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STatiana [176]
3 years ago
11

Six-digit integers will be written using each of the digits 1 through 6 exactly once per six-digit integer. How many different p

ositive 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.)
Mathematics
1 answer:
GenaCL600 [577]3 years ago
3 0

Answer:

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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The complete question is

"One vertex of a polygon is located at (3, –2). After a rotation, the vertex is located at (2, 3). Which transformations could have taken place? Select two options. R0, 90° R0, 180° R0, 270° R0, –90° R0, –270°"

The transformations that could have taken place are R0, 270°or R0, -90°

Let the vertex be V.

So, we have:

V = (3, –2)  before rotation

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The possible transformations from V to V' are;

A 90 degrees counterclockwise rotation (r0, -90) is given :

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So,

(3, –2) --- (2, 3)

A 270 degrees clockwise rotation (r0, 270) is given :

(x, y) --(-y, x)

So,

(3, –2) --- (2, 3)

By comparing the endpoints of the rotations to V' = (2, 3), we can conclude that the transformations that could have taken place are: r0, 270°or r0, -90°.

Read more about transformations at:

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