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​

Answer 1

Answer:

The no .of digits lying between 20,000 and 60,000 is 60,000−20,000=40,000

Since The sum of digits are odd and even in a equal number

So the number of natural numbers having their sum of digits even is 40,000/2=20,000

How satisfied are you with the answer?