Rabbit Puzzle

@pradeep-agrawal-rhdX5z • Oct 27, 2024

Oct 27, 2024

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There is a closed room where we left a rabbit. That rabbit has a unique property. It starts giving birth after 2nd month of its own birth (start of third month), and after that it gives birth in every month. It delivers a single rabbit in each time. And these descendant rabbits inherit the same property.

Let's assume that the room is large enough to hold infinite number of rabbits, and there is enough food, water etc. in the room for infinite number of rabbits. How many rabbits will be there in the room after n months?

-Pradeep

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shalini_goel14

@shalini-goel14-ASmC2J • Jul 2, 2009

Sorry , it will be sum of following series

(n-1) + (n-1)(n-2) +(n-1)(n-2)(n-3)+... [(n-1)(n-2)..{n-(n-1)}]
pradeep_agrawal

@pradeep-agrawal-rhdX5z • Jul 2, 2009

I edited the post to make question more clear.

-Pradeep
Anil Jain

@CrazyBoy • Jul 2, 2009

shalini_goel14
Sorry , it will be sum of following series

(n-1) + (n-1)(n-2) +(n-1)(n-2)(n-3)+... [(n-1)(n-2)..{n-(n-1)}]
I have not calculated till now... but certainly this is not the asnwer..

Consider n= 1
As per your expression, value would be 0

However 1 rabit should be the value....

-CB
shalini_goel14

@shalini-goel14-ASmC2J • Jul 2, 2009

crazyboy
I have not calculated till now... but certainly this is not the asnwer..

Consider n= 1
As per your expression, value would be 0

However 1 rabit should be the value....

-CB
No each rabbit can give birth only in second month so for 1 month[i.e n=1] there will be 0 rabbit.

PS: Correct me if wrong.
pradeep_agrawal

@pradeep-agrawal-rhdX5z • Jul 2, 2009

shalini_goel14
No each rabbit can give birth only in second month so for 1 month[i.e n=1] there will be 0 rabbit.

PS: Correct me if wrong.
Yep, you are wrong here. Even in the first month we have 1 rabbit (the unique rabbit who starts the chain) 😀

And a rabbit can give birth after second month (i.e., start of third month). This is the clarification i have done by editing the puzzle.

-Pradeep
shalini_goel14

@shalini-goel14-ASmC2J • Jul 2, 2009

pradeep_agrawal
Yep, you are wrong here. Even in the first month we have 1 rabbit (the unique rabbit who starts the chain) 😀

And a rabbit can give birth after second month (i.e., start of third month). This is the clarification i have done by editing the puzzle.

-Pradeep
Ok, I got it , it is wrong. Will post later the correct answer. 😀

Bye ! Keep on trying 😉

PS: "Hurry makes curry" -how true 😡
RajdeepCE

@rajdeepce-7UdrG8 • Jul 2, 2009

After n days the number of rabits will be n[sup]th[/sup] fibonacci number. Here is explanation,
1st month = 1 rabit,
2nd month = 1 rabit,(1st rabit cant give birth now)
3rd month = 2 rabit,(1st rabit starts giving birth)
4th month = 3 rabit,(1st rabit gives birth to another)
5th month = 5 rabit,(both rabit start giving birth)
So the answer will be nth element of fibonacci series.
Anil Jain

@CrazyBoy • Jul 2, 2009

And what would be n^th element ?? Thats what he is looking for.. 😀
RajdeepCE

@rajdeepce-7UdrG8 • Jul 2, 2009

I mentioned that in my posts first and last line. And again after n months the number of rabits will be the n[sup]th[/sup] fibonacci number. And the equation of fibonacci series is,
f(n)=f(n-1)+f(n-2), n=2,3,4,... & f(n-1)=f(n-2)=1.
pradeep_agrawal

@pradeep-agrawal-rhdX5z • Jul 3, 2009

RajdeepCE
So the answer will be nth element of fibonacci series.
Yes that the correct answer that the number of rabbits on nth day will be the nth element of the fibonacci series.

RajdeepCE
And the equation of fibonacci series is,
f(n)=f(n-1)+f(n-2), n=2,3,4,... & f(n-1)=f(n-2)=1.
I feel you mean to say "f(1)=f(2)=1" instead of "f(n-1)=f(n-2)=1".

My next query, is there any simple equation through which i can directly calculate the number of rabbits on nth day (i.e., nth element of fibonacci series) instead of going by formula

f(n)=f(n-1)+f(n-2), f(1)=f(2)=1.

-Pradeep
silverscorpion

@silverscorpion-iJKtdQ • Jul 4, 2009

Ok. That's easy.

