n - m - 1} \right)$ natural numbers between $n$ and $m$
So $(324 - 289 - 1)$
$ \Rightarrow 34$
So there are 34 non perfect squares between \[{17^2}\] and ${18^2}$.