# Difference between revisions of "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 $\boxed{501}$.