Proof that $n = 3k + 5l$ for $n > 7$

Question

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?

Answer 1

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 $k\geq3.$

Answer 2

HINT:

• $A_{ 8}=\{3,5\}$
• $A_{ 9}=\{3,3,3\}$
• $A_{10}=\{5,5\}$
• $A_{ n}=A_{n-3}\cup\{3\}$

Answer 3

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.

Answer 4

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.

Answer 5

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ 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.

1. $\delta = 0$ is trivial: $n = 3\ell + 5\times 0$.
2. $\delta = 1 = 10 - 9\quad\imp\quad n = 3\pars{\ell - 3} + 5\times 2$.
3. $\delta = 2 = 5 - 3 \quad\imp\quad n = 3\pars{\ell -1} + 5\times 1$.