Determine all possible triplet(s) (P, Q, N) of duodecimal positive integers, with P<Q and N≥3, such that the base 12 representations of P
N and Q
N will together contain each of the digits 0 to B exactly once. Neither P
N nor Q
N can contain any leading zero.
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Submitted by K Sengupta
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Rating: 5.0000 (1 votes)
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Solution:
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(Hide)
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(P, Q, N) = (6, 763, 3), (12,305, 3), (16, 40B, 3), gives:
(P3, Q3) = (160, 2B5497A83), (1708, 23B469A5), (3460, 578A192B)
For an explanation, refer to the solution submitted by Charlie in this location.
Charlie has also submitted all solutions for the case n=2 in this location. |
