110 lines
2.4 KiB
Markdown
110 lines
2.4 KiB
Markdown
---
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id: 5900f3bc1000cf542c50fecf
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title: 'Problem 80: Square root digital expansion'
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challengeType: 5
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forumTopicId: 302194
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dashedName: problem-80-square-root-digital-expansion
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---
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# --description--
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It is well known that if the square root of a natural number is not an integer, then it is irrational. The decimal expansion of such square roots is infinite without any repeating pattern at all.
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The square root of two is `1.41421356237309504880...`, and the digital sum of the first one hundred decimal digits is `475`.
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For the first `n` natural numbers, find the total of the digital sums of the first one hundred decimal digits for all the irrational square roots.
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# --hints--
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`sqrtDigitalExpansion(2)` should return a number.
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```js
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assert(typeof sqrtDigitalExpansion(2) === 'number');
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```
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`sqrtDigitalExpansion(2)` should return `475`.
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```js
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assert.strictEqual(sqrtDigitalExpansion(2), 475);
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```
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`sqrtDigitalExpansion(50)` should return `19543`.
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```js
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assert.strictEqual(sqrtDigitalExpansion(50), 19543);
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```
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`sqrtDigitalExpansion(100)` should return `40886`.
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```js
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assert.strictEqual(sqrtDigitalExpansion(100), 40886);
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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 sqrtDigitalExpansion(n) {
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return true;
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}
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sqrtDigitalExpansion(2);
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```
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# --solutions--
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```js
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function sqrtDigitalExpansion(n) {
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function sumDigits(number) {
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let sum = 0;
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while (number > 0n) {
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let digit = number % 10n;
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sum += parseInt(digit, 10);
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number = number / 10n;
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}
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return sum;
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}
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function power(numberA, numberB) {
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let result = 1n;
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for (let b = 0; b < numberB; b++) {
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result = result * BigInt(numberA);
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}
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return result;
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}
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// Based on http://www.afjarvis.staff.shef.ac.uk/maths/jarvisspec02.pdf
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function expandSquareRoot(number, numDigits) {
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let a = 5n * BigInt(number);
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let b = 5n;
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const boundaryWithNeededDigits = power(10, numDigits + 1);
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while (b < boundaryWithNeededDigits) {
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if (a >= b) {
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a = a - b;
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b = b + 10n;
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} else {
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a = a * 100n;
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b = (b / 10n) * 100n + 5n;
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}
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}
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return b / 100n;
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}
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let result = 0;
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let nextPerfectRoot = 1;
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const requiredDigits = 100;
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for (let i = 1; i <= n; i++) {
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if (nextPerfectRoot ** 2 === i) {
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nextPerfectRoot++;
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continue;
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}
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result += sumDigits(expandSquareRoot(i, requiredDigits));
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}
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return result;
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}
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```
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