1.7 KiB
id | title | challengeType | forumTopicId | dashedName |
---|---|---|---|---|
5900f4091000cf542c50ff1b | Problem 156: Counting Digits | 5 | 301787 | problem-156-counting-digits |
--description--
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:
n |
f(n, 1) |
---|---|
0 | 0 |
1 | 1 |
2 | 1 |
3 | 1 |
4 | 1 |
5 | 1 |
6 | 1 |
7 | 1 |
8 | 1 |
9 | 1 |
10 | 2 |
11 | 4 |
12 | 5 |
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 \sum{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
.
--hints--
countingDigits()
should return 21295121502550
.
assert.strictEqual(countingDigits(), 21295121502550);
--seed--
--seed-contents--
function countingDigits() {
return true;
}
countingDigits();
--solutions--
// solution required