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## Question

With a tip of the cap to Anza Power, here's a Fibonacci puzzle with smaller numbers.

In today's class, Mr. Bigollo taught his students the Fibonacci series and Golden Ratio.

He noted the series traditionally begins 0, 1, 1, 2, 3, 5, 8, ... , but actually any two

numbers, not necessarily in ascending order, can seed the process.

To make the subject compelling, Mr. Bigollo announces to the class that you have

proven yourself among the Brain Denizens as particularly adept at solving seemingly

impossible mathematical puzzles. Furthermore, you have accepted his invitation to

their session today to participate in a demonstration of the amazing properties of

the Fibonacci series. You stand and smile, graciously, at the adoring stares of the class.

To begin, Mr. B selects ten students to line up in the front of the classroom.

Amazingly, their names are Al, Bob, Chuck, Dave, Ed, Frank, Geoff, Hal, Irv and Jack.

Here's what he asks them to do.

1. Al is to select an integer at random and whisper it to Chuck.
2. Bob also selects an integer at random and whispers it to Chuck.
3. Chuck sums the numbers he hears and whispers his sum and Bob's number to Dave.
4. Dave sums the numbers he hears and whispers his and Chuck's sums to Ed.
5. Ed sums the numbers he hears and whispers his and Dave's sums to Frank.
6. Frank sums the numbers he hears and whispers his and Ed's sums to Geoff.
7. Geoff sums the numbers he hears and whispers his and Frank's sums to Hal.
8. Hal sums the numbers he hears and whispers his and Geoff's sums to Irv.
9. Irv sums the numbers he hears and whispers his and Hal's sums to Jack.
10. Jack sums the numbers he hears.

Finally, each student whispers his own sum [Al and Bob whisper their chosen integers]

to Mr. Bigollo, who writes the sum of the ten numbers on a sheet of paper, which he

then places in an envelope.

To clarify: the envelope contains the sum of the first ten Fibonacci numbers generated

by Al and Bob's initial integer choices. No one but Mr. Bigollo knows this sum.

The students take their seats and Mr. Bigollo invites you to the front of the classroom

and announces to the class that you are now prepared to announce the number

contained in the envelope.

Noting your shock, and before you can protest, Mr, B. smiles and says: Well that

would be impossible of course. So, you may have one piece of information. You may

ask one of the students what his sum was. [That eliminates Al and Bob, they didn't

sum anything, but neither of them alone could supply useful information.]

Now your reputation is on the line.

Which student do you ask, and what is the sum?

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Finally, each student whispers his own sum

Would that be the sum the student hears or the sum the student hears + his number?

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The sum he creates.

Chuck's sum e. g. Is Al's plus Bob's integers.

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the sum Geoff got and multiplied it by 11 to get the final total.

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I was gonna say

Mr B is a student teacher so you can ask him.

but = W = has the right answer.

Let F(x) = Fib(x) correspond to the usual Fibonnaci sequence.

Let G(x) correspond to the sequence seeded by Al and Bob.

G(0) = Al's number, G(1) = Bob's number.

If you draw up a table decomposing G(x) to a combination of G(0) and G(1) you will see

G(6) = F(5) * G(0) + F(6) * G(1)

and

Sum from n = 0 to n = 9 of G(n) = F(11) * G(0) + (F(12) - 1) * G(1)

Now it just so 'happens' that...

<you know>

So you can have the answer even though you cannot determine the sequence.

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this is a trial and error solution, because there is no particular solution at any given "n" number of the series:

Y = F(x)

Y = X1 + X2 + (X1 + X2) +..... (Xn-2 + Xn-1) + Xn

if n=3 then:

Y = X1 + X2 + X3 but X1 + X2 =X3

Y = 2*X3 obviously you ask the 3rd kid....

but if n = 4, this has no solution since you will always arrive with having 2 variable on the right side of the equation.

so I did a tabulation in Excel and had proven =W= 's answer is correct, since: when n=10, the sum of the series "Y" divided by the 7th seed "X7" is always equal to 11. Y = G*11

series X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 variables 11 4 15 19 34 53 87 140 227 367 Qoutient 87 239 63.8 50.4 28.1 18.1 11 6.84 4.22 2.61 Sum 957

Edited by aquila828
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Actually, the answer is in one of the properties of the Fibonacci series.

The sum of any 10 adjacent numbers = 11 times the 7th of the ten.

Edited by Rui Azevedo
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Jack

Let us use a representation A(p,q) where p is A's no and q his sum

Let A's chosen no be a and B's be b

so we have

A(a)

B(b)

C(a+b,b)

D(a+2b,a+b)

E(2a+3b,a+2b)

F(3a+5b,2a+3b)

G(5a+8b,3a+5b)

H(8a+13b,5a+8b)

I(13a+21b,8a+13b)

J(21a+34b)

On looking carefully,you might notice that what each person is actually doing is summing the number of previous 2 persons.nth person's no is the sum to n terms of the series.So the answer should be Jack

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Jack

Let us use a representation A(p,q) where p is A's no and q his sum

Let A's chosen no be a and B's be b

so we have

A(a)

B(b)

C(a+b,b)

D(a+2b,a+b)

E(2a+3b,a+2b)

F(3a+5b,2a+3b)

G(5a+8b,3a+5b)

H(8a+13b,5a+8b)

I(13a+21b,8a+13b)

J(21a+34b)

On looking carefully,you might notice that what each person is actually doing is summing the number of previous 2 persons.nth person's no is the sum to n terms of the series.So the answer should be Jack

Wouldn't Jack's number be the 10th number in the series and not the sum of the 10 numbers as is requested? :|

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