32dbe23f5e
* fix: clean-up Project Euler 301-320 * fix: corrections from review Co-authored-by: Tom <20648924+moT01@users.noreply.github.com> Co-authored-by: Tom <20648924+moT01@users.noreply.github.com>
47 lines
982 B
Markdown
47 lines
982 B
Markdown
---
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id: 5900f49d1000cf542c50ffb0
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title: 'Problem 305: Reflexive Position'
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challengeType: 5
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forumTopicId: 301959
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dashedName: problem-305-reflexive-position
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---
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# --description--
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Let's call $S$ the (infinite) string that is made by concatenating the consecutive positive integers (starting from 1) written down in base 10.
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Thus, $S = 1234567891011121314151617181920212223242\ldots$
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It's easy to see that any number will show up an infinite number of times in $S$.
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Let's call $f(n)$ the starting position of the $n^{\text{th}}$ occurrence of $n$ in $S$. For example, $f(1) = 1$, $f(5) = 81$, $f(12) = 271$ and $f(7780) = 111\\,111\\,365$.
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Find $\sum f(3^k) for 1 ≤ k ≤ 13$.
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# --hints--
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`reflexivePosition()` should return `18174995535140`.
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```js
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assert.strictEqual(reflexivePosition(), 18174995535140);
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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 reflexivePosition() {
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return true;
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}
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reflexivePosition();
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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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