83 lines
2.1 KiB
Markdown
83 lines
2.1 KiB
Markdown
---
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id: 5900f4151000cf542c50ff27
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title: 'Problem 168: Number Rotations'
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challengeType: 5
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forumTopicId: 301802
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dashedName: problem-168-number-rotations
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---
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# --description--
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Consider the number 142857. We can right-rotate this number by moving the last digit (7) to the front of it, giving us 714285.
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It can be verified that $714285 = 5 × 142857$.
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This demonstrates an unusual property of 142857: it is a divisor of its right-rotation.
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For integer number of digits $a$ and $b$, find the last 5 digits of the sum of all integers $n$, $10^a < n < 10^b$, that have this property.
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# --hints--
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`numberRotations(2, 10)` should return `98311`.
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```js
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assert.strictEqual(numberRotations(2, 10), 98311);
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```
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`numberRotations(2, 100)` should return `59206`.
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```js
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assert.strictEqual(numberRotations(2, 100), 59206);
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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 numberRotations(a, b) {
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return 0;
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}
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numberRotations();
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```
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# --solutions--
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```js
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function numberRotations(minDigits, maxDigits) {
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const DIGITS_TO_KEEP = 100000n;
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const powersOfTen = Array(maxDigits).fill(0);
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powersOfTen[0] = 1n;
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for (let i = 1; i < maxDigits; i++) {
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powersOfTen[i] = powersOfTen[i - 1] * 10n;
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}
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// We want numbers of the form xd * m = dx
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// Or more precisely:
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// (x * 10 + d) * m = d*10^(n-1) + x
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// Solving for x:
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// x = d (10^(n-1) - m) / (10 * m - 1)
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let total = 0n;
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for (let numDigits = minDigits; numDigits <= maxDigits; numDigits++) {
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// Check all multiplier - digit pairs to see if a candidate can be built
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// with the correct number of digits
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for (let multiplier = 1n; multiplier < 10n; multiplier++) {
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for (let lastDigit = 1n; lastDigit < 10n; lastDigit++) {
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const numerator = lastDigit * (powersOfTen[numDigits - 1] - multiplier);
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const denominator = (powersOfTen[1] * multiplier - 1n);
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if (numerator % denominator === 0n) {
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const candidate = (numerator / denominator) * 10n + lastDigit;
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if (candidate.toString().length === numDigits) {
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total = (total + candidate) % DIGITS_TO_KEEP;
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}
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}
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}
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}
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}
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return parseInt(total);
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}
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```
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