1984 USAMO Problems/Problem 5
Contents
Problem
is a polynomial of degree such that
Determine .
Solution
By Lagrange Interpolation Formula
and hence
after some calculations we get
Given so we have to find such that
Lemma: If is even
and if is odd
is Using above lemmas we do not get any solution when is odd, but when is even satisfies the required condition, hence This problem needs a solution. If you have a solution for it, please help us out by adding it.
Solution
The (3n+1)th differences of the polynomial are zero. Call it p(x), so we have p(3n+1) - (3n+1)C1 p(3n) + (3n+1)C2 p(3n-1) - ... + (-1)3n+1 p(0) = 0, where rCs is the binomial coefficient. Hence p(3n+1) = 2( (3n+1)C1 - (3n+1)C4 + ... ) + ( (3n+1)C3 - (3n+1)C6 + ... ). Putting n = 1, we get: p(4) = 2( 4C1 - 4C4) + 4C3 = 6 + 4 = 10. So n is not 1. Putting n = 2, we get: p(7) = 2( 7C1 - 7C4 + 7C7) + ( 7C3 - 7C6) = 2( 7 - 35 + 1) + (35 - 7) = -26. So n is not 2. Putting n = 3, we get: p(10) = 2( 10C1 - 10C4 + 10C7 - 10C10) + ( 10C3 - 10C6 + 10C9) = 2(10 - 210 + 120 - 1) + (120 - 210 + 10) = -162 -100 = -262. So n is not 3. Putting n = 4, we get: p(13) = 2( 13C1 - 13C4 + 13C7 - 13C10 + 13C13) + ( 13C3 - 13C6 + 13C9 - 13C12) = 2(13 - 715 + 1716 - 286 + 1) + (286 - 1716 + 715 - 13) = 1458 - 728 = 730. So n = 3 works.
See Also
1984 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
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