Strong Induction: Every natural number $n\geq 8$ can be represented as $n=3k + 5\ell$ Can you please help me and tell, how should I move on? 
Can this be proved by induction?

Every natural number $n\geq 8$ can be represented as $n=3k + 5\ell$.

Thank you in advance

We can avoid an explicit appeal to induction by using the fact that every natural number $n$ has remainder $0$, $1$, or $2$ on division by $3$.  Let $n \ge 8$.
If $n$ has remainder $2$ on division by $3$, then $n-8$ is divisible by $3$, say $n-8=3m$. Represent $8$ using $8=3\cdot 1+5\cdot 1$. Then add $m$ $3$'s. 
If $n$ has remainder $0$ on division by $3$, then $n-9$ is divisible by $3$, say $n-9=3m$. Represent $9$ using $9=3\cdot 3 +5\cdot 0$. Then add $m$ $3$'s. 
If $n$ has remainder $1$ on division by $3$, then $n-10$ is divisible by $3$, say $n-10=3m$. Represent $10$ using $10=3\cdot 0 +5\cdot 2$. Then add $m$ $3$'s. 
The argument for remainder $0$ was a little silly, since if $n$ has remainder $0$ on division by $3$, we can clearly use a bunch of $3$'s to represent $n$. But we wanted the solutions for the three cases to use a single template. 
