Let $m$ be a natural number such that $20000 < m < 60000$ and let $k$ be the sum of all the digits in $m$.…
Let $m$ be a natural number such that $20000 < m < 60000$ and let $k$ be the sum of all the digits in $m$. Then the number of numbers $m$ for which $k$ is even, is
19909
19989
18999
19999
Solution
Let us consider 10 successive five digit numbers
$
\begin{aligned}
& a_1 a_2 a_3 a_4 0 \\
& a_1 a_2 a_3 a_4 1 \\
& a_1 a_2 a_3 a_4 2 \\
& \ldots \ldots \ldots \ldots \ldots . . \\
& a_1 a_2 a_3 a_4 9
\end{aligned}
$
Where, $a_1, a_2, a_3, a_4$ are some digits. We see that half of these 10 numbers i.e. 5 have an even sum of digits.
The first digit $a_1$ can takes $2,3,4,5$ and each of the digits $a_2, a_3, a_4$ can takes 10 different values the units place digit can assume only 5 different values of which the sum of all digits is even.
$\begin{aligned} \text { So, value of } K \text { is } & =4 \times 10^3 \times 5-1 \\ & {[\because 20,000 \text { will not include }] } \\ & =19999 .\end{aligned}$