# Problem 156: Counting Digits
Starting from zero the natural numbers are written down in base 10 like
this: 0 1 2 3 4 5 6 7 8 9 10 11 12.... Consider the digit d=1. After we
write down each number n, we will update the number of ones that have
occurred and call this number f(n,1). The first values for f(n,1), then,
are as follows: nf(n,1) 00 11 21 31 41 51 61 71 81 91 102 114 125 Note
that f(n,1) never equals 3. So the first two solutions of the equation
f(n,1)=n are n=0 and n=1. The next solution is n=199981. In the same
manner the function f(n,d) gives the total number of digits d that have
been written down after the number n has been written. In fact, for
every digit d ≠ 0, 0 is the first solution of the equation f(n,d)=n. Let
s(d) be the sum of all the solutions for which f(n,d)=n. You are given
that s(1)=22786974071. Find ∑ s(d) for 1 ≤ d ≤ 9. Note: if, for some n,
f(n,d)=n for more than one value of d this value of n is counted again
for every value of d for which f(n,d)=n.