Let $f(n)= 10^{n}+3 \times 4^{n+2} +5$

Question

Let $f(n)$= $10^{n}+3*4^{n+2} +5$ ; $ n \in N$. The greatest integer which divides $f(n)$ for all n is:

(a) 27

(b) 9

(c) 3

(d)none of these

Answer 1

Hint:

$$10f(n)-f(n+1)=18*4^{n+2}+45=9[2*4^{n+2}+5]$$ and $$f(1)=207=9*23$$

Answer 2

very easy,if we find one of the answer we are happy,so we will have

$f(n)=10^n+3*4^n*16+5$

right? so we have $f(n)=10^n+48*4^n+5$

if $n=0$ then $f(0)=1+48+5=54$

because $54/9=6$

clearly D and c answer is eliminated,could you continue please?

because $27=9*3$

if we get answer which would be let say

$f(n)=3*f(0)$

if this system has solution,then largest is $27$

let us solve following system

$10^n+48*4^n+5=3*54$

or

$10^n+48*4^n-157=0$

we see that $10^n$ is ended always with $0$,while

for $48*4^n$ for different $n$

$n=0$--> we have $48*4^0=1$

$n=1$ we have $48*4=192$

$n=2$ --> $48*16=768$

$n=3$ --> $48*64=3072$

$n=4$ --> $48*256=12288$

for minus it has not meaning,but for positive do you see pattern?it will never equal any number which is ended with $7$

Answer 3

Every common divisor of $f(n+1)$ and $f(n)$ divides $f(n+1)-f(n)=9(10^n+4^{n+2})$.