The formula for the N[sup]th[/sup] fibonacci number is,

{[(1+sqrt(5))/2][sup]n[/sup] - [(1-sqrt(5))/2][sup]n[/sup]} / sqrt(5)
ie.,

(1+sqrt(5))/2 is nothing but the golden ratio, 1.68. We can call it phi.
So, Nth fibonacci number is,

{phi[sup]n[/sup] - 1/phi[sup]n[/sup]} / sqrt(5);
Am I correct??
pradeep_agrawal

@pradeep-agrawal-rhdX5z • Jul 4, 2009

silverscorpion, not sure how you derived the formula. I tried with different value of n and i am not getting correct answers, e.g.,

For n = 1, answer = 0.485
For n = 2, answer = 1.103
For n = 3, answer = 2.026
For n = 4, answer = 3.506
For n = 5, answer = 5.951
For n = 6, answer = 10.034

I am even not sure if a equation for fibonacci series exist.

-Pradeep
silverscorpion

@silverscorpion-iJKtdQ • Jul 5, 2009

Well, this formula exists and is quite right too. The thing is, you have to round off the answers.
And sorry, I gave the value of phi to be 1.68. It's actually, 1.618.

And another thing I forgot to mention is that, the power should actually be n+1. ie, for getting nth fibonacci number, we have to use phi[sup]n+1[/sup].

So, let's calculate.

For 1, we have

(phi[sup]2[/sup] - 1/phi[sup]2[/sup]) / sqrt(5)
= (1.618[sup]2[/sup] - 0.618[sup]2[/sup]) / sqrt(5) = 1.0

For 2, we have

(phi[sup]3[/sup] - 1/phi[sup]3[/sup]) / sqrt(5)
= (1.618[sup]3[/sup] - 0.618[sup]3[/sup]) / sqrt(5) = 2.0

Try it for other numbers too. It will work. Let me know if you got it!! 😀😀
pradeep_agrawal

@pradeep-agrawal-rhdX5z • Jul 5, 2009

That's cool silverscorpion, this works 😀

-Pradeep
silverscorpion

@silverscorpion-iJKtdQ • Jul 5, 2009

I said it would!!

Happy?? 😀😀
mech_guy

@mech-guy-TMwuj2 • Jul 5, 2009

Hi,

Fibonacci series sum was cool indeed. I got it other way round and my answer is:-
if n >= 3 Total no. of rabbits = 2 + [(n-2)*(n-3)/2]
if n = 1 Total no. of rabbits = 1
if n = 2 Total no. of rabbits = 2
I made following sequence :-

Months:- 1__2__3__4__5__6__7__8__9 ..............
Rabbits:- _1__1__1__1__1__1__1__1__1
__________________1__1__1__1__1__1
_____________________1__1__1__1__1
________________________1__1__1__1
___________________________1__1__1
______________________________1__1
_________________________________1
----------------------------------------
Total rabbits after n months:- 1 + 1 + (1 +2 + 3 + 4 + 5 + 6 + 7 + ....(n-2) terms)
= 2 + SIGMA(n on (n-2) terms)
= 2 + [(n-2)*(n-3)/2] => assuming n>=3 else total = 1 for n = 1 and total = 2 for n = 2

Thanks

PS: my first post, didnt knw how to properly type, so using lots of underscores, hope its comprehendible.
pradeep_agrawal

@pradeep-agrawal-rhdX5z • Jul 5, 2009

That was also a good try mech_guy. But the formula that you have given works correctly till n = 6. When we have n = 7:

Total no. of rabbits = 2 + [(7-2)*(7-3)/2]
= 2 + [5*4/2]
= 2 + 10
= 12

But the actual answer is for n = 7 should be 13.

- Pradeep
mech_guy

@mech-guy-TMwuj2 • Jul 5, 2009

Hi Pradeep,

Formula for SIGMA (1 + 2 + 3....+ n) should be = n*(n+1)/2 while i did calculation for n*(n-1)/2 and that is why got wrong Total.

So total should be 2 + [(n-2)*(n-1)/2]

For n = 7, total number of rabbits after 7 months = 17 (how 13?)

Regards
silverscorpion

@silverscorpion-iJKtdQ • Jul 5, 2009

This is not correct. Using this formula, you get the series,

1, 2, 3, 5, 8, 12, 17, 23, 30,...

This is not fibonacci series.
mech_guy

@mech-guy-TMwuj2 • Jul 5, 2009

Hi SilverScorpion,

Agreed its not a Fibonacci series.
And yes my solution is wrong, i messed it up after 6 month onwards. It was a bad try.

Regards
silverscorpion

@silverscorpion-iJKtdQ • Jul 6, 2009

mech_guy
Hi SilverScorpion,

Agreed its not a Fibonacci series.
And yes my solution is wrong, i messed it up after 6 month onwards. It was a bad try.

Regards

You tried!! That is great in itself!!

And now you know the answer! So, enjoy!😁