How many different $7$ digit numbers are there sum of whose digits is even? The answer will be expressed as $k \times 10^{5}$ . Write the value of $k$

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Solution

Let us consider $10$ successive seven digit numbers
$a_{1} a_{2} a_{3} a_{4} a_{5} a_{6} 0, a_{1} a_{2} a_{3} a_{4} a_{5} a_{6} 1, a_{1} a_{2} a_{3} a_{4} a_{5} a_{6} 2, . . . . . . . . . . . . . . . . a_{1} a_{2} a_{3} a_{4} a_{5} a_{6} 9,$
where $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$ are some digits. We see that half of these $10$ numbers i.e. $5$ numbers have an even sum of digits.
The first digit $a_{1}$ can assume $9$ different values of which the sum of the digits $a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$ can assume $10$ different values.
The last digit $a_{7}$ can assume only $5$ different values of which the sum of all digits is even.
$∴$ There are $9 \times 10^{5} \times 5$
$= 45 \times 10^{5}$ seven digit numbers the sum of whose digits is even.

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