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  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -




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Topic review (newest first)

2005-11-23 04:27:57

The first and second terms define the sequence, so let's call them a and b.

The third term would be a+b.
The fourth term would be a+(a+b) = a+2b.
The fifth term would be (a+b)+(a+2b) = 2a+3b.

The maximum possible value of the first term is when the second term is 1 (because it has to be a positive integer), so 2a = 2004-3. Therefore, the first term must be 1001.5. That's not allowed because it isn't an integer, so let's try when b is 2. Now 2a = 2004-6=1998, so a =999.

Check: 999, 2, 1001, 1003, 2004. It works!

Math Student
2005-11-23 03:11:10

In a sequence of positive integers, every term after the first two terms is the sum of the two previous numbers in the sequence. If the fifth term is 2004, what is the maximum possible value of the first term?

A 399
B 400
C 663
D 999
E 1001


Thanks in advance

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