1.8 KiB
id, title, challengeType, forumTopicId, dashedName
| id | title | challengeType | forumTopicId | dashedName |
|---|---|---|---|---|
| 5900f4ab1000cf542c50ffbd | Problem 318: 2011 nines | 5 | 301974 | problem-318-2011-nines |
--description--
Consider the real number \sqrt{2} + \sqrt{3}.
When we calculate the even powers of \sqrt{2} + \sqrt{3} 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}
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 {(\sqrt{2} + \sqrt{3})}^{2n} approaches 1 for large n.
Consider all real numbers of the form \sqrt{p} + \sqrt{q} with p and q positive integers and p < q, such that the fractional part of {(\sqrt{p} + \sqrt{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 {(\sqrt{p} + \sqrt{q})}^{2n}.
Let N(p,q) be the minimal value of n such that C(p,q,n) ≥ 2011.
Find \sum N(p,q) for p + q ≤ 2011.
--hints--
twoThousandElevenNines() should return 709313889.
assert.strictEqual(twoThousandElevenNines(), 709313889);
--seed--
--seed-contents--
function twoThousandElevenNines() {
return true;
}
twoThousandElevenNines();
--solutions--
// solution required