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>
59 lines
1.7 KiB
Markdown
59 lines
1.7 KiB
Markdown
---
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id: 5900f4a81000cf542c50ffbb
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title: 'Problem 316: Numbers in decimal expansions'
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challengeType: 5
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forumTopicId: 301972
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dashedName: problem-316-numbers-in-decimal-expansions
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---
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# --description--
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Let $p = p_1 p_2 p_3 \ldots$ be an infinite sequence of random digits, selected from {0,1,2,3,4,5,6,7,8,9} with equal probability.
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It can be seen that $p$ corresponds to the real number $0.p_1 p_2 p_3 \ldots$.
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It can also be seen that choosing a random real number from the interval [0,1) is equivalent to choosing an infinite sequence of random digits selected from {0,1,2,3,4,5,6,7,8,9} with equal probability.
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For any positive integer $n$ with $d$ decimal digits, let $k$ be the smallest index such that $p_k, p_{k + 1}, \ldots p_{k + d - 1}$ are the decimal digits of $n$, in the same order.
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Also, let $g(n)$ be the expected value of $k$; it can be proven that $g(n)$ is always finite and, interestingly, always an integer number.
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For example, if $n = 535$, then
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for $p = 31415926\mathbf{535}897\ldots$, we get $k = 9$
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for $p = 35528714365004956000049084876408468\mathbf{535}4\ldots$, we get $k = 36$
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etc and we find that $g(535) = 1008$.
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Given that $\displaystyle\sum_{n = 2}^{999} g\left(\left\lfloor\frac{{10}^6}{n}\right\rfloor\right) = 27280188$, find $\displaystyle\sum_{n = 2}^{999\\,999} g\left(\left\lfloor\frac{{10}^{16}}{n}\right\rfloor\right)$.
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**Note:** $\lfloor x\rfloor$ represents the floor function.
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# --hints--
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`numbersInDecimalExpansion()` should return `542934735751917760`.
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```js
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assert.strictEqual(numbersInDecimalExpansion(), 542934735751917760);
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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 numbersInDecimalExpansion() {
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return true;
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}
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numbersInDecimalExpansion();
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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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