EPTEasier problem. Find values for a, b, and c such that for any integer x, the solution y is also an integer.
Harder problem. Find a general solution for the possible values of a, b, and c that satisfies the above constraints.
Math Problem 1/23
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Slithe
United States985 Posts
Easier problem. Find values for a, b, and c such that for any integer x, the solution y is also an integer. Harder problem. Find a general solution for the possible values of a, b, and c that satisfies the above constraints. | ||
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Muirhead
United States556 Posts
Setting x=-1 we see that -a+b-c is an integer. Adding, we get that 2b is an integer, so b=p/2 for some integer p. Now, setting x=2 we see that 8a+4b+2c is an integer. Thus 8a+2c is an integer. Doubling (1) we have that 2a+2b+2c is an integer, whence 2a+2c is an integer. Hence, 6a is an integer, so a=q/6 for some integer q. Returning to the fact that 2a+2c is integral, we have that 6a+6c is integral. Thus 6c is integral, so c=r/6 for some integer r. Thus all of a,b, and c are expressible as rational numbers with denominators equal to 6. Suppose (a,b,c)=(u/6,v/6,w/6) with u,v, and w integers. We just need to check which values of (u,v,w) work modulo 6. A brute force inspection shows that the exact working possibilities are those for which 3 divides v and 3 divides u+w, giving a complete list of solutions. In short the solutions are all triples (a,b,c)=(u/6,v/6,w/6) where v and u+w are both divisible by 3. | ||
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Muirhead
United States556 Posts
 It isn't combinatorics and therefore I can solve it... | ||
RaGe
Belgium9947 Posts
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Rev0lution
United States1805 Posts
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Polemarch
Canada1564 Posts
I did it almost exactly the same way.A little nagging part that you might've forgotten is that a, b, c are specified to be non-integers. this also makes it not require much brute force.. you can cut down the possibilities a lot immediately since you already deduced that 2b is an integer, v must be 3 mod 6, from which you can deduce u+v+w = 0 mod 6, so u+w = 3 mod 6. | ||
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Slithe
United States985 Posts
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thoraxe
United States1449 Posts
I'm barely in pre-cal, and although we've studied polynomials I just come up with a big '?'. I admire you guys for your intellect, you remind me of those guys of the show "The Big Bang Theory" (btw, good show...damn writer strike) I love teamliquid.net | ||
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Slithe
United States985 Posts
Although I don't think you need much beyond precalc to solve this, that isn't to say that I would expect the average high school student to be able to solve this problem. Just as a point of reference, most of the problems I post here, I get from other college students who have deemed these problems interesting. | ||
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Slithe
United States985 Posts
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gwho
United States632 Posts
On January 24 2008 17:50 Slithe wrote: Speaking of education, I'm wondering what kind of education and experience you other guys have, because you guys seem to be quite good at this. it's usually experience and exposure to doing math riddles, more than a package that some institution gives ya. that's what my intuition tells me. i'm not all that good at these myself though. especially when they're like ... concept restrictions (intergers) than solving or figuring. | ||
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