Question

The total number of seven-digit numbers such that the sum of those digits is even is

• A. $9\times {10}^6$
• B. $45\times {10}^5$
• C. $81\times {10}^5$
• D. $9\times {10}^5$

Answer 1

Answer: (B)

$45\times {10}^5$

Suppose $a_1{a}_2{a}_3{a}_4{a}_5{a}_6{a}_7$ represents a seven digit number. Then $a_1$ taken the value 1, 2, ... 9 and $a_2,a_3\dots a_7$ all take values $0,1,2,\dots$9. lf we keep
$a_1,a_2,\dots .a_6$ fixed then the sum..... $a_1+a_2+\dots .+a_6$ is either even or odd. Since $a_7$ taken 10 values 0, 1, 2 ... 9.

Five of the numbers so formed will be even and 5 odd.
$\therefore$ Required numbers $=$ $\mathrm{9.10.10.10.10.10.5}$$=45x{10}^5$