Solution to problem 45 in Julia

src/Julia/Problem045.jl

@@ -0,0 +1,45 @@
+#=
+Created on 15 Sep 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 45 of Project Euler
+https://projecteuler.net/problem=45
+=#
+
+using BenchmarkTools
+
+function pentagonal(n)
+    return Int(n*(3*n-1)/2)
+end
+
+function hexagonal(n)
+    return Int(n*(2*n-1))
+end
+
+function Problem45()
+    #=
+    Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:
+        Triangle 	  	Tn=n(n+1)/2 	  	1, 3, 6, 10, 15, ...
+        Pentagonal 	  	Pn=n(3n−1)/2 	  	1, 5, 12, 22, 35, ...
+        Hexagonal 	  	Hn=n(2n−1) 	  	    1, 6, 15, 28, 45, ...
+
+    It can be verified that T285 = P165 = H143 = 40755.
+
+    Find the next triangle number that is also pentagonal and hexagonal.
+    =#
+    pentagonal_list = Set(pentagonal(n) for n in 2:100_000)
+    # all hexagonal numbers are also triangle numbers!
+    hexagonal_list = Set(hexagonal(n) for n in 2:100_000)
+
+    ans = sort!(collect(intersect(hexagonal_list, pentagonal_list)))
+    # First one is already known
+    return ans[2]
+end
+
+
+println("Time to evaluate Problem 45:")
+@btime Problem45()
+println("")
+println("Result for Problem 45: ", Problem45())