[puzzle] another little problem

10092017, 03:03 AM
(This post was last modified: 10092017 03:04 AM by AlexFekken.)
Post: #3




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+n31) 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  10062017, 09:03 PM
RE: [puzzle] another little problem  Thomas Okken  10072017, 07:15 PM
RE: [puzzle] another little problem  AlexFekken  10092017 03:03 AM

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