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 -- To reply via email subtract one hundred and four