41 lines
1.0 KiB
Python
41 lines
1.0 KiB
Python
def pd(num):
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factors = []
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for divisor in range(1,1+num//2):
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if num % divisor == 0: factors.append(divisor)
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return factors
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def pdc(num):
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count = 0
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for divisor in range(1,1+num//2):
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if num % divisor == 0: count += 1
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return count
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def fmtres(title, lmt, best, bestc):
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return "The " + title + " number up to and including " + str(lmt) + " with the highest number of proper divisors is " + str(best) + ", which has " + str(bestc)
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def showcount(limit):
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best, bestc, bh, bhc = 0, 0, 0, 0
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for i in range(limit+1):
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divc = pdc(i)
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if divc > bestc: bestc, best = divc, i
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if divc >= bhc: bhc, bh = divc, i
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if best == bh:
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print(fmtres("only", limit, best, bestc))
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else:
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print(fmtres("lowest", limit, best, bestc))
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print(fmtres("highest", limit, bh, bhc))
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print()
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lmt = 10
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for i in range(1, lmt + 1):
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divs = pd(i)
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if len(divs) == 0:
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print("There are no proper divisors of", i)
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elif len(divs) == 1:
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print(divs[0], "is the only proper divisor of", i)
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else:
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print(divs, "are the proper divisors of", i)
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print()
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showcount(20000)
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showcount(25000)
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