1.6 KiB
1.6 KiB
| 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`.
```js
assert.strictEqual(specialPartitions(), 3053105);
```
# --seed--
## --seed-contents--
```js
function specialPartitions() {
return true;
}
specialPartitions();
```
# --solutions--
```js
// solution required
```