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