Added problem 33 in python

1 files changed, 44 insertions, 0 deletions

diff --git a/python/033-digit_cancelling_fractions.py b/python/033-digit_cancelling_fractions.py
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+++ b/python/033-digit_cancelling_fractions.py
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+###
+# Digit cancelling fractions
+# Problem 33
+#
+# The fraction 49/98 is a curious fraction, as an inexperienced mathematician in attempting to simplify
+# it may incorrectly believe that 49/98 = 4/8, which is correct, is obtained by cancelling the 9s.
+#
+# We shall consider fractions like, 30/50 = 3/5, to be trivial examples.
+#
+# There are exactly four non-trivial examples of this type of fraction, less than one in value,
+# and containing two digits in the numerator and denominator.
+#
+# If the product of these four fractions is given in its lowest common terms, find the value of the denominator.
+###
+
+solution_n = 1
+solution_d = 1
+
+for n in range(10, 100):
+    for d in range(10, 100):
+        if n / d >= 1.0:
+            continue
+        if n % 10 == 0 or d % 10 == 0:
+            continue
+        sn = str(n)
+        sd = str(d)
+        if sn[0] in sd:
+            sd = sd.strip(sn[0])
+            sn = sn[1]
+        elif sn[1] in sd:
+            sd = sd.strip(sn[1])
+            sn = sn[0]
+        else:
+            continue
+        if len(sn) == 0 or len(sd) == 0 or float(sd) == 0:
+            continue
+        if int(sn)/ int(sd) == n / d:
+            print(n, "/", d)
+            solution_n *= n
+            solution_d *= d
+
+
+print(solution_n, "/", solution_d)
+print("= 1 / 100")