Prove that all numbers above 12 can be expressed in the form \(7x + 3y\) by any method. Preferably by mathematical induction.

- \(x\) may be equal to \(y\) and,
- \(x\) and \(y\) are positive integers.

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## Comments

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TopNewestThe answer is trivial by Postage Stamp Problem / Chicken McNugget Theorem.

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Yeah! I had thought of this. But I was not that sure to post it :)

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Same here! BTW, did you think of anyway to prove it? If yes do share it.

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-_-

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@Sravanth Chebrolu

Do you know what Chicken-Nugget theorem is? Isn't it has become obvious now? What do you think you need to prove now?@Pi Han Goh Do we require more proof? It seems Chicken Nugget does all the work.

Oh Wait! Sravanth do you want us to prove Chicken Nugget theorem?

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Let \(m\) and \(n\) be positive coprime integers. Then the greatest integer that cannot be written in the form \(am + bn\) for nonnegative integers \(a\) and \(b\) (called the Frobenius number) is \(mn-m-n\).

Here \(m,n=(7,3)\) and substitution gives us answer as \(11\). This is a universally accepted theorem. If you have any problem with this , do tell me. It seems you didn't know this theorem at all.

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Yes sir. I thought of it before posting this note. Actually the question was "You are in a world where, there are only denominations of \($7\) and \($3\). What is the minimum number, above which you can obtain all possible values?"

So, what do you think sir? Can't there be an inductive proof for this?

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"What is the minimum number, above which you can obtain all possible values?"

That is an impossible scenario isn't it?

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I think the most intuitive way to go about this is by expressing the \(10\) numbers after \(12\) (as a set, \(S=\{13,14,15,...,22\}\)). Once that is shown, the proof is essentially over, because we can just add \(10\) to any number in \(S\), which will let us form any \(n>12\) as \(7x+3y\), with \(x\) and \(y\) being non-negative integers.

Clearly, setting \(x=0\) allows us to attain any multiple of \(3\), which in \(S\) will give us \(15,18,\) and \(21\). Similarly, we can set \(y=0\) to attain any multiple of \(7\), which in \(S\) will give us \(14\) and \(21\). So, now we have the in-between cases of \(x,y \neq 0\). We can pluck the right \(x\) and \(y\) rather easily for these. \[13 = 7(1)+3(2)\] \[16= 7(1)+3(3)\] \[17 = 7(2)+3(1)\] \[19=7(1)+3(4)\] \[20=7(2)+3(2)\] \[22=7(1)+3(5)\] So, now we have a corresponding set of pairs \((x,y)\) which give us each number in \(S\), and we'll call this set \(S*=\{(x_{13},y_{13}),(x_{14},y_{14}), ... , (x_{22},y_{22})\}\).

Now, any \(n>12\) can be formed. Say we want \(26\). Then we take the pair that gave us \(16\) and we add one to both \(x\) and \(y\): \(7(1+1)+3(3+1)=26\). So all we have done is add \(10\) to our \(16\). If we want to add \(40\) to an element of \(S\), we just add \(4\) to each \(x,y\) in the appropriate element of \(S*\). In general, we can add any multiple of \(10\) by simply adding \(n\) to each \(x\) and \(y\) of the appropriate pair. Thus we can express any \(n>12\) as \(7x+3y\) for positive \(x\) and \(y\).

It's not the most elegant proof, but it works. I'd love to see a more elegant proof.

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Well done sir! But, this indeed doesn't prove that all numbers can be

surelyexpressed as \(3x+8y\). Anyway it was a very good attempt sir!Log in to reply

What is it lacking?

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Are \(x\) and \(y\) non-negative integers?

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Yeah. Thanks for asking! I've modified it. \(\huge\ddot\smile\)

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If they are, \(15\) is not possible. Do you mean that they are integers (positive or negative)?

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@Sharky Kesa, @Daniel Liu, @Michael Mendrin sir, @Azhaghu Roopesh M sir can you post your own proofs?

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@Calvin Lin sir and @Nihar Mahajan. Can you post your proofs? Please. . .

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