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Suppose that a bank machine can dispense money in either 3\$ or 10\$ bills. Show that any amount over 17\$ could be dispensed with combinations of only the 3\$ or the 10\$ bills

Sarah_47
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    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Nov 21 '20 at 17:28
  • See Frobenius Coin Problem https://en.wikipedia.org/wiki/Coin_problem – player3236 Nov 21 '20 at 17:40
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    Write out expressions for $18,19,20.$ Any number larger than that can be found by adding a multiple of $3$ to one of those expressions. – Will Jagy Nov 21 '20 at 17:40
  • Nitpick. The question in the title is obviously wrong. 18 isn't a multiple of 10, 19 isn't a mulitple of either, 20 isn't a multiple of 3, 21 isn't a multiple of 10, etc. Why not simply so "are a sum of a multiple of 3 and a multiple of 10"? – fleablood Nov 21 '20 at 18:11
  • @fleablood yes, it's true. The question is wrongly stated. I think Sara_47 meant : for all $n \in \mathbb{N}, n>17$ there exist $k_1,k_2 \in \mathbb{N}$ such that $n = 3k_1 + 10k_2$ – M.B. Nov 21 '20 at 19:05
  • It's certain that she meant "How $\color{gray}{\text{should}}$ can I show that all $\color{gray}{\text{numbers}}$ integers $\color{grey}{\text{bigger}}$ larger than 17 can be $\color{red}{\text{shown as multiples of 3 and 10}}$ expressed as the sum of a positive multiple of 3 and a positive multiple of 10?" But I am nitpicky that "shown as multiples of 3 and 10" means something very different than what she intended. (Although the grey stuff is just me putting on my overly fussy editor hat.) – fleablood Nov 21 '20 at 20:01
  • Things for the OP to think about. If $N = 10k + 3m$ then $N+3= 10k + 3(m+1)$. (Also, maybe but not really necessary, if $N=10k + 3m$ and $m\ge 3$ then $N+1= 10(k+1) + 3(m-3)$. And if $N=10k + 3m$ and $k\ge 2$ then $N+1 = 10(k-2) + 3(m + 7)$). – fleablood Nov 21 '20 at 20:09

2 Answers2

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For any $a,b$, such that $g.c.d(a,b)=1$, all numbers greater than $ab-a-b$ can be written in the form $na+mb$. This is is the Fröbenius problem, you can find more information here: The Frobenius Coin Problem

Jorge
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Given a number $n \in \mathbb{N}$, then one of the following can happen:
$n \equiv 0 \ (mod 3)$
$n \equiv 1 \ (mod 3)$
$n \equiv 2 \ (mod 3)$
i.e. $n$ can be of the form $3k$ or $3k+1$ or $3k + 2$.
If $n$ is a multiple of $3$ then you can write $$n = 3k + 10 \cdot 0$$ If $n = 3k + 1$ then $$n = 3k +1 = 3k + 10-9 = 3(k-3) + 10$$ Notice that $k$ is at least $6$ since $n >17$.
If $n = 3k +2$ then $$n = 3k+2 = 3k+20-18 = 3(k-6) + 10 \cdot 2$$ Notice that $k-6$ is non-negative since $n > 17$.

M.B.
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