Average digit sum

Let us first find a general equation for the average of all the digit sums of all nonnegative integers less than $b^n$ written down in base $b$ .

Including zeroes as leading digits, there are $nb^n$ total digits. Since they all appear equally, each digit from $0$ to $b$ appears $\frac{nb^n}{b}$ times, which makes the sum of all the digits $\frac{nb^n(0 + 1 + 2 + ... + (b - 1))}{b}$ and the average of all the digit sums $\frac{nb^n(0 + 1 + 2 + ... + (b - 1))}{b \cdot b^n}$ or $\frac{n(0 + 1 + 2 + ... + (b - 1))}{b}$ .

The sum $0 + 1 + 2 + ... + (b - 1)$ can be expressed as $\frac{b(b - 1)}{2}$ . Substituting this in, the average of all the digit sums is now $\frac{n(\frac{b(b - 1)}{2})}{b}$ or $\frac{n(b - 1)}{2}$ .

In this problem, $b = 25$ and $n = 2018$ , so the average of all the digit sums is $\frac{2018(25 - 1)}{2} = \boxed{24216}$ .

let's make all numbers less than $25^{2018}$ 2018 (base-25) digits long (which does not affect the digit sum). Then we see that on each digit position 0~24 appear exactly the same number of times. The average of 0~24 is 12, so the average digit sum is $2018 \times 12 = 24216$ .

From $00...0 _{(25)}$ to $(24)(24)...(24) _{(25)}$ ,

We can make $\frac{ 25^{2018} -1} {2}$ pairs :
$( (a_{1} ) (a_{2} )... (a_{2018} ) , ( 24 - a_{1} ) ( 24 - a_{2} ) ...(24 - a_{2018} )$ )

and the averages of this pair are same ( $= \frac{ 24 * 2018 }{ 2}$ )

And then, only one number( = (12)(12)...(12) ) is center of them, and the sum of this digit is $\frac{ 24 * 2018 }{ 2}$ , so this doesn't affect average .

So we can know the (Ans) is $\frac{ 24 * 2018 }{ 2}$

Let's try to find a pattern by first considering numbers in base 2:

The average digit sums of all integers less than $2^n$ , denoted by $\varnothing_2(n)$ can now be calculated:

It looks like, $\varnothing_2(n) = \frac 12 n$ . But why exactly is that?

Let's try to average the digit sums in a clever way by grouping them in groups of $2 = 2^1$ and $4 = 2^2$ respectively.

It looks like the average in every group just follows the same pattern as the digit sums. And this pattern will indeed continue. This is because the binary numbers from $2^n$ to $2^{n+1}-1$ are written almost like the numbers from $0$ to $2^n$ , except for another digit 1 in the front, so their digit sum is just one more. Therefore, their average is 1 more than the average of the first half. The average of all numbers is the average of these two (since they both contain $2^n$ numbers ), so $\frac 12$ more than the previous term. We can group the terms again and again until we get to a single number that is the average of all digit sums.

The formula for the average digit sum is indeed

$\varnothing_2(n) = \frac 12 n$

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Let's try to do the same trick for base 3. This time, we can't group 4 numbers, but rather 3.

Again, the pattern repeats self-similarly, and we can write the formula

$\varnothing_3(n) = n$

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Now, it's time to generalize. What we've got so far is

$\varnothing_2(n) = \frac 12 n$

$\varnothing_3(n) = n = \frac 22 n$

The self-similar pattern will continue in all bases and the argument as explained for base 2 will also always work, so the general expreasion for $\varnothing_b(n)$ is

$\boxed{\varnothing_b(n) = \frac {b-1}2 n}$ .

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After all this, we can simply plug in $b=25, n=2018$

$\varnothing_{25}(2018) = \frac {25-1}2 \cdot 2018 = 12 \cdot 2018 = \boxed{24216}$