How many different six-digit numbers are there the sum of whose digits is odd?

Hint: In this particular type of question use the concept that in a particular continuous set of numbers there are half of the numbers whose sum of the digits are odd and half of the numbers whose sum of the digits are even so use these concepts to reach the solution of the question.

Complete step-by-step answer:
As we all know that after every odd number there is an ever number or we can say that after every even number there is an odd number.
For example – that 1 is an odd number and after 1 the next number is 2 which is even.
So we can say that in a particular continuous set of numbers starting from odd and ending on even or vice-versa the number of odd numbers is equal to the number of even numbers.
So if a set contains 1000 numbers, starting from odd and ending on even or vice-versa there are 500 even numbers and 500 odd numbers.
Same case for the sum of the digits half of the numbers have sum of the digits odd and half of the numbers have sum of the digits even.
Now we have to find out different six-digit numbers where the sum of digits is odd.
So the first six digit number is 100000 and the last six digit number is 999999.
So the series becomes, 100000, 100001, 100002, ........, 999999.
As we see that the above series follow the property of A.P, with first term (a) = 100000, common difference (d) = (100001 -100000) = (100002 – 100001) = 1, last term, ${a_n} = 999999$
So the ${n^{th}}$ term of the A.P is given by the formula,
$ \Rightarrow {a_n} = a + \left( {n - 1} \right)d$, where n is the number of terms in the series.
$ \Rightarrow 999999 = 100000 + \left( {n - 1} \right)1$
$ \Rightarrow \left( {n - 1} \right) = 999999 - 100000 = 899999$
$ \Rightarrow n = 899999 + 1 = 900000$
So there are a total 900000 six digits numbers.
So the number of different six-digit numbers whose digits sum is odd =$\dfrac{{900000}}{2} = 450000$.
So this is the required answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the formula of the ${n^{th}}$ term of an A.P which is stated above then simply substitute the values in the formula and calculate the number of six digits terms as above then divide by 2 is the numbers whose sum of digits are odd.