Question

How many three-digit numbers are there ?

Answer 1

Hint: There are total numbers from 1 to 100. Use the symmetricity of the numbers from 101 to 200, 201 to 300 and so on. And three digit numbers are the numbers which have only three digits in them.

Complete Step-by-Step solution:
As we know any number is said to be a three digit number if it consists of only three digit. So, as we know that the smallest three digit number is 100. And the maximum three digit number is 999. So, we need to count the number from 100 to 999. Now, we know that there are 100 numbers from 1 to 100 as we can count them like one, two, three …..hundred.
So, if we follow the same pattern then there should be exactly 100 numbers from 101 to 200 and similarly, there will be 100 numbers from 201 to 300. So, we can get the series as
Numbers from 101 to 200 $=100$
Numbers from 201 to 300 $=100$
Numbers from 301 to 400 $=100$
Numbers from 401 to 500 $=100$
Numbers from 501 to 600 $=100$
Numbers from 601 to 700 $=100$
Numbers from 701 to 800 $=100$
Numbers from 801 to 900 $=100$
Numbers from 901 to 1000 $=99+1 $( 1 for 100)
                                                     $\overline{900}$
Now, we can get a total of three digit numbers by counting the numbers from 100 to 999. There are 900 numbers from 101 to 1000 (including both), since 1000 is a 4-digit number, so we cannot count it for a three digit number. Hence, there are $900-1=899$ numbers from 101 to 999. And since 100 was not included in counting, we can add one more number $\left( 100 \right)$ to the total number of three digit numbers. Hence, there are 899 + 1 $=900$ numbers of three digit.
Hence, the answer is 900.

Note: One may use the other approach that is based on the chapter Arithmetic progression. As we know A.P.(Arithmetic progression) is a series with same successive difference and general term ( ${{n}^{th}}$ term) of A.P. is given as
${{T}_{n}}=a+\left( n-1 \right)d$
So, let the series be $\left( 100,101,102,103......999 \right)$
Put ${{T}_{n}}=999$, $a=$ first term $=100$ and $d=$ common difference $=1$. So, we get
$999=100+\left( n-1 \right)\left( 1 \right)$
$999-100+1=n$
$n=900$
Hence, 999 is the ${{900}^{th}}$ term of the sequence i.e. there are 900 three digit numbers.
One may go wrong if he/she tries to find the total numbers from 100 to 999 by just subtracting 100 and 999. He/she will get an answer as $999-100=899$, which is wrong. As we get there are 900 numbers in the situation. So, one may go wrong with this approach.
So, one may generalised the relation that there are $\left( A-B+1 \right)$ numbers between A and B (including A and B) where A > B.