Solve problem 108 in Python

python/e108.py

@@ -0,0 +1,36 @@
+from fractions import Fraction
+
+
+def get_distinct_solutions(n):
+    """ Get the number of distinct solutions for 1/x + 1/y = 1/n. The idea is
+    that 1/n*2 + 1/n*2 = 1/n is always a solution. So what we do is
+    brute-force all 1/x for x in [2, n * 2]. We have a solution if the
+    difference between 1/n and 1/x has a numerator of 1. """
+    n_inv = Fraction(1, n)
+    n_distinct = 0
+    for x in range(2, n * 2 + 1):
+        x_inv = Fraction(1, x)
+        y_inv = n_inv - x_inv
+        if y_inv.numerator == 1:
+            n_distinct += 1
+    return n_distinct
+
+
+def euler_108():
+    d_max, n_prev = 0, 0
+    # I arrived at the starting values empirically by observing the deltas.
+    for n in range(1260, 1000000, 420):
+        d = get_distinct_solutions(n)
+        if d > d_max:
+            # print("n={} d={} delta={}".format(n, d, n - n_prev))
+            n_prev = n
+            d_max = d
+        if d > 1000:
+            return n
+
+
+if __name__ == "__main__":
+    solution = euler_108()
+    print("e108.py: " + str(solution))
+    assert(solution == 180180)
+