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Old 2016-04-02, 02:15   #1
Xyzzy
 
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Default April 2016

https://www.research.ibm.com/haifa/p...April2016.html
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Old 2016-04-02, 05:05   #2
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A nice variation on the topic of weighing. With constraints ('lives' of robocats vaporized when p(n)>=0, i.e. between two of the roots)...
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Old 2016-04-02, 12:52   #3
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They forgot to mention ** is to the power of. It took me 1/2 hour of reading through past problems to figure that out.
It's not clear if x is integer.

Last fiddled with by a1call on 2016-04-02 at 12:54
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Old 2016-04-03, 14:40   #4
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Quote:
Originally Posted by a1call View Post
They forgot to mention ** is to the power of. It took me 1/2 hour of reading through past problems to figure that out.
It's not clear if x is integer.
Let p(x)=x**3-300*x**2+a*x+b be a cubic polynomial with unknown parameters a and b that has three positive integers roots.
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Old 2016-04-03, 14:49   #5
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Quote:
Originally Posted by WMHalsdorf View Post
Let p(x)=x**3-300*x**2+a*x+b be a cubic polynomial with unknown parameters a and b that has three positive integers roots.
Thanks for that.

Well, then the roots are too messy for me to try anything:

Link

Last fiddled with by a1call on 2016-04-03 at 15:08
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Old 2016-04-03, 17:50   #6
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Hints:

Assume that you know the three roots q, r, and s.

Rewrite the equation as the product of three (x minus root).
Simplify.


Compare to the original equation. What have you now found about q, r, and s?
Now, you will have re-discovered [URL]https://en.wikipedia.org/wiki/Vieta's_formulas[/URL]
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Old 2016-04-07, 04:20   #7
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If we had 14 robocats, we'd only have to "run 14 experiments" (not 16).

"I see dead cats. They don't know they're dead… They're everywhere."
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Old 2016-04-12, 01:02   #8
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Quote:
Originally Posted by Batalov View Post
If we had 14 robocats, we'd only have to "run 14 experiments" (not 16).

"I see dead cats. They don't know they're dead… They're everywhere."
A subcategory exists of unborn cats that do not yet realize they will be alive..epistemic observation.

Last fiddled with by jwaltos on 2016-04-12 at 01:04 Reason: not required
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Old 2016-05-02, 02:45   #9
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https://www.research.ibm.com/haifa/p...April2016.html
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Old 2016-05-05, 05:27   #10
LaurV
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Quote:
Originally Posted by Xyzzy View Post
Quote:
If x**3+300*x**2+a*x+b = (x-x1)(x-x2)(x-x3), then x1+x2+X3=300
I have a problem with that: according with the math I learned in elementary school (grade 8), if x1+x2+X3=300, then the polynomial would be x**3 - 300*x**2.... and not plus...

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Old 2016-05-05, 05:42   #11
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Quote:
Originally Posted by LaurV View Post
I have a big problem with that: according with the math I learned in elementary school (grade 8), if x1+x2+X3=300, then the polynomial would be x**3 - 300*x**2.... and not plus...
Yes. And the problem (https://www.research.ibm.com/haifa/p...April2016.html) calls for p(x)=x**3-300*x**2+a*x+b. The +300 is just a typo in the solution.
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