How old is she?
January 24th, 2014 at 2:59:23 AM
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MathyMiley is the new pole of attraction on the forum.
- How old are you?
- Ya ur so cute!~ U haf to guess. Lemme rite a polynomial P(x), with integer coefficients, of which my age is a root. For each guess I give you the value of P.
Forum member Magician shoots first with a guess.
- Nooo! I'm not that old! The value of P is 299.
You try your luck with a guess of 27. The reply is P(27)=133.
Now with that info you can deduce MathyMiley's age (and as a bonus, the Magician's guess).
- How old are you?
- Ya ur so cute!~ U haf to guess. Lemme rite a polynomial P(x), with integer coefficients, of which my age is a root. For each guess I give you the value of P.
Forum member Magician shoots first with a guess.
- Nooo! I'm not that old! The value of P is 299.
You try your luck with a guess of 27. The reply is P(27)=133.
Now with that info you can deduce MathyMiley's age (and as a bonus, the Magician's guess).
Reperiet qui quaesiverit
January 24th, 2014 at 3:09:24 AM
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She is 39.5 years old? And magician guessed she was 110?
P(x) = 2x + 79
P(x) = 2x + 79
January 24th, 2014 at 3:34:48 AM
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That would yield a negative age of -39.5. It is implicitly assumed that MMiley is already born.
Also: the answer is an integer.
Also: the answer is an integer.
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January 24th, 2014 at 3:36:52 AM
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Hint: there is no way to guess the polynomial. The answer can be found without knowing it.
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January 24th, 2014 at 7:34:38 AM
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If it is not required that her age is integer, then there are exactly two 1st order polynomials that satisfy the conditions:
166x - 4349
83x - 2108
and her age would be x = 26.199 or x = 25.398.
But if her age must be integer then there exists no 1st order polynomial solution and the polynomial must be at least 2nd degree. I guess the solution is not intended to be found this way.
166x - 4349
83x - 2108
and her age would be x = 26.199 or x = 25.398.
But if her age must be integer then there exists no 1st order polynomial solution and the polynomial must be at least 2nd degree. I guess the solution is not intended to be found this way.
January 24th, 2014 at 1:43:14 PM
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Quote: kubikulannMathyMiley is the new pole of attraction on the forum.
- How old are you?
- Ya ur so cute!~ U haf to guess. Lemme rite a polynomial P(x), with integer coefficients, of which my age is a root. For each guess I give you the value of P.
Forum member Magician shoots first with a guess.
- Nooo! I'm not that old! The value of P is 299.
You try your luck with a guess of 27. The reply is P(27)=133.
Now with that info you can deduce MathyMiley's age (and as a bonus, the Magician's guess).
P(y) = 299
P(27) = 133
P(x) = 0
Where x and y are positive integers, and P() is a polynomial. X Is Miley's real age and Y is he Magician's guess.
P(x) will have the form (x+a)(x+b)....
P(y) - P(27) = 166.
P(y) = (y+a)(y+b)....
P(27) = (27+a)(27+b)...
Nope, stuck, I'd have to work out a general form of expanding the polynomials and using the differences. Something will get eliminated somewhere.
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at trente-et-quarante" - Honore de Balzac, 1829
January 24th, 2014 at 2:00:46 PM
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Also y>x
Quote:Nooo! I'm not that old!
January 24th, 2014 at 2:03:38 PM
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also x=-a or -b or ...
January 24th, 2014 at 2:04:18 PM
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what i wrote here was wrong, so I deleted it
January 24th, 2014 at 2:24:45 PM
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Quote: endermikealso x=-a or -b or ...
As a is unknown, I'd say a is the only negative value in the series (x+a)(x+b)... etc. Or otherwise, assume : P(x) = (x-a)(x+b)(x+c)....
Now, does it follow that y > 27 > x?
Also, there is a term Z, which is a constant. I forgot about that (it falls off when we subtract P(y) from P(27)).
Also each X term might have a co-efficient (x-a)(b'x+b)(c'x+c).... so not sure this approach is right after all.
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at trente-et-quarante" - Honore de Balzac, 1829
January 24th, 2014 at 2:49:44 PM
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Ok I have the options narrowed down but I can't quite get it narrowed down all the way to 1 answer (probably I am missing something obvious)
But, from my choices, the numbers that would make the story work are, she is 20, and the magician guessed 33.
(If the answer is that she is 8 and the magician guessed 31... gross)
But, from my choices, the numbers that would make the story work are, she is 20, and the magician guessed 33.
(If the answer is that she is 8 and the magician guessed 31... gross)
January 24th, 2014 at 2:55:20 PM
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The general form should be something like this:
P(x) = k(x-a)(x-b)(x-c)... + C
We know that P(27) = 133. If (and we can't assume this) k=1 and C=0, then the terms (x-a)(x-b)(x-c) must equal 133. The factors of 133 are 1, 7, 19, and 133.
