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Given that $F(1)=1,F(2)=1$ and $F(n)=F(n-1)+F(n-2)$ , $n>2$ for this series, now if $f(n)$ is the remainder if $F(n)$ is divided by $5$ then the value of $f(2000)$ is

  1. 1
  2. 0
  3. 3
  4. 2

For doing this it is obvious that problem is indicating the Fibonacci Series. Now we know that $n th$ term of Fibonacci series can be determined by the formula

$$[ (Phi)^n - (phi)^n ]/\sqrt 5 , Phi=(1+\sqrt5)/2 , phi=(1-\sqrt5)/2 $$

I dont no whether this is needed for this problem or not but I observed that $F(5)=5, F(10)=55, F(15)=610$ so I guess $F(2000)$ must be divisible by 5 so answer is 0.

I'm not satisfy with this and need better explanation or analytic solution. help me...

Sourav
  • 281

3 Answers3

2

To prove your statement "$F(5)=5,F(10)=55,F(15)=610$ so I guess $F(2000)$ must be divisible by $5$ so answer is $0$", you may use mathematical induction to prove "$F(5n)$ is divisible by $5$".

The base case, $F(5) = 5$, is true.

Assume the case $n = k$ is true, therefore we can write $F(5k) = 5p$ for some positive integer $k$ and $p$, now for $n = k+1,$

$F(5(k+1)) = F(5k+5) = F(5k+4) + F(5k+3) = 2 \times F(5k+3) + F(5k+2) $

$ = 2 \times (F(5k+2) + F(5k+1)) + F(5k+1) + F(5k) $

$ = 2 \times F(5k+2) + 3 \times F(5k+1) + F(5k) $

$ = 2 \times (F(5k+1) + F(5k)) + 3 \times F(5k+1) + F(5k) $

$ = 5 \times F(5k+1) + 3 \times F(5k) $

$ = 5 \times F(5k+1) + 15p$

$ = 5 \times (F(5k+1) + 3p)$

Hence by mathematical induction, $F(5n)$ is divisible by $5$ for all positive integers $n$.

From the proof, you can also see that $F(n+5) = 5\times F(n+1) + 3\times F(n)$. Therefore if $F(n)$ is divisible by 5, so is $F(n+5)$. Applying this repeatedly you will get $F(n)$ is divisible by $5$ when $n$ is divisible by $5$. This is a simple explanation of your observation.

LaBird
  • 1,018
0

From wiki, $F_{5m}$ is divisible by $F_5=5$ where $m$ is any integer

See also : on the divisibility properties of fibonacci numbers and Fibonacci GCD

lab bhattacharjee
  • 274,582
0

Compute the recurrence, $\bmod5$. As there are just $25$ possible pairs of input values, you quickly find the period.

$$\color{blue}{1, 1, 2, 3, 0, 3, 3, 1, 4, 0, 4, 4, 3, 2, 0, 2, 2, 4, 1, 0,} 1, 1, 2, 3, 0, 3, 3, 1, 4, 0, \cdots$$