Prove every natural number greater than $13$ can be express as sum of $3$ and $8$ (regardless of order)
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What? $3+8 = 11$ so from where I sit only $11$ can be expressed as a sum of $3$ and $8$. – fleablood Mar 21 '21 at 02:32
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1I think he means a linear combination. Perhaps like 20 = 43 + 18. – DOUGLAS BRUNSON Mar 21 '21 at 02:33
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Should I assume you mean that ever natural number greater than $13$ can be written as $3n + 8m$ for $n,m \in \mathbb N$ (including $0$)? – fleablood Mar 21 '21 at 02:35
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"I think he means a linear combination. " It that is what the OP meant he should have made a bit of effort to be clearer. – fleablood Mar 21 '21 at 02:35
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This answer may answer your question. There, it shows that any number greater than or equal to $(3-1)(8-1)=14$ can be written as a non-negative linear combination of $3$ and $8$. – robjohn Mar 21 '21 at 02:36
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exactly, 3n + 8n – daniel perez maximo Mar 21 '21 at 02:37
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1just sshow you can get 14 15 and 16 – hunter Mar 21 '21 at 02:55
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@hunter: that is sufficient. – robjohn Mar 21 '21 at 02:59
2 Answers
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If you apply a well-known Euclidean algorithm, you will get
$$3m+8n=k$$
$$m = 8 z + 3 k, ~n = -3 z - k, ~ z\in\mathbb Z$$
Then, we need
$$\begin{cases} 8z+3k ≥0 \\-3z-k≥0\\ k>13\end{cases}$$
Finally, we have
$$k>13, ~ \frac{3k}{8}≥z≥\frac k3$$
For $k>13$, there always exist an integer $z$, such that $\frac{3k}{8}≥z≥\frac k3$ holds.
lone student
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@BillDubuque How do you find that these problems are dupe? Sometimes I can't find it when I search. Thus, my work becomes meaningless. – lone student Mar 21 '21 at 03:12
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First prove this for $14,15,16$ because they are the first numbers greater than $13$.
$14=3+3+8$, $15=3+3+3+3+3$, $16 = 8+8$.
Now, if you add $3$ to each of these, you get
$17 = 3+3+3+8$, $18=3+3+3+3+3+3$, $19=8+8+3$
keep going, adding $3$, and you get
$20 = 3+3+3+3+8$, $21=3+3+3+3+3+3+3$, $22=8+8+3+3$
You can clearly see from these examples that all numbers greater than $13$ can be expressed as the sum of $3$s and $8$ because we can just keep adding $3$ to our original trio to get all the numbers.
Some Guy
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