Finding the average of a long series of integers For each integer from 0 to 999, Max wrote the sum of its digits down.What is the average of the numbers that Max wrote down?
I started to work this out by taking in mind that for the two-digit numbers there are two sums the same for all numbers but ten,but how do I proceed?
 A: The average of the sum of the three digits is three times the average of the third (or second, or first) digit.
The average third digit is the average of $0,1,2,3,4,5,6,7,8,9.$ Pairing these off from both ends of the sequence we see that there are 5 pairs having average $4.5$ each. Thus the average third digit is 4.5, and the average of the numbers that Max wrote down is 3 times that number.
A: For clarity let us say that all 1000 numbers have been written with 3 digits, padding with zeroes on the left if necessary.
Computing the average is no more difficult that computing the sum of the numbers that Max wrote down.
This you can do by computing the sum of the first digits of all 1000 numbers, then the sum of the second digits and the third digits (which is twice the same as the sum of the first digits).
Finally, note that the sum of the first 9 positive integers (also the number of clinking sounds at a toast with 10 participants) is equal to the number of choices ('combinations') of 2 elements out of 10.
$$C^{10}_2=\frac{10!}{2!(10-2)!}=\frac{9.10}2.$$
A: We can write each number using exactly three digits, if we allow leading zeroes (which don't change the sum of a number's digits). If we do this, then each digit occurs the same number of times in the total sum, so the answer is just $3$ times the average value of a decimal digit, which is $3 \times 4\frac12 = 13\frac12$.
