* 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>
57 lines
1.3 KiB
Markdown
57 lines
1.3 KiB
Markdown
---
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id: 5900f4531000cf542c50ff65
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title: 'Problem 230: Fibonacci Words'
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challengeType: 5
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forumTopicId: 301874
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dashedName: problem-230-fibonacci-words
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---
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# --description--
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For any two strings of digits, A and B, we define FA,B to be the sequence (A,B,AB,BAB,ABBAB,...) in which each term is the concatenation of the previous two.
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Further, we define DA,B(n) to be the nth digit in the first term of FA,B that contains at least n digits.
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Example:
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Let A=1415926535, B=8979323846. We wish to find DA,B(35), say.
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The first few terms of FA,B are: 1415926535 8979323846 14159265358979323846 897932384614159265358979323846 14159265358979323846897932384614159265358979323846
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Then DA,B(35) is the 35th digit in the fifth term, which is 9.
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Now we use for A the first 100 digits of π behind the decimal point: 14159265358979323846264338327950288419716939937510 58209749445923078164062862089986280348253421170679
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and for B the next hundred digits:
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82148086513282306647093844609550582231725359408128 48111745028410270193852110555964462294895493038196 .
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Find ∑n = 0,1,...,17 10n× DA,B((127+19n)×7n) .
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# --hints--
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`euler230()` should return 850481152593119200.
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```js
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assert.strictEqual(euler230(), 850481152593119200);
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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 euler230() {
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return true;
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}
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euler230();
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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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