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How many non-perfect square numbers are there between ${15^2}$ and ${16^2}$?

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Answer
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Hint: First we have to define what the terms we need to solve the problem are. First of all, we just need to know such things about the perfect numbers, which are the numbers that obtain by multiplying any whole number (zero to infinity) twice, or the square of the given number yields a whole number like $\sqrt 9 = 3$ or $9 = {3^2}$.

Complete step-by-step solution:
Since the given question is about non-perfect square numbers, there are $2n$ natural numbers that are lying between two consecutive perfect square (as per hint) numbers ${n^2}$ and ${(n + 1)^2}$. so, we will count only natural numbers except for the not perfect squares.
From the given question we have, ${15^2}$ and ${16^2}$, since we can able to write ${16^2}$ as ${(15 + 1)^2}$
Now we will need to find the non-perfect numbers between ${15^2}$ and ${(15 + 1)^2}$, thus $n = 15$ and there are $2n$ natural numbers. now substitute that n in the perfect square equation which is $2(15)$, the natural number for the non-perfect square. Hence further solving we get count $30$ numbers of non-perfect square numbers that are between ${15^2}$ and ${16^2}$
We can also able to solve this problem by comparing the two square numbers and subtracted by one
Which is ${15^2}$,${16^2}$ can be written as ${15^2} = 225$ and ${16^2} = 256$
The formula for this method is $n - m - 1$ (m is the smallest among the both and n is the greater one) and n is always less than m. Hence, we get $n - m - 1 = 256 - 225 - 1$ solving this we get $256 - 225 - 1 = 30$
Hence there are $30$ non-perfect square numbers between ${15^2}$ and ${16^2}$

Note: Since the equation is about non-perfect square numbers between ${15^2}$ and ${16^2}$, so that only the substrate one occurs $n - m - 1$ if not then like non-perfect square numbers from ${15^2}$ and ${16^2}$, we can simply find by using the $n - m$. So, between is the key for the different formulas.