Problem 120

Comments

I’m guessing there was some mathmematical trick here that I was missing. I wasn’t sure how to find out how many values of n to test for each value of a. I first tried 1000 and that wasn’t the right answer. I noticed that 500 gave me a smaller value. I tried 2000 and got the right answer, so I guess what was good enough!

Code

count = 0

for a in range(3, 1001):
	Rmax = 0
	print a
	for n in range(2000):
		r = ((a-1)**n + (a+1)**n) % a**2
		if r > Rmax:
			Rmax = r
	count += Rmax

print count