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+# Ref4ctory
+
+This challenge involved finding the factors of multiple composite
+numbers.
+
+``` python
+def check_factors(a,b,ab):
+    if abs(a)<=1 or abs(b)<=1:
+        print("too easy")
+        return False
+    if type(a*b) == float:
+        print("no floats please")
+        return False
+    return a*b == ab
+```
+
+The `check_factors` function disallows floats and factors $-1\le a\le 1$
+and $-1\le b\le 1$.
+
+Our quick-and-dirty strategy was therefore to use the factor $a=2$ and
+$b=ab/2$ if $a\pmod2=0$. If $a\pmod2\ne0$, we would use wolframalpha's
+prime factorization (because primes are awesome). While we had some
+problems with typing fast enough (`ncat` timeouted), we got the flag
+using this approach in the end.