Move Project Euler under puzzles.

puzzles/project_euler/p009_SpecialPythagoreanTriplet.py

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+#
+#Special Pythagorean triplet
+#Problem 9
+#
+#A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
+#
+#a^2 + b^2 = c^2
+#For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.
+#
+#There exists exactly one Pythagorean triplet for which a + b + c = 1000.
+#Find the product abc.
+#
+#Solution:
+#
+
+# a^2+b^2=c^2=(1000-a-b)^2= 1000^2-2000(a+b)+(a+b)^2=
+# = 1000^2-2000(a+b)+a^2+2ab+b^2 =>
+# => 0=1000^2-2000(a+b)+2ab =>
+# => 500000=1000(a+b)-ab
+# => 500=a+b - ab/1000
+# => ab is dividable to 1000=2*2*2*5*5*5
+# => a dividable at least to 5 and b to 2
+#
+#for a in range(5,1000,5):
+#    for b in range(2,1000-a,2):
+#        i+=1
+#        if (a**2+b**2) == (1000-a-b)**2:
+#            print( a, b, 1000-a-b, '->', a**2, '+', b**2, '=', a**2+b**2 )
+#            print( ' while (1000-a-b)**2 =', (1000-a-b)**2 )
+#            print( 'product =', a*b*(1000-a-b) )
+#            print( 'product =', (a+b)*(5*10**5+10**3) - (a+b)**2 - 5*10**8 )
+#
+# a better solution
+# => 500000=1000(a+b)-ab
+# => (500000-1000a)/(1000-a)=b
+# => 1000(500-a)/(1000-a)=b
+
+for a in range(1,500):
+    r=1000*(500-a)
+    q=1000-a
+    b=r//q
+    if (a**2+b**2) == (1000-a-b)**2:
+        print( a, b, 1000-a-b, '->', a**2, '+', b**2, '=', a**2+b**2 )
+        print( ' while (1000-a-b)**2 =', (1000-a-b)**2 )
+        print( 'product =', a*b*(1000-a-b) )
+        print( 'product =', (a+b)*(5*(10**5)+10**3) - (a+b)**2 - 5*(10**8) )