Question

How many numbers are there containing 2 digits?

Answer 1

Hint:
Here, we will use the concept of arithmetic progression to solve the question. The number of 2 digit numbers can be written in the form of an A.P. We will use the formula for $$n^{\mathrm{t}\mathrm{h}}$$ term of an A.P. and simplify the equation to get the number of terms in the A.P. and hence, the number of 2 digit numbers.
Formula Used: The $$n^{\mathrm{t}\mathrm{h}}$$ term of an A.P. is given by the formula $$a_n=a+\left(n-1\right)d$$, where $$a$$ is the first term of the A.P., $$d$$ is the common difference, and $$n$$ is the number of terms in the A.P.

Complete step by step solution:
An arithmetic progression is a sequence where the difference of every two consecutive numbers is a fixed constant.
We will use the formula for $$n^{\mathrm{t}\mathrm{h}}$$ term of an arithmetic progression to find the number of 2 digits numbers.
Let the number of 2 digit numbers be $$n$$.
We know that the first 2 digit number is 10 and the last 2 digit number is 99.
The second 2 digit number is 11.
The third 2 digit number is 12.
Therefore, we get the sequence of 2 digit numbers as
10, 11, 12, ……, 99
We can observe that the sequence forms an arithmetic progression with first term 10 and the last term 99.
First, we will find the common difference of the A.P.
The difference between any two consecutive terms of an A.P. is said to be its common difference.
Therefore, we get
Common difference, $$d=11-10=1$$
Now, the $$n^{\mathrm{t}\mathrm{h}}$$ term of an A.P. is given by the formula $$a_n=a+\left(n-1\right)d$$, where $$a$$ is the first term of the A.P., $$d$$ is the common difference, and $$n$$ is the number of terms in the A.P.
Substituting $$a=10$$, $$d=1$$, and $$a_n=99$$ in the formula, we get
$$99=10+\left(n-1\right)1$$
We will simplify this equation to get the number of terms in the A.P.
Subtracting 10 from both sides of the equation, we get
$$\begin{array}{l}\Rightarrow 99-10=10+\left(n-1\right)1-10\\ \Rightarrow 89=n-1\end{array}$$
Adding 1 on both sides of the equation, we get
$$\begin{array}{l}\Rightarrow 89+1=n-1+1\\ \Rightarrow 90=n\end{array}$$
Thus, the number of terms in the A.P. 10, 11, 12, ……, 99 is 90.

$$\therefore$$ There are 90 numbers containing 2 digits.

Note:
We can also solve the question using an alternate method.
The last 2 digit number is 99.
We know that there are 99 numbers between 1 and 99, both included.
The numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 are 1 digit numbers.
Therefore, the number of 2 digit numbers is the difference between the numbers between 1 and 99, and the number of 1 digit numbers.
Thus, we get
Total number of 2 digit numbers $$=99-9=90$$
Therefore, there are 90 numbers containing 2 digits.