Oliver Eyton-Williams ee1e8abd87

feat(curriculum): restore seed + solution to Chinese (#40683 )

* 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>

2021-01-12 19:31:00 -07:00

1.2 KiB

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id, title, challengeType, forumTopicId, dashedName

id	title	challengeType	forumTopicId	dashedName
5900f3f51000cf542c50ff08	Problem 137: Fibonacci golden nuggets	5	301765	problem-137-fibonacci-golden-nuggets

--description--

Consider the infinite polynomial series AF(x) = xF1 + x2F2 + x3F3 + ..., where Fk is the kth term in the Fibonacci sequence: 1, 1, 2, 3, 5, 8, ... ; that is, Fk = Fk−1 + Fk−2, F1 = 1 and F2 = 1.

For this problem we shall be interested in values of x for which AF(x) is a positive integer.

Surprisingly AF(1/2)

(1/2).1 + (1/2)2.1 + (1/2)3.2 + (1/2)4.3 + (1/2)5.5 + ...

= 1/2 + 1/4 + 2/8 + 3/16 + 5/32 + ...

= 2 The corresponding values of x for the first five natural numbers are shown below.

xAF(x) √2−11 1/22 (√13−2)/33 (√89−5)/84 (√34−3)/55

We shall call AF(x) a golden nugget if x is rational, because they become increasingly rarer; for example, the 10th golden nugget is 74049690. Find the 15th golden nugget.

--hints--

euler137() should return 1120149658760.

assert.strictEqual(euler137(), 1120149658760);

--seed--

--seed-contents--

function euler137() {

  return true;
}

euler137();

--solutions--

// solution required

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