Let m be a natural number such that 20000

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(D) Consider the range of numbers from $20001$ to $59999$. The total number of integers in this range is $59999 - 20001 + 1 = 39999$. We can group the

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$19999$

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(D) Consider the range of numbers from $20001$ to $59999$. The total number of integers in this range is $59999 - 20001 + 1 = 39999$. We can group these numbers into sets of $10$ consecutive integers of the form $(a_1 a_2 a_3 a_4 0, a_1 a_2 a_3 a_4 1, \dots, a_1 a_2 a_3 a_4 9)$. In any such set of $10$ consecutive integers,the sum of the digits of $9$ numbers will have the same parity as the sum of the digits of the $10^{th}$ number,except for the last digit which cycles through $0$ to $9$. Specifically,in any set of $10$ consecutive integers,exactly $5$ will have an even sum of digits and $5$ will have an odd sum of digits. Since we have $39999$ numbers,we can form $3999$ full groups of $10$ (total $39990$ numbers) and there are $9$ remaining numbers ($59991$ to $59999$). For the $39990$ numbers,exactly half have an even sum of digits: $39990 / 2 = 19995$. For the remaining $9$ numbers ($59991$ to $59999$),the sum of the first four digits $(5+9+9+9 = 32)$ is even. Thus,the sum of all digits is even if the last digit is even $(0, 2, 4, 6, 8)$. In the range $59991$ to $59999$,the last digits are $1, 2, 3, 4, 5, 6, 7, 8, 9$. The even last digits are $2, 4, 6, 8$ (total $4$ numbers). Total count $= 19995 + 4 = 19999$.

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

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