Question

If $n\in N,f\left(n\right)={37}^{n+2}+{16}^{n+1}+{30}^n$, then

• A. $f\left(n\right)+1$ is divisible by $3$
• B. $f\left(n\right)$ is divisible by $3$
• C. $f\left(n\right)$ is divisible by $7$
• D. $f\left(4\right)$ is divisible by $5$

Answer 1

Answer: (A), (C), (D)

$f\left(4\right)$ is divisible by $5$

${37}^{n+2}=(36+1{)}^{n+2}=36k_1+1{16}^{n+1}=(15+1{)}^{n+1}=15k_2+1{30}^n$
Where $k_1,k_2\in N$
Now,
$f\left(n\right)+1=36k_1+15k_2+{30}^n+3$
Therefore, this is divisible by $3$.

${37}^{n+2}=(35+2{)}^{n+2}=35k_3+2^{n+2}{16}^{n+1}=(14+2{)}^{n+1}=14k_4+2^{n+1}{30}^n=(28+2{)}^n=28k_5+2^n$
Where $k_3,k_4,k_5\in N$
Now,
$f\left(n\right)=35k_3+14k_4+28k_5+2^n(2^2+2+1)\Rightarrow f\left(n\right)=35k_3+14k_4+28k_5+2^n\left(7\right)$
Therefore, this is divisible by $7$.

${37}^{n+2}=(35+2{)}^{n+2}=35k_3+2^{n+2}{16}^{n+1}=(15+1{)}^{n+1}=15k_2+1{30}^n$
Now,
$f\left(n\right)=35k_3+15k_2+{30}^n+2^{n+2}+1\Rightarrow f\left(4\right)=35k_3+15k_2+{30}^n+65$
Therefore, $f\left(4\right)$ is divisible by $5$.