92 lines
		
	
	
		
			1.8 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
			
		
		
	
	
			92 lines
		
	
	
		
			1.8 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
| ---
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| id: 5900f3931000cf542c50fea6
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| title: 'Problem 39: Integer right triangles'
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| challengeType: 5
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| forumTopicId: 302054
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| dashedName: problem-39-integer-right-triangles
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| ---
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| 
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| # --description--
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| 
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| If `p` 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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| 
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| {20,48,52}, {24,45,51}, {30,40,50}
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| 
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| For which value of `p` ≤ `n`, is the number of solutions maximized?
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| 
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| # --hints--
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| 
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| `intRightTriangles(500)` should return a number.
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| 
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| ```js
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| assert(typeof intRightTriangles(500) === 'number');
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| ```
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| 
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| `intRightTriangles(500)` should return 420.
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| 
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| ```js
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| assert(intRightTriangles(500) == 420);
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| ```
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| 
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| `intRightTriangles(800)` should return 720.
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| 
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| ```js
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| assert(intRightTriangles(800) == 720);
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| ```
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| 
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| `intRightTriangles(900)` should return 840.
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| 
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| ```js
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| assert(intRightTriangles(900) == 840);
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| ```
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| 
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| `intRightTriangles(1000)` should return 840.
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| 
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| ```js
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| assert(intRightTriangles(1000) == 840);
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| ```
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| 
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| # --seed--
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| 
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| ## --seed-contents--
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| 
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| ```js
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| function intRightTriangles(n) {
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| 
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|   return n;
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| }
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| 
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| intRightTriangles(500);
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| ```
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| 
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| # --solutions--
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| 
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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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| 
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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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| 
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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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