Demonstrate that it is divisible by 9 and 13

Question

a) $10^{n+3}.4^{n+2}+5$ divisible by 9

b) $2^{12n+9}-5^{4n+1}$ by 13

What I tried:

a) ${{10}^{n+4}}\cdot {{4}^{n+3}}+5=40\cdot \left( {{10}^{n+3}}\cdot {{4}^{n+2}}+5 \right)-200+5$

b) ${{2}^{12n+21}}-{{5}^{4n+5}}={{2}^{12}}\cdot {{2}^{12n+9}}-{{5}^{4}}\cdot {{5}^{4n+1}}=4096\cdot \left( M13+{{5}^{4n+1}} \right)-625\cdot {{5}^{4n+1}}=M13+3471\cdot {{5}^{4n+1}}$

Answer 1

If you are familiar with mod, then (b) can be done fairly easily. \begin{align*} 2^{12n+9} & \equiv (2^{3})^{4n+3} \pmod{13}\\ & \equiv (-5)^{4n+3} \pmod{13}\\ & \equiv -(5)^{4n+3} \pmod{13}\\ 2^{12n+9}-5^{4n+1} & \equiv -(5)^{4n+3}-5^{4n+1} \pmod{13}\\ & \equiv -(5)^{4n+1}[25+1] \pmod{13}\\ & \equiv 0 \pmod{13}. \end{align*}

Answer 2

For a) take $n=1$ then we get $$640005$$ which is not divisible by $9$

Answer 3

a) Doesn't look right. Double check the expression.
b) Notice $3471 = 13\times 267$

Answer 4

first one by polynomial remainder theorem ( treating 10 as x), goes to $4^{n+2}+5$ which then is equivalent mod 9 to : $4^n\equiv 7$ which fails.

The second via Fermat's little theorem, and the square of 5 being -1 mod 13,which squares to 1,falls to $2^9-5\equiv 507\equiv 0\bmod 13$