59 lines
1.7 KiB
Markdown
59 lines
1.7 KiB
Markdown
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---
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id: 5900f3a61000cf542c50feb9
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title: 'Problem 58: Spiral primes'
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challengeType: 5
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forumTopicId: 302169
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dashedName: problem-58-spiral-primes
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---
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# --description--
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Starting with 1 and spiralling anticlockwise in the following way, a square spiral with side length 7 is formed.
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<div style='text-align: center;'>
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<strong><span style='color: red;'>37</span></strong> 36 35 34 33 32 <strong><span style='color: red;'>31</span></strong><br>
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38 <strong><span style='color: red;'>17</span></strong> 16 15 14 <strong><span style='color: red;'>13</span></strong> 30<br>
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39 18 <strong><span style='color: red;'>5</span></strong> 4 <strong><span style='color: red;'>3</span></strong> 12 29<br>
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40 19 6 1 2 11 28<br>
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41 20 <strong><span style='color: red;'>7</span></strong> 8 9 10 27<br>
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42 21 22 23 24 25 26<br>
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<strong><span style='color: red;'>43</span></strong> 44 45 46 47 48 49<br>
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</div>
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It is interesting to note that the odd squares lie along the bottom right diagonal, but what is more interesting is that 8 out of the 13 numbers lying along both diagonals are prime; that is, a ratio of 8/13 ≈ 62%.
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If one complete new layer is wrapped around the spiral above, a square spiral with side length 9 will be formed. If this process is continued, what is the side length of the square spiral for which the ratio of primes along both diagonals first falls below 10%?
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# --hints--
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`spiralPrimes()` should return a number.
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```js
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assert(typeof spiralPrimes() === 'number');
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```
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`spiralPrimes()` should return 26241.
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```js
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assert.strictEqual(spiralPrimes(), 26241);
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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 spiralPrimes() {
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return true;
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}
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spiralPrimes();
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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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