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Let Q(x)=ax2+bx+c and put that into P. Plug in x =1,2,3 and now you have a system of three linear equations with three possibilities each. Now it's just combinatorics! Calculate the inverse matrix and it is easy to see that out of the 27, 5 do not work. This the answer is 22.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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2^{34}
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What are your thoughts? What have you tried?
Must the polynomial have complex, real or integer coefficients?
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What I have tried is :
Degree of P(x) is 3 and R(x) is also 3
So degree of P(x).R(x) is 6.
Degree of Q(x) must be 2 so that degree of P(Q(x)) is 6.
P(Q(x)) = P(x).R(x)
(Q(x)-1)(Q(x)-2)(Q(x)-3) = (x-1)(x-2)(x-3).R(x)
Now what I have to do ?
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Great job. A key piece of information is that the degree of Q(x) is 6.
What can we say about the value of Q(1)? Q(2)? Q(3)?
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(Q(x)-1)(Q(x)-2)(Q(x)-3) = (x-1)(x-2)(x-3).R(x)
[Q(1)-1][Q(1)-2][Q(1)-3] =0
Q(1)-1 = 0. So Q(1) =1
Q(1)-2 = 0. So Q(1) = 2
Q(1)-3 = 0. So Q(1) = 3
Hence Q(1) = 1or 2or 3
Similarly Q(2) = 1or 2or 3
And Q(3) = 1or 2or 3
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Q(1),Q(2),Q(3).
Great, so we now know the possible values ofLets say that Q(1)=1,Q(2)=2,Q(3)=2. How many polynomials Q(x) of degree 2 satisfy that condition?
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lagrange interpolation
Check outHint: If we have n equations of the form f(xi)=yi, then there is a unique polynomial of degree at most n−1 which satisfies the conditions.
I think the answer is 2
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Answer is 22
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Yup, that's the numerical answer. The more interesting part is the actual solution.
Let Q(x)=ax2+bx+c and put that into P. Plug in x =1,2,3 and now you have a system of three linear equations with three possibilities each. Now it's just combinatorics! Calculate the inverse matrix and it is easy to see that out of the 27, 5 do not work. This the answer is 22.