KingGeorge
  • KingGeorge
[UNSOVED] What is the largest number (in terms of a, b, c) that can not be expressed in the form \[ax+by+cz\] where \[x,y,z≥0\]\[gcd(a,b,c)=1\]and all values are non-negative integers? As an example, take \[6x+10y+15z\]Here, the largest integer that can't be expressed in this form given \[x, y, z \geq 0\]is 29. If instead you're solving the same problem, only with \[ax+by\]instead of\[ax+by+cz\] with the same conditions that every value is a non-negative integer, the largest integer not expressible in this form is given by the formula\[ab-a-b\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
IMO problem?
anonymous
  • anonymous
A similar problem appear in IMO 1983, Let \( a, b,\) and \(c\) be positive integers, no two of which have a common divisor greater than \(1\). Show that \(2abc − ab − bc − ca\) is the largest integer that cannot be expressed in the form \(xab+yca+zab\), where\( x, y,\) and \(z\) are non-negative integers.
KingGeorge
  • KingGeorge
no, the very last part of a homework problem from our textbook. I've figured out the two variable case without too much trouble, but for some reason, I just don't know how to generalize it.

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KingGeorge
  • KingGeorge
Well, that appears to be almost the generalization I'm looking for...
anonymous
  • anonymous
Yes, I think we still need to do some work, btw what level of mathematics is this?
anonymous
  • anonymous
Like high-school , undergrad, etc...
KingGeorge
  • KingGeorge
undergrad - intro to number theory
anonymous
  • anonymous
Cool, I thought so.
KingGeorge
  • KingGeorge
I've worked through a few other cases while varying c. So far I've gotten 25 if a=6, b=10, c=11. And 23 if a=6, b=10, c=9.
KingGeorge
  • KingGeorge
Here a few more cases 11 if (a, b, c) = (4, 5, 6) also 11 if (a, b, c) = (4, 5, 7) also 11 if (a, b, c) = (4, 6, 9)
KingGeorge
  • KingGeorge
My best guess currently, for those of you looking at this, is that the solution is the smallest integer that satisfies these equations\[\begin{matrix} ab-a-b-d = cx \\ ac-a-c-d=by \\ bc-b-c-d=az \end{matrix}\]where d is the solution we're looking for, and x, y, z are still all positive.

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