ee1e8abd87
* feat(tools): add seed/solution restore script * chore(curriculum): remove empty sections' markers * chore(curriculum): add seed + solution to Chinese * chore: remove old formatter * fix: update getChallenges parse translated challenges separately, without reference to the source * chore(curriculum): add dashedName to English * chore(curriculum): add dashedName to Chinese * refactor: remove unused challenge property 'name' * fix: relax dashedName requirement * fix: stray tag Remove stray `pre` tag from challenge file. Signed-off-by: nhcarrigan <nhcarrigan@gmail.com> Co-authored-by: nhcarrigan <nhcarrigan@gmail.com>
69 lines
1.4 KiB
Markdown
69 lines
1.4 KiB
Markdown
---
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id: 5900f4ab1000cf542c50ffbd
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title: 'Problem 318: 2011 nines'
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challengeType: 5
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forumTopicId: 301974
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dashedName: problem-318-2011-nines
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---
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# --description--
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Consider the real number √2+√3.
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When we calculate the even powers of √2+√3
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we get:
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(√2+√3)2 = 9.898979485566356...
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(√2+√3)4 = 97.98979485566356...
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(√2+√3)6 = 969.998969071069263...
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(√2+√3)8 = 9601.99989585502907...
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(√2+√3)10 = 95049.999989479221...
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(√2+√3)12 = 940897.9999989371855...
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(√2+√3)14 = 9313929.99999989263...
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(√2+√3)16 = 92198401.99999998915...
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It looks like that the number of consecutive nines at the beginning of the fractional part of these powers is non-decreasing. In fact it can be proven that the fractional part of (√2+√3)2n approaches 1 for large n.
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Consider all real numbers of the form √p+√q with p and q positive integers and p<q, such that the fractional part of (√p+√q)2n approaches 1 for large n.
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Let C(p,q,n) be the number of consecutive nines at the beginning of the fractional part of (√p+√q)2n.
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Let N(p,q) be the minimal value of n such that C(p,q,n) ≥ 2011.
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Find ∑N(p,q) for p+q ≤ 2011.
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# --hints--
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`euler318()` should return 709313889.
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```js
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assert.strictEqual(euler318(), 709313889);
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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 euler318() {
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return true;
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}
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euler318();
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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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