Solution to problem 35 in Julia

src/Julia/Problem035.jl

@@ -0,0 +1,72 @@
+#=
+Created on 02 Sep 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 35 of Project Euler
+https://projecteuler.net/problem=35
+=#
+
+using BenchmarkTools
+using Combinatorics
+using Primes
+
+# function circular_number(n)
+#     result = []
+#     perms = collect(permutations(digits(n), length(n)))
+#     for i in perms
+#         push!(result, parse(Int, join(i)))
+#     end
+#     return result
+# end
+
+function circular_number(n)
+    if n <10
+        return [n]
+    end
+    digs=digits(n)
+    d=length(digs)
+    cyc=zeros(Int,d)
+    x=[10^i for i in 0:d-1]
+    for i in 1:d
+        cyc[i]=sum(x .* digs[vcat(i:d,1:i-1)])
+    end
+    return cyc
+end
+
+function Problem35()
+    #=
+    The number, 197, is called a circular prime because all rotations of the
+    digits: 197, 971, and 719, are themselves prime.
+
+    There are thirteen such primes below 100:
+    2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97.
+
+    How many circular primes are there below one million?
+    =#
+    circular_primes = []
+    cnt = 0
+    for i in 2:1_000_000
+        if isprime(i)
+            all_primes = true
+            for j in circular_number(i)
+                if !isprime(j)
+                    all_primes = false
+                    break
+                end
+            end
+            if all_primes
+               cnt +=1 #push!(circular_primes, i)
+            end
+        end
+    end
+
+    return cnt #length(circular_primes)
+end
+
+
+println("Time to evaluate Problem 35:")
+@btime Problem35()
+println("")
+println("Result for Problem 35: ", Problem35())