# Project Euler | Problem 9 | Special Pythagorean Triplet

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__Problem Description__ :

For example, 32 + 42 = 9 + 16 = 25 = 52. There exists exactly one Pythagorean triplet for which a + b + c = 1000. Find the product abc.

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__Concept__ :

We have following two equations.

1) a^2+b^2=c^2

2) a+b+c=1000

on solving these equations mathematically.

1<=a<=999

b=(1000^2-(2*a*1000))/(2*(1000-a))

c=1000-b-a

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__Java Program__ :

package com.javamultiplex.projecteuler; /** * * @author Rohit Agarwal * @category Project Euler Problems * @problem Problem 9 * */ public class Problem9 { public static void main(String[] args) { int limit = 1000, a, b, c, result; for (int i = 1; i < limit; i++) { a = i; /* * if a+b+c=n and c^2=a^2+b^2 on solving above equations we can find * value of b in terms of a and n b=(n^2-2an)/2(n-a) * */ b = (int) ((Math.pow(limit, 2) - 2 * a * limit) / (2 * (limit - a))); c = limit - b - a; if (isPythagoreanTriplet(a, b, c)) { if (a + b + c == limit) { result = a * b * c; System.out.println("Pythagorean triplet for which a + b + c = 1000 is : " + result); break; } } } } //a,b and c are Pythagorean Triplets if c>b>a and c^2=a^2+b^2 private static boolean isPythagoreanTriplet(int a, int b, int c) { boolean result = false; if (c > b && b > a) { if (c * c == a * a + b * b) { result = true; } } return result; } }

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__Output__ :

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__References__ :

https://en.wikipedia.org/wiki/Pythagorean_triple

https://docs.oracle.com/javase/7/docs/api/java/lang/Math.html

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**Tags : Project Euler Problem 9 solution in Java, Mathematics problems, Special Pythagorean Triplet, for loop, if else statement.**

Project Euler | Problem 9 | Special Pythagorean Triplet
Reviewed by Rohit Agarwal
on
3/10/2017
Rating:

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