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#1 | ||||||||||||||
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Join Date: May 2011
Posts: 83
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In a sequence of positive integers, every term after the first two terms is the sum of the two previous terms in the sequence. If the fifth term is 2012, what is the maximum possible value of the first term? (This problem is #13 from this year's state Mathcounts sprint round)
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#2 | |||||||||||||||
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Join Date: Dec 2008
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Quote:
A, B, C, D, 2012 Since every term after the first two terms is the sum of the previous terms then C + D = 2012 B + C = D Substituting gives: C + B + C = 2012 B + 2C = 2012 A + B = C Substituting gives: B + 2(A + B) = 2012 2A + 3B = 2012 So for A to be the maximum then B must be a minimum. The smallest positive integer is 1. 2A + 3(1) = 2012 2A = 2009 A = 1004.5 So does A = 1004 or 1005? Say A = 1004: 1004, 1, 1005, 1006, 2011 Nope. Say A = 1005: 1005, 1, 1006, 1007, 2013 Nope. What? Well let's try B = 2, the next smallest positive integer. 2A + 3(2) = 2012 2A = 2006 A = 1003 1003, 2, 1005, 1007, 2012 Yeah! So the max. value for the first term is 1003. |
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#3 | ||||||||||||||
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Join Date: May 2011
Posts: 83
Thanks: 227
Thanked 223 Times in 83 Posts
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Correct!
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