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>
1.5 KiB
id, title, challengeType, forumTopicId, dashedName
| id | title | challengeType | forumTopicId | dashedName |
|---|---|---|---|---|
| 5900f3a51000cf542c50feb8 | Problem 57: Square root convergents | 5 | 302168 | problem-57-square-root-convergents |
--description--
It is possible to show that the square root of two can be expressed as an infinite continued fraction.
By expanding this for the first four iterations, we get:
1 + \\frac 1 2 = \\frac 32 = 1.5
1 + \\frac 1 {2 + \\frac 1 2} = \\frac 7 5 = 1.4
1 + \\frac 1 {2 + \\frac 1 {2+\\frac 1 2}} = \\frac {17}{12} = 1.41666 \\dots
1 + \\frac 1 {2 + \\frac 1 {2+\\frac 1 {2+\\frac 1 2}}} = \\frac {41}{29} = 1.41379 \\dots
The next three expansions are \\frac {99}{70}, \\frac {239}{169}, and \\frac {577}{408}, but the eighth expansion, \\frac {1393}{985}, is the first example where the number of digits in the numerator exceeds the number of digits in the denominator.
In the first one-thousand expansions, how many fractions contain a numerator with more digits than denominator?
--hints--
squareRootConvergents() should return a number.
assert(typeof squareRootConvergents() === 'number');
squareRootConvergents() should return 153.
assert.strictEqual(squareRootConvergents(), 153);
--seed--
--seed-contents--
function squareRootConvergents() {
return true;
}
squareRootConvergents();
--solutions--
// solution required