Question

The number of different six digit numbers, the sum of whose digits is odd is

• A. ${10}^6$
• B. ${5.10}^5$
• C. ${9.10}^5$
• D. ${45.10}^5$

Answer 1

The correct option is D ${45.10}^5$

The number of different six digit numbers are $9\times {10}^5$ in which half of them will have sum of digits even and half of them odd and hence required number is $\frac{9\times {10}^5}{2}=45\times {10}^4$