ThatDonGuy
ThatDonGuy
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January 24th, 2014 at 6:45:03 PM permalink
Quote: AxiomOfChoice

Aren't there infinitely many solutions to that?


Yes - in fact, I'm pretty sure that there is a solution P(x) = ax3 + bx2 + cx + d for every integer a.

I used a differences method on P(29) - P(28) and P(28) - P(27) to get:

216 x2 - 12013 x + 167020
kubikulann
kubikulann
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January 25th, 2014 at 6:42:00 AM permalink
Good answer, Don.

Quote: ThatDonGuy

Yes - in fact, I'm pretty sure that there is a solution P(x) = ax3 + bx2 + cx + d for every integer a.

I used a differences method on P(29) - P(28) and P(28) - P(27) to get:

216 x2 - 12013 x + 167020

Indeed, one of many possibilities. Anyway, each P(x) is valid up to a multiplicative constant.

The beauty of the problem, I think, is in the answer NOT requiring to know the P(x).

Bravo, Don!
Reperiet qui quaesiverit
kubikulann
kubikulann
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January 25th, 2014 at 7:02:57 AM permalink
A general solution
For every polynomial P(x) with integer coefficients, it is true that

P(a) - P(b) = (a-b) Q(a,b) where Q(x,y) is a polynomial with integer coefficients.

Knowing the value of P(a) - P(b), you can infer that (a-b) is a factor of it.

Using Don's notation (X=her age, M=Magician's guess, Y=Your guess)

P(M) - P(X) = 299 = 13 x 23 of which (M-X) is a factor (1, 13, 23, or 299)
P(X)- P(27) = 133 = 7 x 19 of which (X - 27) is a factor (1, 7, 19, or 133)
P(M) - P(27) = 166 = 2 * 83 of which (27 - M) is a factor (1, 2, 83, or 166)

This all put together has only two solutions, of which one respects the information that M was greater than X.

Reperiet qui quaesiverit
ThatDonGuy
ThatDonGuy
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January 25th, 2014 at 12:07:24 PM permalink
And here's a generic solution for P(x) as a cubic:

Let P(x) = ax3 + bx2 + cx + d

P(29) = 24389a + 841b + 29c + d = 299
P(28) = 21952a + 784b + 28c + d = 0
P(27) = 19683a + 729b + 27c + d = 133

P(29) - P(28) = 2437a + 57b + c = 299
P(28) - P(27) = 2269a + 55b + c = -133
(P(29) - P(28)) - (P(28) - P(27)) = 168a + 2b = 432

For any value a:
b = 216 - 84a
c = 299 - 2437a - 57 * (216 - 84a)
= 299 - 2437a - 12312 + 4788a
= 2351a - 12013
d = -21952a - 784b - 28c
= -21952a - 784 * (216 - 84a) - 28 * (2351a - 12013)
= a * (-21952 + 784 * 84 - 28 * 2351) + (-784 * 216 + 28 * 12013)
= -21924 a + 167020

P(x) = ax3 + (216 - 84a) x2 + (2351a - 12013) x + (167020 - 21924a)
If a is an integer, then so are the other three coefficients

paisiello
paisiello
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January 25th, 2014 at 12:27:35 PM permalink
Did we ever find out old she was?
beachbumbabs
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beachbumbabs
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January 25th, 2014 at 12:35:23 PM permalink
So you're saying that she's 28 and the Magician's guess was 29. That seems to be the only answer that satisfies all the criteria and your expanded formulae. Phooey, kind of; 1 did not seem valid to me in trying to suss it out. Oh, well.
If the House lost every hand, they wouldn't deal the game.
Buzzard
Buzzard
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January 25th, 2014 at 1:30:56 PM permalink
Babs is at that awkward age. Old enough to know better, too young to resist the temptation.
Shed not for her the bitter tear Nor give the heart to vain regret Tis but the casket that lies here, The gem that filled it Sparkles yet

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