Proof that $n = 3k + 5l$ for $n > 7$ 
Show that for every n greater than $7$, there are non-negative integers $k$ and $l$ such that $$n = 3k+ 5l.$$

So induction seems like a possibility.
$n = 3k + 5l$ and so $n + 1 = 3k + 5l + 1$. 
What can be done in such a case? Can I get a hint?
 A: Obviously every multiple of $3$ can be written in this form taking $l=0
 $. So assume $n=3k+1$. we have $$n=3k+1=3k+10-9=3\left(k-3\right)+5\cdot2
 $$ and if $n=3k+2
 $ we have $$n=3k+2=3k+5-3=3\left(k-1\right)+5
 $$ so it is sufficient to note that $8=5+3
 $ and we have done, since for $n\geq9
 $ we have $k\geq3.$
A: HINT:


*

*$A_{ 8}=\{3,5\}$

*$A_{ 9}=\{3,3,3\}$

*$A_{10}=\{5,5\}$

*$A_{ n}=A_{n-3}\cup\{3\}$

A: By strong induction:  We can write
\begin{align*}
8 & = 1 \cdot 3 + 1 \cdot 5\\
9 & = 3 \cdot 3 + 0 \cdot 5\\
10 & = 0 \cdot 3 + 2 \cdot 5
\end{align*}
Let $n \geq 10$.  Assume that we can write each integer $m$ such that $8 \leq m \leq n$ in the form $m = 3k + 5l$ for some non-negative integers $k, l$.  We wish to show that $n + 1$ can also be written in this form.  Since $n + 1 \geq 10 + 1 = 11$, $n + 1 - 3 = n - 2 \geq 8$, by the induction hypothesis, there exist non-negative integers $k'$ and $l'$ such that $n - 2 = 3k' + 5l'$.  Therefore, $$n + 1 = n - 2 + 3 = 3k' + 5l' + 3 = 3(k' + 1) + 5l'$$  Hence, each integer $n > 7$ can be expressed in the form $n = 3k + 5l$, where $k$ and $l$ are non-negative integers.     
A: This uses  a variant of induction that ascends by $j\,$ from $\,j\,$ base cases  (here $\,j = 3).\,$  Notice $\,P(n)\,\Rightarrow\,P(n\!+\!3)\,$ by adding $\,1\,$ to $\,k.\ $ Write $\,P\,$ for the set of  $\,n\ge 8$ where $\,P\,$ holds true. Therefore, by the Theorem below, the truth of $\,P\,$ at $\,8,9,10\,$ ascends to all $\,n\ge 8.$ 
Theorem $\ $ Suppose $\,P\subseteq \Bbb N\,$ satisfies $\,n\in P\,\Rightarrow\, n\!+\!3\in P,\ $ for all $\,n\ge a.\ $ Then
$$\,a,a\!+\!1,a\!+\!\color{#c00}2\in P\,\Rightarrow\,n\in P{\rm\, \ for\ all\,\ } n\ge a$$
Proof $\ $ If not there is a least counterexample $\,\ell\not\in P.\,$ By our base hypothesis $\,\ell \ge a\!+\!\color{#c00}3\,$ so $\,\ell\!-\!3\ge a.\,$ Hence by our shift-closure hypothesis $\,\ell = (\ell\! -\! 3)+ 3\in P,\,$ contradiction.
Remark $\ $ Clearly the proof generalizes from the shift increment $\,j=3\,$ to arbitrary $\,j\ge 1,\,$ with $\,j\,$ consecutive integers $\,a,a\!+\!1,\ldots,a\!+\!j\!-\!1\,$ serving as the base cases, i.e. the foundation of the induction. Notice that the case $\,j=1\,$ is simply ordinary induction.
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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Any integer number $n > 7$ can be written as $n = 3\ell + \delta$ where $\delta \in \braces{0,1,2}$ and $\ell \geq 2$. $\ell$ is an integer.


*

*$\delta = 0$ is trivial: $n = 3\ell + 5\times 0$.

*$\delta = 1 = 10 - 9\quad\imp\quad n = 3\pars{\ell - 3} + 5\times 2$.

*$\delta = 2 = 5 - 3 \quad\imp\quad n = 3\pars{\ell -1} + 5\times 1$.

