chore(i18n,learn): processed translations (#45333)

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---
id: 5900f52a1000cf542c51003c
title: 'Problem 445: Retractions A'
title: 'Problema 445: Retrazioni A'
challengeType: 5
forumTopicId: 302117
dashedName: problem-445-retractions-a
@ -8,20 +8,26 @@ dashedName: problem-445-retractions-a
# --description--
For every integer n>1, the family of functions fn,a,b is defined
Per ogni intero $n > 1$, la famiglia di funzioni $f_{n, a, b}$ è definita da:
by fn,a,b(x)≡ax+b mod n for a,b,x integer and 0
$f_{n, a, b}(x) ≡ ax + b\bmod n$ per $a, b, x$ interi e $0 \lt a \lt n$, $0 \le b \lt n$, $0 \le x \lt n$.
You are given that ∑ R(c) for c=C(100 000,k), and 1 ≤ k ≤99 999 ≡628701600 (mod 1 000 000 007). (C(n,k) is the binomial coefficient).
Chiameremo $f_{n, a, b}$ una retrazione se $f_{n, a, b}(f_{n, a, b}(x)) \equiv f_{n, a, b}(x)\bmod n$ per ogni $0 \le x \lt n$.
Find ∑ R(c) for c=C(10 000 000,k), and 1 ≤k≤ 9 999 999. Give your answer modulo 1 000 000 007.
Sia $R(n)$ il numero di retrazioni per $n$.
Ti è dato che
$$\sum_{k = 1}^{99\\,999} R(\displaystyle\binom{100\\,000}{k}) \equiv 628\\,701\\,600\bmod 1\\,000\\,000\\,007$$
Trova $$\sum_{k = 1}^{9\\,999\\,999} R(\displaystyle\binom{10\\,000\\,000}{k})$$ Dai la tua risposta modulo $1\\,000\\,000\\,007$.
# --hints--
`euler445()` should return 659104042.
`retractionsA()` dovrebbe restituire `659104042`.
```js
assert.strictEqual(euler445(), 659104042);
assert.strictEqual(retractionsA(), 659104042);
```
# --seed--
@ -29,12 +35,12 @@ assert.strictEqual(euler445(), 659104042);
## --seed-contents--
```js
function euler445() {
function retractionsA() {
return true;
}
euler445();
retractionsA();
```
# --solutions--