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Re: Challenge problem
*To*: mathgroup at smc.vnet.net
*Subject*: [mg71210] Re: Challenge problem
*From*: Bill Rowe <readnewsciv at sbcglobal.net>
*Date*: Fri, 10 Nov 2006 06:38:18 -0500 (EST)
On 11/9/06 at 3:37 AM, Mark.Coleman at LibertyMutual.com (Coleman, Mark)
wrote:
>I recently received the annual newsletter from the Math-Stats
>department of my undergraduate alma mater. In part of the newsletter
>they posed the following challenge problem:
>"For which rational numbers q is 3q^3 + 10q^2 + 3q an integer?"
>The problem comes from an annual mathematics competition the school
>sponsors.
Surely, there is more to this problem. First, there is the
trivial answer that q is an integer since the set of rationals
includes the set of integers.
To get a somewhat less trivial answer, consider the following
In[42]:=
expr = Together[3*q^3 + 10*q^2 + 3*q /. q -> a/b]
Out[42]=
(3*a^3 + 10*b*a^2 + 3*b^2*a)/b^3
In[43]:=
Reduce[Mod[Numerator[expr /. b -> 2],
Denominator[expr /. b -> 2]] == 0, a, Integers]
Out[43]=
C[1] â?? Integers && (a == 8*C[1] ||
a == 8*C[1] + 2 ||
a == 8*C[1] + 4 ||
a == 8*C[1] + 6)
That is there are an infinite number of solutions of the form q=
m/2 for integer m
A similar result can be obtained for q = m/3 for integer m
--
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