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@Sravanth C.
–
Do you know what Chicken-Nugget theorem is? Isn't it has become obvious now? What do you think you need to prove now? @Sravanth Chebrolu

@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?

@Nihar Mahajan
–
No! Okay, you say it's become obvious that no such minimum number exists. Well then, why don't you tell me maximum the number which cannot be expressed as $3x+7y$ using Chicken Nugget theorem; I am not much familiar with it. :P

@Nihar Mahajan
–
Then, it's by that theorem that we've proved that $3x+7y$ can be expressed by any number above $12$ isn't it?(Or, is it? maybe I'm mistaken)

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.

@Nihar Mahajan
–
Not until today. Anyways, thanks for co-operating with me(of course thanks for teaching me a new theorem!) . I think I should leave now, bye!(I heard rumors that you're active on nights too! So, enjoy your learning. . . . :P)

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?

@Pi Han Goh
–
Yeah, I thought so sir. But taking a second look thought it's quite possible. When I tried by hit and trial I found that $12$ is such number, but I wasn't able to prove it. . . Can you please help me?

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.

@Ryan Tamburrino
–
There is no proof that combining the results will give surely give out all the numbers. But, that is not a big issue sir, you've done very well. Can you try it using induction?

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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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$12$ can be expressed in the form $7x + 3y$

I want you to prove that all numbers above-_-

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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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$3x+7y$ using Chicken Nugget theorem; I am not much familiar with it. :P

No! Okay, you say it's become obvious that no such minimum number exists. Well then, why don't you tell me maximum the number which cannot be expressed asLog in to reply

$11$! -_- (if x,y are non- negative)

It is of courseLog in to reply

$3x+7y$ can be expressed by any number above $12$ isn't it?(Or, is it? maybe I'm mistaken)

Then, it's by that theorem that we've proved thatLog in to reply

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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$12$ is such number, but I wasn't able to prove it. . . Can you please help me?

Yeah, I thought so sir. But taking a second look thought it's quite possible. When I tried by hit and trial I found thatLog in to reply

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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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$15=7\times0+3\times5$

Why not?Log in to reply

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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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