How many $10$-digit numbers are there, such that the sum of the digits is divisible by $2$?

Complete question is;

How many 10 - digit numbers are there, such that the sum of the digits is divisible by 2?

There are 4500000000 ten digit numbers whose sum is divisible by 2.

Step-by-step explanation:

Since we are dealing with 10 digit numbers, the first 10 digit number whose sum is divisible by 2 is; 1000000001.

While the last 10 digit number whose sum is divisible by 2 is 9999999999.

Now,to find how many 10 digit numbers whose sum are divisible by 2, we will simply divide the difference of both numbers by 2 and then add 1 to the answer. This is because subtracting the minimum from the maximum gives us the number of 10 digit numbers we have. Then dividing by 2 gives the number of ten digit numbers with sum divisible by 2 in between. Then adding 1 to the result to cover the number not included gives;

((9999999999 - 1000000001)/2) + 1 = 4500000000

huh? im pretty sure it is in 199

step-by-step explanation:

step-by-step explanation: