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---
id: 5900f4b91000cf542c50ffcc
title: 'Problem 333: Special partitions'
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challengeType: 5
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forumTopicId: 301991
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dashedName: problem-333-special-partitions
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---
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# --description--
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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 $i, j ≥ 0$.
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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 $8 + 9 = (2^3 \times 3^0 + 2^0 \times 3^2)$.
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Many integers have more than one valid partition, the first being 11 having the following two partitions.
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$$\begin{align}
& 11 = 2 + 9 = (2^1 \times 3^0 + 2^0 \times 3^2) \\\\
& 11 = 8 + 3 = (2^3 \times 3^0 + 2^0 \times 3^1)
\end{align}$$
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Let's define $P(n)$ as the number of valid partitions of $n$. For example, $P(11) = 2$.
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Let's consider only the prime integers $q$ which would have a single valid partition such as $P(17)$.
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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 $P(q) = 1$.
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# --hints--
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`specialPartitions()` should return `3053105` .
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```js
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assert.strictEqual(specialPartitions(), 3053105);
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```
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# --seed--
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## --seed-contents--
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```js
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function specialPartitions() {
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return true;
}
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specialPartitions();
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```
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# --solutions--
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```js
// solution required
```