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What is the greatest possible value of the greatest common divisor of two consecutive Fibonacci numbers?

GCD of two consecutive Fibonacci numbers, ALWAYS = 1

1st post walks    5 m + 5 m = 10m

2nd  post walks     10m + 10m = 20

Arithmetic sequence   d = 10   a₁=10  a₂₀₀ = 2000

sum    n(a₁+a₂₀₀)/2  = 200 (10+2000)/2 = 201000      In reality, she doesn't need to walk back after the last post is placed

 so subtract 1000     200 000 = 200 km

See the details of the correct answer here:  https://web2.0calc.com/questions/if-greg-rolls-four-fair-six-sided-dice-what-is-the

sqrt{25z^14y^10} =   5 z⁷ y⁵

"Our assumptions and expectations may give us a perceptual set - a predisposition to interpret an ambiguous stimulus one way to rather than another" Myers and Jeeves (p. 61). In light of this statement, what are two examples of how you and another person have differed in your perception of the same event.   

(1)  Pasquale and I looked up in the sky.  He thought it was a bird.  I thought it was a plane.  We were both wrong. 

(2)  Enrique and I saw a man holding a firearm.  He thought it was a criminal.  I thought it was a plainclothes cop. 

small prize possibility    3 aces  + 3 kings + 11 remaing hearts   =  17 out of 52 =   17/52

Yo are looking for  dollars per ounce     or   dollars/ounce

first one   $ 3.49 / 25 oz  = $ .1396 per ounce

  now do the second one to find which is lower (smaller number)

Example   1:1000           35 mm x 1000      by    45 mm x 10000

                                       35000 mm            by    45 000 mm

                                        35 meters            by    45 meters

    You should be able to do this with the others......

b) 0.008333

There are two ways to fix it

the large squares are too big.....make them the same width as the two smallest squares

   or the smaller squares are too small ....make then as tall as the largest two squares....

An amendment to the terms of service require users to be at least as smart as the Calculator, else the Calculator has the option to refuse service. 

productfor(n, 1, 100, 7^n) % 10^10 = 6696251249 - These are the last 10 digits.

Product =4  x  9 = 36

1 - Add them together, and that is the "maximum"

2 - Subtract the smaller from the larger, and that is the "minimum"

Each die has 6 faces, and the odds of a 6 or 1 showing up are 1 in 6. So, there are several possible victory conditions:

1) One of the 4 die show a 1, and none show 6.

2) Two of the die show one and one or none show 6.

3) Three of the die show one, and 2 or less show 6.

All 4 dice show 6. There are 1296 possible combinations of dice that could be rolled.

The probabilities are as follows.

4 in 81 1 of the 4 dice shows a 1, and none show 6.

1 in 81 2 of the dice show one and one or none show 6.

5 in 1296 3 of the dice show one, and the 4th die result doesn't matter, as long as it isn't 1. 1 in 1269 - All 4 dice show 6.

So, there is a 43 in 648 chance of this occurring.

Best answer I could come up with. Had to do a little googling to confirm my answer and I found this explination. More clearer than what I came up with but same answer.

My computer says the 420th term of 6! = 432651.

Melody or somebody else will give you a "combinatoric" explanation.

thanks you were kind, I had solved it like that too; however it seems strange that a clacolatrice does not solve such a simple calculation

Do you need help with anything?

As  ranges n over the positive integers, what is the maximum possible value that the greatest common divisor of 13n + 8 and 5n +3  can take?

A short computer code shows that no matter what the value of "n" is, the GCD of (13n + 8) and (5n +3) is ALWAYS = 1. 

Sorry, I cannot explain it algebraically, but maybe Alan, Melody, EP....can!.

Possible derivation:
d/dx(log(abs(cos(π/x))))

Using the chain rule, d/dx(log(abs(cos(π/x)))) = ( dlog(u))/( du) ( du)/( dx), where u = abs(cos(π/x)) and d/( du)(log(u)) = 1/u:
 = (d/dx(abs(cos(π/x))))/abs(cos(π/x))

Using the chain rule, d/dx(abs(cos(π/x))) = ( dabs(u))/( du) ( du)/( dx), where u = cos(π/x) and d/( du)(abs(u)) = u/abs(u):
 = ((cos(π/x) d/dx(cos(π/x)))/(abs(cos(π/x))))/abs(cos(π/x))

Simplify the expression:
 = (cos(π/x) (d/dx(cos(π/x))))/abs(cos(π/x))^2

Using the chain rule, d/dx(cos(π/x)) = ( dcos(u))/( du) ( du)/( dx), where u = π/x and d/( du)(cos(u)) = -sin(u):
 = -d/dx(π/x) sin(π/x) cos(π/x)/abs(cos(π/x))^2

Factor out constants:
 = -(cos(π/x) sin(π/x))/abs(cos(π/x))^2 π d/dx(1/x)

Use the power rule, d/dx(x^n) = n x^(n - 1), where n = -1.
d/dx(1/x) = d/dx(x^(-1)) = -x^(-2):
 = -(π cos(π/x) sin(π/x))/abs(cos(π/x))^2 (-1)/(x^2)

Simplify the expression:
 = (π cos(π/x) sin(π/x))/(x^2 abs(cos(π/x))^2)

Simplifying powers, abs(cos(π/x))^2 = cos^2(π/x):

= (π tan(π/x))/x^2     [Courtesy of Mathematica 11 Home Edition]

Before the first bounce it travels 12

  the next bounce is up and down 3   = 6

    the next bounces is up and down   3/4 =  6/4

       geometric sequence   a_{1 = 6}     r = 1/4

sum =   12     + a(1/(1-r))  = 12 + 6 (1/(1-1/4))  = 20 m

You can also verify it this way:

1 - Calculate ALL the "combinations" first, then convert the "combinations" into "permutations":

The combinations are: 112223,  112224,  112225,  112226.

Each combination has this many permutations: 6! / [2!.3!] =720 / 12 =60 permutations each.

60  x  4 = 240 / 6^6 = 5 /972

It appears the output is just the input +2

n+2

Will be this

(x+7)(x-5-3i)(x-5+3i)     

  or 

   Multiply this to get

x³-3x²-36x+238

6C2 * 4C3 * 4 == 240 / 6^6 == 5 / 972

Just multiply the two numbers:

90  x  50  = 4,500 pounds - weight of the rhinoceros.