53 lines
1.1 KiB
Markdown
53 lines
1.1 KiB
Markdown
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---
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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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forumTopicId: 302124
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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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The modular inverses of these numbers modulo 15 are: 1, 8, 4, 13, 2, 11, 7, 14 because
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$$\begin{align} & 1 \times 1\bmod 15 = 1 \\\\ & 2 \times 8 = 16\bmod 15 = 1 \\\\ & 4 \times 4 = 16\bmod 15 = 1 \\\\ & 7 \times 13 = 91\bmod 15 = 1 \\\\ & 11 \times 11 = 121\bmod 15 = 1 \\\\ & 14 \times 14 = 196\bmod 15 = 1 \end{align}$$
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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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