Co-authored-by: Oliver Eyton-Williams <ojeytonwilliams@gmail.com> Co-authored-by: Kristofer Koishigawa <scissorsneedfoodtoo@gmail.com> Co-authored-by: Beau Carnes <beaucarnes@gmail.com>
		
			
				
	
	
		
			100 lines
		
	
	
		
			2.1 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
			
		
		
	
	
			100 lines
		
	
	
		
			2.1 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
---
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id: 5900f3931000cf542c50fea6
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challengeType: 5
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isHidden: false
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title: 'Problem 39: Integer right triangles'
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forumTopicId: 302054
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---
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## Description
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<section id='description'>
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If <var>p</var> is the perimeter of a right angle triangle with integral length sides, {a,b,c}, there are exactly three solutions for p = 120.
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{20,48,52}, {24,45,51}, {30,40,50}
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For which value of <var>p</var> ≤ `n`, is the number of solutions maximized?
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</section>
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## Instructions
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<section id='instructions'>
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</section>
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## Tests
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<section id='tests'>
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```yml
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tests:
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  - text: <code>intRightTriangles(500)</code> should return a number.
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    testString: assert(typeof intRightTriangles(500) === 'number');
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  - text: <code>intRightTriangles(500)</code> should return 420.
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    testString: assert(intRightTriangles(500) == 420);
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  - text: <code>intRightTriangles(800)</code> should return 720.
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    testString: assert(intRightTriangles(800) == 720);
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  - text: <code>intRightTriangles(900)</code> should return 840.
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    testString: assert(intRightTriangles(900) == 840);
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  - text: <code>intRightTriangles(1000)</code> should return 840.
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    testString: assert(intRightTriangles(1000) == 840);
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```
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</section>
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## Challenge Seed
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<section id='challengeSeed'>
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<div id='js-seed'>
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```js
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function intRightTriangles(n) {
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  // Good luck!
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  return n;
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}
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intRightTriangles(500);
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```
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</div>
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</section>
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## Solution
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<section id='solution'>
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```js
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// Original idea for this solution came from
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// https://www.xarg.org/puzzle/project-euler/problem-39/
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function intRightTriangles(n) {
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  // store the number of triangles with a given perimeter
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  let triangles = {};
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  // a is the shortest side
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  for (let a = 3; a < n / 3; a++)
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  // o is the opposite side and is at least as long as a
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    for (let o = a; o < n / 2; o++) {
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      let h = Math.sqrt(a * a + o * o); // hypotenuse
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      let p = a + o + h;  // perimeter
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      if ((h % 1) === 0 && p <= n) {
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        triangles[p] = (triangles[p] || 0) + 1;
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      }
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    }
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  let max = 0, maxp = null;
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  for (let p in triangles) {
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    if (max < triangles[p]) {
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      max = triangles[p];
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      maxp = parseInt(p);
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    }
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  }
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  return maxp;
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}
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```
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</section>
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