Answer: For the first positive integer, when \(n=1\),

\(10^{1}+3\times 4^{1+2}+5=10+3\times 4^{3}+5=10+192+5=207=9\times 23\),

so the expression is divisible by \(9\).

Now assume that for some positive integer \(k\),

\(10^{k}+3\times 4^{k+2}+5\)

is divisible by \(9\). We will show that the case \(n=k+1\) is also divisible by \(9\).

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

\(f(n+1)-f(n)=\bigl(10^{n+1}-10^n\bigr)+3\bigl(4^{n+3}-4^{n+2}\bigr)\).

So

\(f(n+1)-f(n)=10^n(10-1)+3\cdot 4^{n+2}(4-1)=9\bigl(10^n+4^{n+2}\bigr)\),

which is divisible by \(9\).

Since \(f(k)\) is divisible by \(9\) by the inductive hypothesis, and \(f(k+1)-f(k)\) is divisible by \(9\), it follows that \(f(k+1)\) is also divisible by \(9\).

Therefore, by mathematical induction, \(10^{n}+3\times 4^{n+2}+5\) is divisible by \(9\) for all positive integers \(n\).

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

Base case: when \(n=1\),

\(f(1)=10+3\times 4^3+5=10+192+5=207=9\times 23\).

Hence \(f(1)\) is divisible by \(9\).

Inductive step: assume that for some positive integer \(k\), \(f(k)\) is divisible by \(9\). We must show that \(f(k+1)\) is divisible by \(9\).

Calculate the difference:

\(f(k+1)-f(k)=\bigl(10^{k+1}+3\times 4^{k+3}+5\bigr)-\bigl(10^k+3\times 4^{k+2}+5\bigr)\).

This simplifies to

\(f(k+1)-f(k)=10^k(10-1)+3\times 4^{k+2}(4-1)\)

\(=9\bigl(10^k+4^{k+2}\bigr)\).

So \(f(k+1)-f(k)\) is divisible by \(9\).

Since both \(f(k)\) and \(f(k+1)-f(k)\) are divisible by \(9\), their sum \(f(k+1)\) is also divisible by \(9\).

Therefore \(f(k)\) divisible by \(9\) implies \(f(k+1)\) divisible by \(9\).

As the base case is true, it follows by mathematical induction that \(10^{n}+3\times 4^{n+2}+5\) is divisible by \(9\) for all positive integers \(n\).