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66
ProjectEuler/Problem12.java
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66
ProjectEuler/Problem12.java
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package ProjectEuler;
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/**
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* The sequence of triangle numbers is generated by adding the natural numbers.
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* So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28.
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* The first ten terms would be:
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* <p>
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* 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
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* <p>
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* Let us list the factors of the first seven triangle numbers:
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* <p>
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* 1: 1
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* 3: 1,3
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* 6: 1,2,3,6
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* 10: 1,2,5,10
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* 15: 1,3,5,15
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* 21: 1,3,7,21
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* 28: 1,2,4,7,14,28
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* We can see that 28 is the first triangle number to have over five divisors.
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* <p>
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* What is the value of the first triangle number to have over five hundred divisors?
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* <p>
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* link: https://projecteuler.net/problem=12
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*/
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public class Problem12 {
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/**
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* Driver Code
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*/
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public static void main(String[] args) {
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assert solution1(500) == 76576500;
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}
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/* returns the nth triangle number; that is, the sum of all the natural numbers less than, or equal to, n */
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public static int triangleNumber(int n) {
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int sum = 0;
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for (int i = 0; i <= n; i++)
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sum += i;
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return sum;
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}
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public static int solution1(int number) {
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int j = 0; // j represents the jth triangle number
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int n = 0; // n represents the triangle number corresponding to j
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int numberOfDivisors = 0; // number of divisors for triangle number n
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while (numberOfDivisors <= number) {
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// resets numberOfDivisors because it's now checking a new triangle number
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// and also sets n to be the next triangle number
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numberOfDivisors = 0;
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j++;
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n = triangleNumber(j);
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// for every number from 1 to the square root of this triangle number,
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// count the number of divisors
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for (int i = 1; i <= Math.sqrt(n); i++)
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if (n % i == 0)
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numberOfDivisors++;
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// 1 to the square root of the number holds exactly half of the divisors
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// so multiply it by 2 to include the other corresponding half
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numberOfDivisors *= 2;
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}
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return n;
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}
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}
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