1.5 KiB
| id | title | challengeType | forumTopicId | dashedName |
|---|---|---|---|---|
| 5900f4e51000cf542c50fff6 | Problem 374: Maximum Integer Partition Product | 5 | 302036 | problem-374-maximum-integer-partition-product |
--description--
An integer partition of a number n is a way of writing n as a sum of positive integers.
Partitions that differ only in the order of their summands are considered the same. A partition of n into distinct parts is a partition of n in which every part occurs at most once.
The partitions of 5 into distinct parts are:
5, 4 + 1 and 3 + 2.
Let f(n) be the maximum product of the parts of any such partition of n into distinct parts and let m(n) be the number of elements of any such partition of n with that product.
So f(5) = 6 and m(5) = 2.
For n = 10 the partition with the largest product is 10 = 2 + 3 + 5, which gives f(10) = 30 and m(10) = 3. And their product, f(10) \times m(10) = 30 \times 3 = 90
It can be verified that \sum f(n) \times m(n) for 1 ≤ n ≤ 100 = 1\\,683\\,550\\,844\\,462.
Find \sum f(n) \times m(n) for 1 ≤ n ≤ {10}^{14}. Give your answer modulo 982\\,451\\,653, the 50 millionth prime.
--hints--
maximumIntegerPartitionProduct() should return 334420941.
assert.strictEqual(maximumIntegerPartitionProduct(), 334420941);
--seed--
--seed-contents--
function maximumIntegerPartitionProduct() {
return true;
}
maximumIntegerPartitionProduct();
--solutions--
// solution required