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1985 IMO Problems/Problem 3

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Problem

For any polynomial P(x) = a_0 + a_1 x + \cdots + a_k x^k with integer coefficients, the number of coefficients which are odd is denoted by w(P). For i = 0, 1, \ldots, let Q_i (x) = (1+x)^i. Prove that if i_1, i_2, \ldots , i_n are integers such that 0 \leq i_1 < i_2 < \cdots < i_n, then

w(Q_{i_1} + Q_{i_2} + \cdots + Q_{i_n}) \ge w(Q_{i_1}).

Solution

We first observe that (1+x)^{2^m} \equiv 1 + x^{2^m} \pmod{2}, so for any polynomial P of degree less than 2^m, w(P\cdot Q_{2^m}) = 2w (P).

Let k be the largest power of 2 that is less than or equal to i_n. We proceed by induction on k.

For k = 0, the problem is trivial.

Now assume that the problem holds for k-1. We now have two cases: i_1 \ge k, and i_1 < k.

In the first case, we note that w \left( \sum_{j=1}^{n}Q_{i_j} \right) = w \left( (1+x)^k \sum_{j=1}^{n}Q_{i_j - k} \right) = 2 w \left( \sum_{j=1}^{n}Q_{i_j..., which is greater than or equal to w( Q_{i_1} ) by assumption.

In the second case, we use the division algorithm to note that

\sum_{j=1}^{n}Q_{i_j} = \sum_{j=0}^{k-1}a_j x^j + (1+x)^k \sum_{j=0}^{k-1} b_j x^j \equiv \sum_{j=0}^{k-1}\left[ (a_j + b_j)x....

But by assumption, w \left( \sum_{j=0}^{k-1}a_j x^j \right) \ge w( Q_{i_1}), and for each odd a_j, at least one of (a_j + b_j ) and b_j must be odd, Q.E.D.


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

1985 IMO (Problems)
Preceded by
Problem 2
1 2 3 4 5 6 Followed by
Problem 4
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