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The question as posed is: What is the largest bet that cannot be made with chips worth $7.00$ and $9.00?$ To prove my conjecture, I am supposed to use both forms of induction.

After research, I found this quantity to be the Frobenius number, $g(a_1,a_2),$ if $\gcd(a_1,a_2)=1$. Through experimentation, or a formula, one finds this 'largest' bet to $47.$

Thus our problem can be stated as follows:

Show that every integer $n\geq48$ can be written $$n=7a+9b$$ for $a,b\in\mathbb{Z}^+\cup\{0\}$.

My attempt via induction:

Let $S:=\{7a+9b:7a+9b\geq48\text{ and }\ a,b\in\mathbb{Z}^+\cup\{0\}\}$. First, we confirm that $48\in S$. Take $a=3$ and $b=3$ to see $48\in S$.

Now assume for some $n>48 $ that it can be written $$n=7a+9b$$ for positive (or zero) integers and thus $n\in S$. We need to show $n+1\in S.$ Thus $$n=7a+9b\in S\implies$$ $$n+1=7a+9b+1=7a+9b+4(7)-3(9)=7(a+4)+9(b-3)\in S$$

But here is where I get confused. In the book the author justifies confusingly that $a+4$ and $b-3$ are always positive (or zero) integers, and argues that this process shows you can always bet $1$ dollar more by adding four $7$ dollar chips and simultaneously taking away three $9$ dollar chips, but how can we always be sure there's enough $9$ dollar chips to take away, or, how can we be sure $a+4$ and $b-3$ are always positive?

I'm not sure about the strong induction case either.

Here is the process I'm trying to reproduce enter image description here

The part I am failing to understand and rigorously accept is from the paragraph "Let S be..." and down.

  • The first proof is explained in answers in the linked dupe (e.g. hmakhom's and mine). The second proof lifts the representations of the $\color{#c00}5$ base case integers $,28,29,\ldots 32,$ up to all greater integers by simply $,\color{#0a0}{{\rm incrementing}\ x}$ to add $,\color{#c00}5,$ i.e. $,5x+8y = n,\Rightarrow, 5(\color{#0a0}{x!+!1})+8y = n+\color{#c00}5.,$ So doing this once to our base $5$ integers yields reps for $,5+{28,29,\ldots 32} = 33,34,\ldots,37,,$ and continuing this inductively yields reps for all larger integers. – Bill Dubuque Jan 12 '20 at 17:32
  • If you have further questions please post comments either here or there. – Bill Dubuque Jan 12 '20 at 17:35
  • Oops, I meant to refer to this answer for the first proof (dupe list now updated). – Bill Dubuque Jan 12 '20 at 18:10

2 Answers2

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Here is a different way of looking at this.

We can make any multiple of $7$

With one $9$ we can make any positive integer equivalent to $2$ modulo $7$ provided it is at least $9$

With two nines we get all $\equiv 4 \bmod 7$ provided they are at least $18$

With three nines we get $\equiv 6$ provided at least $27$

Four nines gives $\equiv 1$ provided at least $36$

Five nines gives $\equiv 3$ provided at least $45$

Six nines gives $\equiv 5$ provided at least $54$

This covers all the equivalence classes modulo $7$ and the last number you can't do is $54-7=47$ - the largest in the equivalence class of $5$ which can't be done.

This argument can be adapted to give a general result - if the two numbers have no factor in common you can exhaust the equivalence classes in this way.

For your example, and your method, suppose you have a total $\ge 48$ and fewer than three $9s$ then you have at least $30=48-18$ made up of sevens and therefore have at least four sevens in the total.

On the other hand if you have fewer than four sevens you have at least $48-21=27$ made up of nines and therefore at least three nines in the total.

Mark Bennet
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You are right that justifying we can have $b\ge3$ is not obvious. In fact the argument would seem to have $b$ decreasing by $3$ on each induction step. The justification is that supposedly it's obvious that you can always have $b\ge3$ though... I honestly don't have any intuitive explanation why on this part.

In my opinion though, the strong induction proof is much nicer.

Assume we have $\{n,n+1,n+2,n+3,n+4,n+5,n+6\}\subseteq S$.

One can then easily verify that $\{n+1,n+2,n+3,n+4,n+5,n+6,n+7\}\subseteq S$ by adding $7$ to $n$.

By induction, $\{n,n+1,n+2,n+3,n+4,n+5,n+6\}\subseteq S$ for any $n\ge48$. As an immediate consequence, $n\in S$ for any $n\ge48$.

Note that this procedure requires the stronger base case of $\{48,49,50,51,52,53,54\}\subseteq S$, but now avoids fiddling with the coefficients.

In fact, the strong induction follows more obviously from how the Frobenius number can be found, which is more reason to proving it in that direction.

Simply Beautiful Art
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