Again, including those assumptions, each term could evaluate to +- any of those factors, and it would follow that the age is either 26, 28, 20, 34, 8, 46, or 160.
If you use the same logic for the wizards guess, the number 299 factors into 1, 13, 23, and 299. It would follow that the wizard's guess is either 1 high, 1 low, 13 high, 13 low, 23 high, 23 low, or 299 high. (being a wizard, an age of 299+ might exist to him)
edit: the wizard didn't guess low, so he can only be 1, 13, 23, or 299 high (given the assumptions).
I only submit this because it might help someone else get on the right track - it hasn't helped me so far, probably because the assumptions are incorrect, but I had to hang it up and start working.
P(x) = k(x-a)(x-b)(x-c)... + C
We know that P(27) = 133. If (and we can't assume this) k=1 and C=0, then the terms (x-a)(x-b)(x-c) must equal 133. The factors of 133 are 1, 7, 19, and 133.
Again, including those assumptions, each term could evaluate to +- any of those factors, and it would follow that the age is either 26, 28, 20, 34, 8, 46, or 160.
If you use the same logic for the wizards guess, the number 299 factors into 1, 13, 23, and 299. It would follow that the wizard's guess is either 1 high, 1 low, 13 high, 13 low, 23 high, 23 low, or 299 high. (being a wizard, an age of 299+ might exist to him)
edit: the wizard didn't guess low, so he can only be 1, 13, 23, or 299 high (given the assumptions).
I only submit this because it might help someone else get on the right track - it hasn't helped me so far, probably because the assumptions are incorrect, but I had to hang it up and start working.
I heart Crystal Math.
January 24th, 2014 at 3:03:33 PM
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Quote: CrystalMathThe general form should be something like this:
P(x) = k(x-a)(x-b)(x-c)... + C
We know that P(27) = 133. If (and we can't assume this) k=1 and C=0, then the terms (x-a)(x-b)(x-c) must equal 133. The factors of 133 are 1, 7, 19, and 133.
Again, including those assumptions, each term could evaluate to +- any of those factors, and it would follow that the age is either 26, 28, 20, 34, 8, 46, or 160.
If you use the same logic for the wizards guess, the number 299 factors into 1, 13, 23, and 299. It would follow that the wizard's guess is either 1 high, 1 low, 13 high, 13 low, 23 high, 23 low, or 299 high. (being a wizard, an age of 299+ might exist to him)
I only submit this because it might help someone else get on the right track - it hasn't helped me so far, probably because the assumptions are incorrect, but I had to hang it up and start working.
We both used the same logic, but I forgot to include 1 and n in my numbers. So I had her age "narrowed down" to 20, 34, 8, 46. I picked 20 for obvious reasons :)
But, yeah, I forgot that 27-age can also be +/- 1 or +/- 133. So 26 and 28 are in play as well (assuming reasonable human lifespans). I'm still going with 20 (total guess on my part)
January 24th, 2014 at 3:25:21 PM
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Quote: CrystalMathIf (and we can't assume this) k=1 and C=0, then the terms (x-a)(x-b)(x-c) must equal 133.
(emphasis mine)
Damn, that's right too. Ok back to the drawing board.
January 24th, 2014 at 4:12:29 PM
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Neither.Quote: AxiomOfChoiceBut, from my choices, the numbers that would make the story work are, she is 20, and the magician guessed 33.
(If the answer is that she is 8 and the magician guessed 31... gross)
Reperiet qui quaesiverit
January 24th, 2014 at 4:14:54 PM
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You don't need to develop that far.Quote: CrystalMathThe general form should be something like this:
P(x) = k(x-a)(x-b)(x-c)... + C
Try factorizing P(y) - P(x).
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January 24th, 2014 at 4:16:46 PM
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But then the Magician's guess would not be realistic.Quote: Jufo81If it is not required that her age is integer, then there are exactly two 1st order polynomials that satisfy the conditions:
166x - 4349
83x - 2108
Reperiet qui quaesiverit
January 24th, 2014 at 4:31:50 PM
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I think she's 21 and the Magician guessed 34.
If the House lost every hand, they wouldn't deal the game.
January 24th, 2014 at 6:05:12 PM
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If I can assume that she's less than 100...
Now the hard part: seeing if I can figure out what the coefficients of P are.
She is 28, and the guess was 29
Let X be her age, and M the magician's guess
Also, let P(n) = a0 + a1 n + a2 n2 + a3 n3 + ...
