Question

The number of $7$ digit numbers the sum of whose digits is even is

• A. $35\times {10}^5$
• B. $45\times {10}^5$
• C. $50\times {10}^5$
• D. None of these

Answer 1

Answer: (B)

$45\times {10}^5$

A number of the 7 digit will be of the form $a{x}_1{x}_2{x}_3{x}_4{x}_{5}0,a{x}_1{x}_2{x}_3{x}_4{x}_{5}1,a{x}_1{x}_2{x}_3{x}_4{x}_{5}2,a{x}_1{x}_2{x}_3{x}_4{x}_{5}3,...a{x}_1{x}_2{x}_3{x}_4{x}_{5}9$$

Where $a\in \{1,2,3,\mathrm{4...8},9\}$ and $x_1{,x}_2,x_3{,x}_4{,x}_5\in \{0,1,2,3,\mathrm{4...8},9\}$

Since the sum of digits is even, thus if $a+x_1{+x}_2{+x}_{3+}{x}_4+x_5$ is an even number, then the digit at unit's place must be 0,2,4,6,8. and if $a+x_1{+x}_2{+x}_{3+}{x}_4+x_5$ is an odd number, then the number at unit's place must be 1,3,5,7,9.

So required numbers$=9\times 10\times 10\times 10\times 10\times 10\times 5=45\times {10}^5$