51 lines
1.6 KiB
Markdown
51 lines
1.6 KiB
Markdown
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---
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id: 5900f46d1000cf542c50ff7f
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title: 'Problem 255: Rounded Square Roots'
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challengeType: 5
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forumTopicId: 301903
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dashedName: problem-255-rounded-square-roots
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---
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# --description--
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We define the rounded-square-root of a positive integer n as the square root of n rounded to the nearest integer.
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The following procedure (essentially Heron's method adapted to integer arithmetic) finds the rounded-square-root of n: Let d be the number of digits of the number n. If d is odd, set x0 = 2×10(d-1)⁄2. If d is even, set x0 = 7×10(d-2)⁄2. Repeat:
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until xk+1 = xk.
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As an example, let us find the rounded-square-root of n = 4321.n has 4 digits, so x0 = 7×10(4-2)⁄2 = 70. Since x2 = x1, we stop here. So, after just two iterations, we have found that the rounded-square-root of 4321 is 66 (the actual square root is 65.7343137…).
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The number of iterations required when using this method is surprisingly low. For example, we can find the rounded-square-root of a 5-digit integer (10,000 ≤ n ≤ 99,999) with an average of 3.2102888889 iterations (the average value was rounded to 10 decimal places).
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Using the procedure described above, what is the average number of iterations required to find the rounded-square-root of a 14-digit number (1013 ≤ n < 1014)? Give your answer rounded to 10 decimal places.
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Note: The symbols ⌊x⌋ and ⌈x⌉ represent the floor function and ceiling function respectively.
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# --hints--
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`euler255()` should return 4.447401118.
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```js
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assert.strictEqual(euler255(), 4.447401118);
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```
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# --seed--
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## --seed-contents--
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```js
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function euler255() {
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return true;
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}
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euler255();
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```
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# --solutions--
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```js
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// solution required
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```
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