Since the woman said that the guess was too high, M > X
P(M) = a0 +a1 M + a2 M2 + ... = 299
P(X) = a0 + a1 X + a2 X2 + ... = 0
P(M) - P(X) = a1 (M - X) + a2 (M2 - X2) + a2 (M3 - X3) + ... = 299 = 13 x 23
(M - X) (a1 + a2 (M + X) + a3 (M2 + MX + X2) = 299
Thus M - X must be 1, 13, 23, or 299
Similarly, using X in place of M, 27 in place of X, and 133 (= 299 - 166) in place of 299:
(X - 27) (a1 + a2 (X + 27) + a3 (X2 + 27X + 729) = 133 = 7 * 19
Note that (27 - X) (a1 + a2 (X + 27) + a3 (X2 + 27X + 729) = -133 = -1 * 7 * 19
Thus, |X - 27| = 1, 7, 19, or 133, which means X is 26, 20, 8, 28, 34, 46, or 160
Doing the same with M and 27:
(M - 27) (a1 + a2 (M + 27) + a3 (M2 + 27M + 729) = 166 = 2 * 83
(27 - M) (a1 + a2 (M + 27) + a3 (M2 + 27M + 729) = -166 = -1 * 2 * 83
Thus, |M - 27| = 1, 2, 83, or 166, which means M is 26, 25, 27, 28, 110, or 193
Since M > X:
If M = 25, the X values < 25 (8 and 20) have M-X values of 13 and 5
If M = 26, the X values < 26 (8 and 20) have M-X values of 14 and 6
If M = 28, the X values < 28 (8, 20, and 26) have M-X values of 20, 8, and 2
If M = 29, the X values < 29 (8, 20, 26, and 28) have M-X values of 21, 9, 3, and 1
Only M = 29 and X = 28 have an M-X value in {1, 13, 23, 299}
Let X be her age, and M the magician's guess
Also, let P(n) = a0 + a1 n + a2 n2 + a3 n3 + ...
Since the woman said that the guess was too high, M > X
P(M) = a0 +a1 M + a2 M2 + ... = 299
P(X) = a0 + a1 X + a2 X2 + ... = 0
P(M) - P(X) = a1 (M - X) + a2 (M2 - X2) + a2 (M3 - X3) + ... = 299 = 13 x 23
(M - X) (a1 + a2 (M + X) + a3 (M2 + MX + X2) = 299
Thus M - X must be 1, 13, 23, or 299
Similarly, using X in place of M, 27 in place of X, and 133 (= 299 - 166) in place of 299:
(X - 27) (a1 + a2 (X + 27) + a3 (X2 + 27X + 729) = 133 = 7 * 19
Note that (27 - X) (a1 + a2 (X + 27) + a3 (X2 + 27X + 729) = -133 = -1 * 7 * 19
Thus, |X - 27| = 1, 7, 19, or 133, which means X is 26, 20, 8, 28, 34, 46, or 160
Doing the same with M and 27:
(M - 27) (a1 + a2 (M + 27) + a3 (M2 + 27M + 729) = 166 = 2 * 83
(27 - M) (a1 + a2 (M + 27) + a3 (M2 + 27M + 729) = -166 = -1 * 2 * 83
Thus, |M - 27| = 1, 2, 83, or 166, which means M is 26, 25, 27, 28, 110, or 193
Since M > X:
If M = 25, the X values < 25 (8 and 20) have M-X values of 13 and 5
If M = 26, the X values < 26 (8 and 20) have M-X values of 14 and 6
If M = 28, the X values < 28 (8, 20, and 26) have M-X values of 20, 8, and 2
If M = 29, the X values < 29 (8, 20, 26, and 28) have M-X values of 21, 9, 3, and 1
Only M = 29 and X = 28 have an M-X value in {1, 13, 23, 299}
Now the hard part: seeing if I can figure out what the coefficients of P are.
January 24th, 2014 at 6:08:25 PM
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Quote: ThatDonGuyNow the hard part: seeing if I can figure out what the coefficients of P are.
Aren't there infinitely many solutions to that? Or is there some chance that the "integer coefficients" restriction makes it unique?
January 24th, 2014 at 6:45:03 PM
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Quote: AxiomOfChoiceAren'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
January 25th, 2014 at 6:42:00 AM
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Good answer, Don.
The beauty of the problem, I think, is in the answer NOT requiring to know the P(x).
Bravo, Don!
Indeed, one of many possibilities. Anyway, each P(x) is valid up to a multiplicative constant.Quote: ThatDonGuyYes - 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
The beauty of the problem, I think, is in the answer NOT requiring to know the P(x).
Bravo, Don!
Reperiet qui quaesiverit
January 25th, 2014 at 7:02:57 AM
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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.
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
January 25th, 2014 at 12:07:24 PM
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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
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
January 25th, 2014 at 12:27:35 PM
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Did we ever find out old she was?
January 25th, 2014 at 12:35:23 PM
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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.
January 25th, 2014 at 1:30:56 PM
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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
