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.

f(x) has value 150 at x= 0

and 0 at x= 3

so d(f(x))/dt = slope = -150/3 = -50

a 4/6 = 9/12   not proportional

step-by-step explanation:

to find which are not proportional, we can use cross products

if it is true, they are proportional.   if it is not true, they are not proportional.

a 4/6 = 9/12

4* 12 = 6*9

48 = 54

not this is not proportional  

b 2/3 = 12/18  

2* 18 = 3 * 12

36 = 36

this is proportional

c 5/4 = 10/8  

5*8 = 4*10

40 = 40

this is proportional

d 7/5 = 35/25

7*25 = 5 * 35

175 = 175

is proportional