2.9 KiB
id | title | challengeType | forumTopicId | dashedName |
---|---|---|---|---|
5900f3c51000cf542c50fed6 | Problem 88: Product-sum numbers | 5 | 302203 | problem-88-product-sum-numbers |
--description--
A natural number, N
, that can be written as the sum and product of a given set of at least two natural numbers, \\{a_1, a_2, \ldots , a_k\\}
is called a product-sum number: N = a_1 + a_2 + \cdots + a_k = a_1 × a_2 × \cdots × a_k
.
For example, 6 = 1 + 2 + 3 = 1 × 2 × 3.
For a given set of size, k
, we shall call the smallest N with this property a minimal product-sum number. The minimal product-sum numbers for sets of size, k
= 2, 3, 4, 5, and 6 are as follows.
k=3: 6 = 1 × 2 × 3 = 1 + 2 + 3
k=4: 8 = 1 × 1 × 2 × 4 = 1 + 1 + 2 + 4
k=5: 8 = 1 × 1 × 2 × 2 × 2 = 1 + 1 + 2 + 2 + 2
k=6: 12 = 1 × 1 × 1 × 1 × 2 × 6 = 1 + 1 + 1 + 1 + 2 + 6
Hence for 2 ≤ k
≤ 6, the sum of all the minimal product-sum numbers is 4 + 6 + 8 + 12 = 30; note that 8
is only counted once in the sum.
In fact, as the complete set of minimal product-sum numbers for 2 ≤ k
≤ 12 is \\{4, 6, 8, 12, 15, 16\\}
, the sum is 61
.
What is the sum of all the minimal product-sum numbers for 2 ≤ k
≤ limit
?
--hints--
productSumNumbers(6)
should return a number.
assert(typeof productSumNumbers(6) === 'number');
productSumNumbers(6)
should return 30
.
assert.strictEqual(productSumNumbers(6), 30);
productSumNumbers(12)
should return 61
.
assert.strictEqual(productSumNumbers(12), 61);
productSumNumbers(300)
should return 12686
.
assert.strictEqual(productSumNumbers(300), 12686);
productSumNumbers(6000)
should return 2125990
.
assert.strictEqual(productSumNumbers(6000), 2125990);
productSumNumbers(12000)
should return 7587457
.
assert.strictEqual(productSumNumbers(12000), 7587457);
--seed--
--seed-contents--
function productSumNumbers(limit) {
return true;
}
productSumNumbers(6);
--solutions--
function productSumNumbers(limit) {
function getProductSums(curProduct, curSum, factorsCount, start) {
const k = curProduct - curSum + factorsCount;
if (k <= limit) {
if (curProduct < minimalProductSums[k]) {
minimalProductSums[k] = curProduct;
}
for (let i = start; i < Math.floor((limit / curProduct) * 2) + 1; i++) {
getProductSums(curProduct * i, curSum + i, factorsCount + 1, i);
}
}
}
const minimalProductSums = new Array(limit + 1).fill(2 * limit);
getProductSums(1, 1, 1, 2);
const uniqueProductSums = [...new Set(minimalProductSums.slice(2))];
let sum = 0;
for (let i = 0; i < uniqueProductSums.length; i++) {
sum += uniqueProductSums[i];
}
return sum;
}