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| − | == Problem ==
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| − | For any polynomial <math> P(x) = a_0 + a_1 x + \cdots + a_k x^k </math> with integer coefficients, the number of coefficients which are odd is denoted by <math>w(P) </math>. For <math> i = 0, 1, \ldots </math>, let <math>Q_i (x) = (1+x)^i </math>. Prove that if <math> i_1, i_2, \ldots , i_n </math> are integers such that <math> 0 \leq i_1 < i_2 < \cdots < i_n </math>, then
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| − | <center>
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| − | <math>
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| − | w(Q_{i_1} + Q_{i_2} + \cdots + Q_{i_n}) \ge w(Q_{i_1})
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| − | </math>.
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| − | </center>
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| − | == Solution ==
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| − | We first observe that <math>(1+x)^{2^m} \equiv 1 + x^{2^m} \pmod{2}</math>, so for any polynomial <math>P</math> of degree less than <math>2^m </math>, <math> w(P\cdot Q_{2^m}) = 2w (P)</math>.
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| − | Let <math>k</math> be the largest power of 2 that is less than or equal to <math>i_n</math>. We proceed by induction on <math>k</math>.
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| − | For <math>k = 0</math>, the problem is trivial.
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| − | Now assume that the problem holds for <math>k-1</math>. We now have two cases: <math> i_1 \ge k</math>, and <math>i_1 < k</math>.
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| − | In the first case, we note that <math> 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 - k} \right)</math>, which is greater than or equal to <math> w( Q_{i_1} )</math> by assumption.
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| − | In the second case, we use the division algorithm to note that
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| − | <center>
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| − | <math>
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| − | \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^j + b_j x^{j+k} \right] \pmod{2}
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| − | </math>.
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| − | </center>
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| − | But by assumption, <math> w \left( \sum_{j=0}^{k-1}a_j x^j \right) \ge w( Q_{i_1})</math>, and for each odd <math>a_j </math>, at least one of <math>(a_j + b_j ) </math> and <math>b_j </math> must be odd, Q.E.D.
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| − | {{alternate solutions}}
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| − | {{IMO box|year=1985|num-b=2|num-a=4}}
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| − | [[Category:Olympiad Algebra Problems]]Suuuuuuuuuuuuuuuuuuuuuu
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