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curriculum/challenges/italian/10-coding-interview-prep/project-euler/problem-318-2011-nines.md
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---
id: 5900f4ab1000cf542c50ffbd
title: 'Problem 318: 2011 nines'
title: 'Problema 318: 2011 nove'
challengeType: 5
forumTopicId: 301974
dashedName: problem-318-2011-nines
@ -8,44 +8,28 @@ dashedName: problem-318-2011-nines
# --description--
Consider the real number √2+√3.
Considera il numero reale $\sqrt{2} + \sqrt{3}$.
When we calculate the even powers of √2+√3
Quando calcoliamo le potenze pari di $\sqrt{2} + \sqrt{3}$ troviamo:
we get:
$$\begin{align} & {(\sqrt{2} + \sqrt{3})}^2 = 9.898979485566356\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^4 = 97.98979485566356\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^6 = 969.998969071069263\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^8 = 9601.99989585502907\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^{10} = 95049.999989479221\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^{12} = 940897.9999989371855\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^{14} = 9313929.99999989263\ldots \\\\ & {(\sqrt{2} + \sqrt{3})}^{16} = 92198401.99999998915\ldots \\\\ \end{align}$$
(√2+√3)2 = 9.898979485566356...
Sembra che il numero di nove consecutivi all'inizio della parte frazionaria di queste potenze non diminuisca. In realtà si può dimostrare che la parte frazionaria di ${(\sqrt{2} + \sqrt{3})}^{2n}$ si avvicina 1 per $n$ di grandi dimensioni.
(√2+√3)4 = 97.98979485566356...
Considera tutti i numeri reali del modulo $\sqrt{p} + \sqrt{q}$ con $p$ e $q$ interi positivi e $p < q$, tali che la parte frazionaria di ${(\sqrt{p} + \sqrt{q})}^{2n}$ si avvicina 1 per $n$ di grandi dimensioni.
(√2+√3)6 = 969.998969071069263...
Sia $C(p,q,n)$ il numero di nove consecutivi all'inizio della parte frazionaria di ${(\sqrt{p} + \sqrt{q})}^{2n}$.
(√2+√3)8 = 9601.99989585502907...
Sia $N(p,q)$ il valore minimo di $n$ tale che $C(p,q,n) ≥ 2011$.
(√2+√3)10 = 95049.999989479221...
(√2+√3)12 = 940897.9999989371855...
(√2+√3)14 = 9313929.99999989263...
(√2+√3)16 = 92198401.99999998915...
It looks like that the number of consecutive nines at the beginning of the fractional part of these powers is non-decreasing. In fact it can be proven that the fractional part of (√2+√3)2n approaches 1 for large n.
Consider all real numbers of the form √p+√q with p and q positive integers and p<q, such that the fractional part of (√p+√q)2n approaches 1 for large n.
Let C(p,q,n) be the number of consecutive nines at the beginning of the fractional part of (√p+√q)2n.
Let N(p,q) be the minimal value of n such that C(p,q,n) ≥ 2011.
Find ∑N(p,q) for p+q ≤ 2011.
Trova $\sum N(p,q)$ per $p + q ≤ 2011$.
# --hints--
`euler318()` should return 709313889.
`twoThousandElevenNines()` dovrebbe restituire `709313889`.
```js
assert.strictEqual(euler318(), 709313889);
assert.strictEqual(twoThousandElevenNines(), 709313889);
```
# --seed--
@ -53,12 +37,12 @@ assert.strictEqual(euler318(), 709313889);
## --seed-contents--
```js
function euler318() {
function twoThousandElevenNines() {
return true;
}
euler318();
twoThousandElevenNines();
```
# --solutions--