1.6 KiB
1.6 KiB
id | challengeType | title |
---|---|---|
5900f4ab1000cf542c50ffbd | 5 | Problem 318: 2011 nines |
Description
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.
Instructions
Tests
tests:
- text: <code>euler318()</code> should return 709313889.
testString: assert.strictEqual(euler318(), 709313889, '<code>euler318()</code> should return 709313889.');
Challenge Seed
function euler318() {
// Good luck!
return true;
}
euler318();
Solution
// solution required