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---
id: 5900f5311000cf542c510042
title: 'Problem 451: Modular inverses'
challengeType: 5
forumTopicId: 302124
dashedName: problem-451-modular-inverses
---
# --description--
Consider the number 15.
There are eight positive numbers less than 15 which are coprime to 15: 1, 2, 4, 7, 8, 11, 13, 14.
The modular inverses of these numbers modulo 15 are: 1, 8, 4, 13, 2, 11, 7, 14 because
$$\begin{align}
& 1 \times 1\bmod 15 = 1 \\\\
& 2 \times 8 = 16\bmod 15 = 1 \\\\
& 4 \times 4 = 16\bmod 15 = 1 \\\\
& 7 \times 13 = 91\bmod 15 = 1 \\\\
& 11 \times 11 = 121\bmod 15 = 1 \\\\
& 14 \times 14 = 196\bmod 15 = 1
\end{align}$$
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.
So $I(15) = 11$.
Also $I(100) = 51$ and $I(7) = 1$.
Find $\sum I(n)$ for $3 ≤ n ≤ 2 \times {10}^7$
# --hints--
`modularInverses()` should return `153651073760956`.
```js
assert.strictEqual(modularInverses(), 153651073760956);
```
# --seed--
## --seed-contents--
```js
function modularInverses() {
return true;
}
modularInverses();
```
# --solutions--
```js
// solution required
```