Lukas Werner
23 lines
358 B
Python
23 lines
358 B
Python
num = 2520
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factors = [i for i in range(1, 20+1)]
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def perf_num(num):
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for f in factors:
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if num % f == 0:
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pass
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else:
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return False
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else:
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return True
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i = 1
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while True:
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if i % 1000 == 0:
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print("status", i, end='\r')
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if perf_num(i):
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print("found num", i)
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break
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i += 1
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