6cfd0fc503
* fix: improve Project Euler descriptions and test case Improve formatting of Project Euler test descriptions. Also add poker hands array and new test case for problem 54 * feat: add typeof tests and gave functions proper names for first 100 challenges * fix: continue fixing test descriptions and adding "before test" sections * fix: address review comments * fix: adjust grids in 18 and 67 and fix some text that reference files rather than the given arrays * fix: implement bug fixes and improvements from review * fix: remove console.log statements from seed and solution
2.5 KiB
2.5 KiB
id, challengeType, title, forumTopicId
| id | challengeType | title | forumTopicId |
|---|---|---|---|
| 5900f39c1000cf542c50feae | 5 | Problem 47: Distinct primes factors | 302145 |
Description
The first two consecutive numbers to have two distinct prime factors are:
14 = 2 × 7
15 = 3 × 5
15 = 3 × 5
The first three consecutive numbers to have three distinct prime factors are:
644 = 22 × 7 × 23
645 = 3 × 5 × 43
646 = 2 × 17 × 19
645 = 3 × 5 × 43
646 = 2 × 17 × 19
Find the first four consecutive integers to have four distinct prime factors each. What is the first of these numbers?
Instructions
Tests
tests:
- text: <code>distinctPrimeFactors(2, 2)</code> should return a number.
testString: assert(typeof distinctPrimeFactors(2, 2) === 'number');
- text: <code>distinctPrimeFactors(2, 2)</code> should return 14.
testString: assert.strictEqual(distinctPrimeFactors(2, 2), 14);
- text: <code>distinctPrimeFactors(3, 3)</code> should return 644.
testString: assert.strictEqual(distinctPrimeFactors(3, 3), 644);
- text: <code>distinctPrimeFactors(4, 4)</code> should return 134043.
testString: assert.strictEqual(distinctPrimeFactors(4, 4), 134043);
Challenge Seed
function distinctPrimeFactors(targetNumPrimes, targetConsecutive) {
// Good luck!
return true;
}
distinctPrimeFactors(4, 4);
Solution
function distinctPrimeFactors(targetNumPrimes, targetConsecutive) {
function numberOfPrimeFactors(n) {
let factors = 0;
// Considering 2 as a special case
let firstFactor = true;
while (n % 2 == 0) {
n = n / 2;
if (firstFactor) {
factors++;
firstFactor = false;
}
}
// Adding other factors
for (let i = 3; i < Math.sqrt(n); i += 2) {
firstFactor = true;
while (n % i == 0) {
n = n / i;
if (firstFactor) {
factors++;
firstFactor = false;
}
}
}
if (n > 1) { factors++; }
return factors;
}
let number = 0;
let consecutive = 0;
while (consecutive < targetConsecutive) {
number++;
if (numberOfPrimeFactors(number) >= targetNumPrimes) {
consecutive++;
} else {
consecutive = 0;
}
}
return number - targetConsecutive + 1;
}