feat(problem): 12 - highly divisible triangle number

readme.md

@@ -9,7 +9,7 @@ All of the solutions here are written in rust. [main.rs](src/main.rs) is the hom
 
 ## Challenge Completion
 
-### <!-- completed -->11<!-- completed --> out of 100 public challenges completed.
+### <!-- completed -->12<!-- completed --> out of 100 public challenges completed.
 
 - [x] 1 - [Multiples of 3 or 5](src/bin/1.rs)
 - [x] 2 - [Even Fibonacci numbers](src/bin/2.rs)
@@ -22,7 +22,7 @@ All of the solutions here are written in rust. [main.rs](src/main.rs) is the hom
 - [x] 9 - [Special Pythagorean triplet](src/bin/9.rs)
 - [x] 10 - [Summation of primes](src/bin/10.rs)
 - [x] 11 - [Largest product in a grid](src/bin/11.rs)
-- [ ] 12 - Highly divisible triangular number
+- [x] 12 - [Highly divisible triangular number](src/bin/12.rs)
 - [ ] 13 - Large sum
 - [ ] 14 - Longest Collatz sequence
 - [ ] 15 - Lattice paths

src/bin/12.rs

@@ -0,0 +1,47 @@
+/*
+Problem 12 - Highly divisible triangular number
+
+The sequence of triangle numbers is generated by adding the natural numbers. So the 7 th  triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:
+1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
+Let us list the factors of the first seven triangle numbers:
+1 : 1   3 : 1,3   6 : 1,2,3,6  10 : 1,2,5,10  15 : 1,3,5,15  21 : 1,3,7,21  28 : 1,2,4,7,14,28
+We can see that 28 is the first triangle number to have over five divisors.
+What is the value of the first triangle number to have over five hundred divisors?
+*/
+
+fn factors(number: usize) -> Vec<Vec<usize>> {
+    let max = (number as f64).sqrt() as usize;
+    let mut factors: Vec<Vec<usize>> = vec![vec![1, number]];
+    let is_even = number % 2 == 0;
+
+    for current_factor in (if is_even {2} else {3})..(max + 1) {
+        if number % current_factor != 0 {
+            continue;
+        }
+
+        factors.push(vec![current_factor, number / current_factor]);
+    }
+
+    return factors;
+}
+
+fn first_triangle_with_n_divisors(n: usize) -> usize {
+    let mut i = 1;
+
+    loop {
+        let triangle = (i * (i + 1)) / 2;
+        let factors = factors(triangle);
+
+        if factors.len() * 2 > n {
+            return triangle;
+        }
+
+        i += 1;
+    }
+}
+
+fn main() {
+    let value = first_triangle_with_n_divisors(500);
+
+    println!("The first triangle number with over 500 divisors is {value}");
+}