ThatDonGuy Joined: Jun 22, 2011
• Posts: 4194
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 Joined: Jun 28, 2011
• Posts: 905
January 25th, 2014 at 6:42:00 AM permalink

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 Joined: Jun 28, 2011
• Posts: 905
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 Joined: Jun 22, 2011
• Posts: 4194
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 Joined: Oct 30, 2011
• Posts: 546
January 25th, 2014 at 12:27:35 PM permalink
Did we ever find out old she was?
beachbumbabs Joined: May 21, 2013
• Posts: 14227
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 Joined: Oct 28, 2012