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[puzzle] another little problem
10-09-2017, 03:03 AM (This post was last modified: 10-09-2017 03:04 AM by AlexFekken.)
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RE: [puzzle] another little problem
Looks correct me to me.

And in general, if n1 is the number of distinct prime factors that the gcd and lcm have in common with the same exponent, n2 the number that they have in common with a different exponent, and n3 the number in the lcm but not the gcd, then the answer would be:

2^(n2+n3-1) if (n2+n3) > 0
1 if n2=n3=0, i.e. gcd = lcm = x = y

That the gcd and lcm are factorials is irrelevant, but makes it easy to count the prime factors.
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[puzzle] another little problem - pier4r - 10-06-2017, 09:03 PM
RE: [puzzle] another little problem - AlexFekken - 10-09-2017 03:03 AM

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