Question

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

• A. 19909
• B. 19989
• C. 18999
• D. 19999

Answer 1

Answer: (D)

Let us consider $10$ successive five digit numbers
$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 \ldots $
$a_{1} a_{2} a_{3} a_{4} 9$
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.
So, value of $K$ is $=4 \times 10^{3} \times 5-1$
$[\because 20,000$ will not include ]
$=19999$