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								id: 5900f5311000cf542c510042
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								title: 'Problem 451: Modular inverses'
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								challengeType: 5
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											2019-08-05 09:17:33 -07:00
										 
									 
								 
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								forumTopicId: 302124
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											2021-01-13 03:31:00 +01:00
										 
									 
								 
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								dashedName: problem-451-modular-inverses
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								---
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								# --description--
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								Consider the number 15.
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								There are eight positive numbers less than 15 which are coprime to 15: 1, 2, 4, 7, 8, 11, 13, 14.
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											2021-07-30 17:20:31 +02:00
										 
									 
								 
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								The modular inverses of these numbers modulo 15 are: 1, 8, 4, 13, 2, 11, 7, 14 because
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								$$\begin{align}
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								  & 1  \times 1\bmod 15 = 1 \\\\
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								  & 2  \times 8  = 16\bmod 15 = 1 \\\\
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								  & 4  \times 4  = 16\bmod 15 = 1 \\\\
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								  & 7  \times 13 = 91\bmod 15 = 1 \\\\
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								  & 11 \times 11 = 121\bmod 15 = 1 \\\\
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								  & 14 \times 14 = 196\bmod 15 = 1
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								\end{align}$$
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								Let $I(n)$ be the largest positive number $m$ smaller than $n - 1$ such that the modular inverse of $m$ modulo $n$ equals $m$ itself.
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								So $I(15) = 11$.
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								Also $I(100) = 51$ and $I(7) = 1$.
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								Find $\sum I(n)$ for $3 ≤ n ≤ 2 \times {10}^7$
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								# --hints--
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								`modularInverses()` should return `153651073760956`.
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								```js
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								assert.strictEqual(modularInverses(), 153651073760956);
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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 modularInverses() {
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								  return true;
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								}
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								modularInverses();
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								```
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								# --solutions--
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								```js
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								// solution required
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								```
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