Question

How many integer numbers are lying between the squares of the numbers 12 and 13?

Answer 1

Hint: Integers are the numbers which do not have any decimal part after the decimal point.
There are exactly n integers from 1 to n, including both 1 and n, where n is positive.
The value of $ {{12}^{2}}=144 $ and $ {{13}^{2}}=169 $ .
Notice that the integers "between" a and b, do not include the integers a and b themselves.

Complete step-by-step answer:
The squares of 12 and 13 are $ {{12}^{2}}=144 $ and $ {{13}^{2}}=169 $ respectively.
Since 169 and 144 do not have to be counted, we can count the number of integers from 1 to 168 and subtract the number of integers from 1 to 144 from them, to give the number of integers between 144 and 169.
There are 168 integers from 1 to 168.
There are 144 integers from 1 to 144.
So, the number of integers from 145 to 168 (including both 145 and 168) will be: $ 168-144=24 $ .
So, the correct answer is “24”.

Note: An integer (from the Latin "integer" meaning "whole") is defined as a number that can be written without a fractional component. Integers can be positive, negative or neutral (0).
e.g. $ -1,\ 24,\ 0 $ etc. are integers, while $ 3.75,\ 2\dfrac{1}{2},\ \sqrt{2} $ are not.
There are $ b-a+1 $ integers from integer a to integer b, including both of them.
The numbers which are not integers are either rational (can be expressed as a ratio of two integers.) or irrational (cannot be expressed as a ratio of two integers. e.g. $ \sqrt 3 $ ).
The combined set of Rational and Irrational numbers is known as Real numbers. We can represent any Real Number on the number line.
We can also count the number of integers between them as $ {{(n+1)}^{2}}-{{n}^{2}}-2 $ .
  $ {{(a\pm b)}^{2}}={{a}^{2}}\pm 2ab+{{b}^{2}} $
  $ {{(a\pm b)}^{3}}={{a}^{3}}\pm 3ab(a\pm b)\pm {{b}^{3}} $