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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

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    Same: http://math.stackexchange.com/questions/11812222015-04-12

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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.

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    We are finding non-negative solutions of linear diophantine equations. In general, if $a$ and $b$ are relatively prime positive integers, for any n > ab-a-b there are non-negative integers $x$, $y$ such that $ax+by=n$.2011-10-15
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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.

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    @MartianInvader: Oh, sigh. Quite right.2011-10-14
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Hint $ $ If $\rm\,n\,$ is representable so too is $\rm\,n\!+\!\color{#c00}3$ (by adding $1$ to $\rm k)$ so if the $\,\color{#c00} 3\,$ consecutive integers $\,8,9,10\,$ are representable then induction $\Rightarrow$ so too are all larger integers $\,3\!+\![8,9,10],\,$ $\, 6\!+\![8,9,10]\,\ldots$

Remark $ $ 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: $\rm\quad n\ =\ 3\bigg[\dfrac{n-5\:k}{3}\bigg] + 5\:k\ \ $ for $\rm\ \ k\: = \ {-}n\:\ mod\:\ 3,\ \ k \in \{0,1,2\}$

Thus $\rm\ \ 3\:m + \ \ 8\ =\ 3\:m + 3\cdot 1 + 5\cdot 1\ =\ 3\:(m+1) + 5\cdot 1$
and $\rm\ \ \ \ 3\:m + \ \ 9\ =\ 3\:m + 3\cdot 3 + 5\cdot 0\ =\ 3\:(m+3) + 5\cdot 0$
and $\rm\ \ \ \ 3\:m + 10\ =\ 3\:m + 3\cdot 0 + 5\cdot 2\ =\ 3\:(m+0) + 5\cdot 2$

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

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    @Lissa I've added some examples in the answer.2011-10-17