# 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 $\boxed{501}$.

## See Also

 2007 iTest (Problems) Preceded by:Problem 52 Followed by:Problem 54 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • TB1 • TB2 • TB3 • TB4
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