69 lines
1.7 KiB
Markdown
69 lines
1.7 KiB
Markdown
---
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id: 5900f4091000cf542c50ff1b
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title: 'Problem 156: Counting Digits'
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challengeType: 5
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forumTopicId: 301787
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dashedName: problem-156-counting-digits
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---
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# --description--
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Starting from zero the natural numbers are written down in base 10 like this:
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0 1 2 3 4 5 6 7 8 9 10 11 12....
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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:
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| $n$ | $f(n, 1)$ |
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|-----|-----------|
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| 0 | 0 |
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| 1 | 1 |
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| 2 | 1 |
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| 3 | 1 |
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| 4 | 1 |
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| 5 | 1 |
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| 6 | 1 |
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| 7 | 1 |
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| 8 | 1 |
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| 9 | 1 |
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| 10 | 2 |
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| 11 | 4 |
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| 12 | 5 |
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Note that $f(n, 1)$ never equals 3.
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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.
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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 $f(n, d) = n$.
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You are given that $s(1) = 22786974071$. Find $\sum{s(d)}$ for $1 ≤ d ≤ 9$.
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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 $f(n, d) = n$.
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# --hints--
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`countingDigits()` should return `21295121502550`.
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```js
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assert.strictEqual(countingDigits(), 21295121502550);
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```
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# --seed--
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## --seed-contents--
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```js
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function countingDigits() {
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return true;
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}
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countingDigits();
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```
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# --solutions--
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```js
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// solution required
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```
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