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

Question

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

Answer 1

Clearly, $8$ can be represented as $3k+5\ell$, by taking $k=1$ and $\ell=1$.

Likewise, $9=3(3) + 5(0)$; and $10=3(0) + 5(2)$. So we can represent $8$, $9$, and $10$.

Now, assume that for $m\gt 10$, and you can represent all numbers strictly smaller than $m$ that are $8$ or larger (the induction hypothesis). To show that you can represent $m$, consider $m-3$ first.

Answer 2

Hint $ \,n\,$ representable $\Rightarrow$ so too is $\,n\!+\!\color{#c00}3$ (by adding $1$ to $ k)$ so if $\,\color{#c00} 3\,$ consecutive integers $\,8,9,10\,$ are representable then induction $\Rightarrow$ so too are all larger integers $\,3\!+\![8,9,10],\,$ $\, 6\!+\![8,9,10]\,\ldots$

Rigorously: $\!\bmod 3\!:\ 8,9,10\equiv 2,0,1$ is a complete set of residues (remainders) $\!\bmod 3,\,$ so every natural $n$ is congruent to one of them, say $\,n\equiv r,\,$ so $\,n = 3q+r,\,$ and $\,n\ge 8\Rightarrow q\ge 0,\,$ so we can add three $q$ times to $r$ to get $n$, i.e. $\, r = 3k+5\ell\Rightarrow n = 3q+3k+5\ell = 3(q+k)+5\ell$.

Remark $ $ Above the induction is encapsulated in the Division with Remainder Algorithm (the special case $\,n\div 3,\,$ i.e. every natural $n = 3q+r$ for $\,0\le r\le 2)$. The same idea works for any number of integers (or "coins", "stamps", "McNuggets", etc). For some underlying geometric intuition see this post on this so-called Frobenius problem (which includes a list of many of its related names to aid in literature searches).

Explicitly: $\quad n\ =\ 3\bigg[\dfrac{n-5\:\ell}{3}\bigg] + 5\:\!\ell\ \ $ for $\ \ \ell\: = \ {-}n\bmod 3,\ \ \ell \in \{0,1,2\}$

yields the desired representations for all integers $\ge 8.$

This answer gives a general form of the above inductive proof.

Answer 3

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.