Six-digit integers will be written using each of the digits 1 through 6 exactly once per six-digit integer. How many different positive integers can be written such that all pairs of consecutive digits of each integer are relatively prime? (Note: 1 is relatively prime
to all integers.)

If 6 is an interior entry then it has to appear in one of the subwords 165 or 561. If all three of {2,3,4} are on the same side of these the digit 3 has to be in the middle of the three (gives 4 ways). If one of {2,3,4} stands alone it has to be one of 2 and 4 (gives another 8 ways). In all there are 24 admissible words of this kind.

If 6 is the first or last entry the entry next to 6 has to be 1 or 5. The remaining four digits could be arranged in 4!=24 ways, but we have to exclude the 2⋅3!=12 ways containing 24 or 42 as a subword. It follows that there are 2⋅2⋅(24−12)=48 admissible words of this kind.

The total number of admissible words therefore is 72.

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step-by-step explanation:

answer: c

(2, -4)

goes 7 to the

(9, 4)

easy