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.

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.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestThe answer is trivial by Postage Stamp Problem / Chicken McNugget Theorem. – Pi Han Goh · 1 year, 10 months ago

Log in to reply

– Nihar Mahajan · 1 year, 10 months ago

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

– Sravanth Chebrolu · 1 year, 10 months ago

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

– Nihar Mahajan · 1 year, 10 months ago

I am not able to understand what exactly you want us to prove now?Log in to reply

-_- – Sravanth Chebrolu · 1 year, 10 months ago

Log in to reply

@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? – Nihar Mahajan · 1 year, 10 months ago

Log in to reply

– Sravanth Chebrolu · 1 year, 10 months ago

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. :PLog in to reply

– Nihar Mahajan · 1 year, 10 months ago

It is of course \(11\)! -_- (if x,y are non- negative)Log in to reply

– Sravanth Chebrolu · 1 year, 10 months ago

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)Log 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. – Nihar Mahajan · 1 year, 10 months ago

Log in to reply

– Sravanth Chebrolu · 1 year, 10 months ago

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)Log in to reply

So, what do you think sir? Can't there be an inductive proof for this? – Sravanth Chebrolu · 1 year, 10 months ago

Log in to reply

That is an impossible scenario isn't it? – Pi Han Goh · 1 year, 10 months ago

Log in to reply

– Sravanth Chebrolu · 1 year, 10 months ago

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?Log in to reply

– Pi Han Goh · 1 year, 10 months ago

I've already given you the answer above.Log in to reply

– Sravanth Chebrolu · 1 year, 10 months ago

So, you mean it's impossible to prove it?Log in to reply

– Pi Han Goh · 1 year, 10 months ago

It's impossible to prove it because there is no maximum value. Please read the Wiki page.Log in to reply

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 · 1 year, 10 months ago

Log in to reply

surelyexpressed as \(3x+8y\). Anyway it was a very good attempt sir! – Sravanth Chebrolu · 1 year, 10 months agoLog in to reply

– Ryan Tamburrino · 1 year, 10 months ago

What is it lacking?Log in to reply

– Sravanth Chebrolu · 1 year, 10 months ago

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?Log in to reply

– Ryan Tamburrino · 1 year, 10 months ago

Ah! Chicken McNugget Theorem! How could I forget. There you go.Log in to reply

@Sharky Kesa, @Daniel Liu, @Michael Mendrin sir, @Azhaghu Roopesh M sir can you post your own proofs? – Sravanth Chebrolu · 1 year, 10 months ago

Log in to reply

Log in to reply

– Sravanth Chebrolu · 1 year, 10 months ago

Yes. They must be integers.Log in to reply

Are \(x\) and \(y\) non-negative integers? – Ryan Tamburrino · 1 year, 10 months ago

Log in to reply

– Sravanth Chebrolu · 1 year, 10 months ago

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

– Andrei Golovanov · 1 year, 10 months ago

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

– Ryan Tamburrino · 1 year, 10 months ago

Non-negative includes 0.Log in to reply

– Sravanth Chebrolu · 1 year, 10 months ago

Why not? \(15=7\times0+3\times5\)Log in to reply

– Andrei Golovanov · 1 year, 10 months ago

Oh, sorry! I forgot about 0.Log in to reply

@Calvin Lin sir and @Nihar Mahajan. Can you post your proofs? Please. . . – Sravanth Chebrolu · 1 year, 10 months ago

Log in to reply