2009 AIME I Problem 14

Below is the professionally curated solution for Problem 14 of the 2009 AIME I, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AIME I solutions, or check the answer key.

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Concepts:system of equationsDiophantine Equationoptimization

Difficulty rating: 3060

14.

For t=1,2,3,4,t = 1, 2, 3, 4, define St=i=1350ait,S_t = \sum_{i=1}^{350} a_i^t, where ai{1,2,3,4}.a_i \in \{1, 2, 3, 4\}. If S1=513S_1 = 513 and S4=4745,S_4 = 4745, find the minimum possible value for S2.S_2.

Solution:

For j=1,2,3,4,j = 1, 2, 3, 4, let mjm_j be the number of aia_i equal to j.j. Then m1+m2+m3+m4=350,m1+2m2+3m3+4m4=513,m1+16m2+81m3+256m4=4745.m_1 + m_2 + m_3 + m_4 = 350, \quad m_1 + 2m_2 + 3m_3 + 4m_4 = 513, \quad m_1 + 16m_2 + 81m_3 + 256m_4 = 4745.

Subtracting the first equation from the other two gives m2+2m3+3m4=163m_2 + 2m_3 + 3m_4 = 163 and 15m2+80m3+255m4=4395;15m_2 + 80m_3 + 255m_4 = 4395; subtracting 1515 times the former from the latter leaves 50m3+210m4=1950,50m_3 + 210m_4 = 1950, that is, 5m3+21m4=195.5m_3 + 21m_4 = 195. Hence m4m_4 is a nonnegative multiple of 5,5, and only m4=0m_4 = 0 and m4=5m_4 = 5 keep everything nonnegative, giving (m1,m2,m3,m4)=(226,85,39,0)(m_1, m_2, m_3, m_4) = (226, 85, 39, 0) or (215,112,18,5).(215, 112, 18, 5).

These yield S2=m1+4m2+9m3+16m4=917S_2 = m_1 + 4m_2 + 9m_3 + 16m_4 = 917 and 905905 respectively, so the minimum is 905.905.

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