* fix: clean-up Project Euler 441-460 * fix: corrections from review Co-authored-by: Tom <20648924+moT01@users.noreply.github.com> Co-authored-by: Tom <20648924+moT01@users.noreply.github.com>
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id, title, challengeType, forumTopicId, dashedName
id | title | challengeType | forumTopicId | dashedName |
---|---|---|---|---|
5900f5311000cf542c510042 | Problem 451: Modular inverses | 5 | 302124 | 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
.
assert.strictEqual(modularInverses(), 153651073760956);
--seed--
--seed-contents--
function modularInverses() {
return true;
}
modularInverses();
--solutions--
// solution required