freeCodeCamp/curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-333-special-partitions.md

id, title, challengeType, forumTopicId, dashedName

id	title	challengeType	forumTopicId	dashedName
5900f4b91000cf542c50ffcc	Problem 333: Special partitions	5	301991	problem-333-special-partitions

--description--

All positive integers can be partitioned in such a way that each and every term of the partition can be expressed as 2^i \times 3^j, where i, j ≥ 0.

Let's consider only those such partitions where none of the terms can divide any of the other terms. For example, the partition of 17 = 2 + 6 + 9 = (2^1 \times 3^0 + 2^1 \times 3^1 + 2^0 \times 3^2) would not be valid since 2 can divide 6. Neither would the partition 17 = 16 + 1 = (2^4 \times 3^0 + 2^0 \times 3^0) since 1 can divide 16. The only valid partition of 17 would be 8 + 9 = (2^3 \times 3^0 + 2^0 \times 3^2).

Many integers have more than one valid partition, the first being 11 having the following two partitions.

$$\begin{align} & 11 = 2 + 9 = (2^1 \times 3^0 + 2^0 \times 3^2) \\ & 11 = 8 + 3 = (2^3 \times 3^0 + 2^0 \times 3^1) \end{align}$$

Let's define P(n) as the number of valid partitions of n. For example, P(11) = 2.

Let's consider only the prime integers q which would have a single valid partition such as P(17).

The sum of the primes q <100 such that P(q) = 1 equals 233.

Find the sum of the primes q < 1\\,000\\,000 such that P(q) = 1.

--hints--

specialPartitions() should return 3053105.

assert.strictEqual(specialPartitions(), 3053105);

--seed--

--seed-contents--

function specialPartitions() {

  return true;
}

specialPartitions();

--solutions--

// solution required