- 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(x) = 2x + 79

Also: the answer is an integer.

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.

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.

Quote:Nooo! I'm not that old!

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.

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)

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.

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)

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.

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)

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

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

Let X be her age, and M the magician's guess

Also, let P(n) = a

_{0}+ a

_{1}n + a

_{2}n

^{2}+ a

_{3}n

^{3}+ ...

Since the woman said that the guess was too high, M > X

P(M) = a

_{0}+a

_{1}M + a

_{2}M

^{2}+ ... = 299

P(X) = a

_{0}+ a

_{1}X + a

_{2}X

^{2}+ ... = 0

P(M) - P(X) = a

_{1}(M - X) + a

_{2}(M

^{2}- X

^{2}) + a

_{2}(M

^{3}- X

^{3}) + ... = 299 = 13 x 23

(M - X) (a

_{1}+ a

_{2}(M + X) + a

_{3}(M

^{2}+ MX + X

^{2}) = 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) (a

_{1}+ a

_{2}(X + 27) + a

_{3}(X

^{2}+ 27X + 729) = 133 = 7 * 19

Note that (27 - X) (a

_{1}+ a

_{2}(X + 27) + a

_{3}(X

^{2}+ 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) (a

_{1}+ a

_{2}(M + 27) + a

_{3}(M

^{2}+ 27M + 729) = 166 = 2 * 83

(27 - M) (a

_{1}+ a

_{2}(M + 27) + a

_{3}(M

^{2}+ 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.

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?

Quote:AxiomOfChoiceAren't there infinitely many solutions to that?

Yes - in fact, I'm pretty sure that there is a solution P(x) = ax

^{3}+ bx

^{2}+ cx + d for every integer a.

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

^{2}- 12013 x + 167020

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) = ax

^{3}+ bx^{2}+ cx + d for every integer a.

I used a differences method on P(29) - P(28) and P(28) - P(27) to get:216 x^{2}- 12013 x + 167020

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

Bravo, Don!

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.

Let P(x) = ax

^{3}+ bx

^{2}+ 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) = ax

^{3}+ (216 - 84a) x

^{2}+ (2351a - 12013) x + (167020 - 21924a)

If a is an integer, then so are the other three coefficients