For $n \ge 0$, let $f_n$ denote the $n$th fibonacci number (that is, $f_0 = 0, f_1 = 1$, and $f_n = f_{n-1} + f_{n-2}$ for all $n \ge 2$). what is the greatest possible value of the greatest common divisor of two consecutive fibonacci numbers?

2425

step-by-step explanation:

a.) list the factors of 40.

1/2/4/5/8/10/20/40

b.) list the factors of 40 that represent the number of cans that could be on each shelf.

5

c.) how many different arrangements of cans are possible?

just one because if you put more on one shelf you will go way over 40 and the cabinet will probably break.

d.) what are the possible arrangements of the cans?

put 5 cans on each self to reach 40 cans in total.

e.) which arrangement uses all the shelves?

5 because you can put 5 cans on each shelf, and since there are 8 shelfs you are equally going to get 40 cans.

-1   1/6.   you dont subtract the denominators

step-by-step explanation:

b. b < 1 might be the correct answer