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gikf 7907f62337 fix(curriculum): clean-up Project Euler 121-140 (#42731)
* fix: clean-up Project Euler 121-140

* fix: corrections from review

Co-authored-by: Sem Bauke <46919888+Sembauke@users.noreply.github.com>

* fix: missing backticks

Co-authored-by: Kristofer Koishigawa <scissorsneedfoodtoo@gmail.com>

* fix: corrections from review

Co-authored-by: Tom <20648924+moT01@users.noreply.github.com>

* fix: missing delimiter

Co-authored-by: Sem Bauke <46919888+Sembauke@users.noreply.github.com>
Co-authored-by: Kristofer Koishigawa <scissorsneedfoodtoo@gmail.com>
Co-authored-by: Tom <20648924+moT01@users.noreply.github.com>
2021-07-16 21:38:37 +02:00

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---
id: 5900f3f21000cf542c50ff04
title: 'Problem 133: Repunit nonfactors'
challengeType: 5
forumTopicId: 301761
dashedName: problem-133-repunit-nonfactors
---
# --description--
A number consisting entirely of ones is called a repunit. We shall define $R(k)$ to be a repunit of length $k$; for example, $R(6) = 111111$.
Let us consider repunits of the form $R({10}^n)$.
Although $R(10)$, $R(100)$, or $R(1000)$ are not divisible by 17, $R(10000)$ is divisible by 17. Yet there is no value of n for which $R({10}^n)$ will divide by 19. Remarkably, 11, 17, 41, and 73 are the only four primes below one-hundred that can be a factor of $R({10}^n)$.
Find the sum of all the primes below one-hundred thousand that will never be a factor of $R({10}^n)$.
# --hints--
`repunitNonfactors()` should return `453647705`.
```js
assert.strictEqual(repunitNonfactors(), 453647705);
```
# --seed--
## --seed-contents--
```js
function repunitNonfactors() {
return true;
}
repunitNonfactors();
```
# --solutions--
```js
// solution required
```