2007 iTest Problems/Problem 53

The following problem is from the Ultimate Question of the 2007 iTest, where solving this problem required the answer of a previous problem. When the problem is rewritten, the T-value is substituted.

Problem

Three distinct positive Fibonacci numbers, all greater than $1536$ , are in arithmetic progression. Let $N$ be the smallest possible value of their sum. Find the remainder when $N$ is divided by $2007$ .

Solution

By definition, for a Fibonacci number, $a_1 = a_2 = 1$ and $a_n = a_{n-1} + a_{n-2}$ . From the definition, $a_{n+1} = a_n + a_{n-1}$ . That means the numbers $a_{n-2}$ , $a_n$ , and $a_{n+1}$ are in arithmetic progression with common difference $a_{n-1}$ .

Writing out the Fibonacci numbers, the first numbers that come after $1536$ are $1597$ , $2584$ , $4181$ , and $6765$ . That means the desired three numbers are $1597$ , $4181$ , and $6765$ . The sum of the three numbers is $12543$ , and the remainder after dividing by $2007$ is $\boxed{501}$ .

2007 iTest (Problems)
Preceded by: Problem 52	Followed by: Problem 54
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2007 iTest Problems/Problem 53

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