curriculum/challenges/english/blocks/project-euler-problems-1-to-100/5900f3c51000cf542c50fed8.md
The smallest number expressible as the sum of a prime square, prime cube, and prime fourth power is 28. In fact, there are exactly four numbers below fifty that can be expressed in such a way:
33 = 3<sup>2</sup> + 2<sup>3</sup> + 2<sup>4</sup>
49 = 5<sup>2</sup> + 2<sup>3</sup> + 2<sup>4</sup>
47 = 2<sup>2</sup> + 3<sup>3</sup> + 2<sup>4</sup>
</div>How many numbers below n can be expressed as the sum of a prime square, prime cube, and prime fourth power?
primePowerTriples(50) should return a number.
assert(typeof primePowerTriples(50) === 'number');
primePowerTriples(50) should return 4.
assert.strictEqual(primePowerTriples(50), 4);
primePowerTriples(10035) should return 684.
assert.strictEqual(primePowerTriples(10035), 684);
primePowerTriples(500000) should return 18899.
assert.strictEqual(primePowerTriples(500000), 18899);
primePowerTriples(5000000) should return 138932.
assert.strictEqual(primePowerTriples(5000000), 138932);
primePowerTriples(50000000) should return 1097343.
assert.strictEqual(primePowerTriples(50000000), 1097343);
function primePowerTriples(n) {
return true;
}
primePowerTriples(50);
function primePowerTriples(n) {
function getSievePrimes(max) {
const primes = [];
const primesMap = new Array(max).fill(true);
primesMap[0] = false;
primesMap[1] = false;
for (let i = 2; i <= max; i += 2) {
if (primesMap[i]) {
primes.push(i);
for (let j = i * i; j <= max; j = j + i) {
primesMap[j] = false;
}
}
if (i === 2) {
i = 1;
}
}
return primes;
}
function getPowersSummed(numbers, powers, limit, curSum) {
if (curSum >= limit) {
return [];
} else if (powers.length === 0) {
return [curSum];
}
const powersSummed = [];
const curPower = powers[0];
const powersLeft = powers.slice(1);
for (let i = 0; i < numbers.length; i++) {
const curNumber = numbers[i];
const nextSum = curSum + curNumber ** curPower;
if (nextSum >= limit) {
return powersSummed;
}
const result = getPowersSummed(
numbers,
powersLeft,
limit,
curSum + curNumber ** curPower
);
powersSummed.push(...result);
}
return powersSummed;
}
const maximumBaseNumber = Math.floor(Math.sqrt(n - 2 ** 3 - 2 ** 4)) + 1;
const primes = getSievePrimes(maximumBaseNumber);
const uniqueSums = new Set(getPowersSummed(primes, [2, 3, 4], n, 0));
return uniqueSums.size;
}