Difference between revisions of "2007 iTest Problems/Problem 53"

(Solution to Problem 53)
 
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''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.''
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== Problem ==
 
== Problem ==
Let <math>T=\text{TNFTPP}</math>. Three distinct positive Fibonacci numbers, all greater than <math>T</math>, are in arithmetic progression. Let <math>N</math> be the smallest possible value of their sum. Find the remainder when <math>N</math> is divided by <math>2007</math>.
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Three distinct positive Fibonacci numbers, all greater than <math>1536</math>, are in arithmetic progression. Let <math>N</math> be the smallest possible value of their sum. Find the remainder when <math>N</math> is divided by <math>2007</math>.
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==Solution==
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By definition, for a Fibonacci number, <math>a_1 = a_2 = 1</math> and <math>a_n = a_{n-1} + a_{n-2}</math>.  From the definition, <math>a_{n+1} = a_n + a_{n-1}</math>.  That means the numbers <math>a_{n-2}</math>, <math>a_n</math>, and <math>a_{n+1}</math> are in arithmetic progression with common difference <math>a_{n-1}</math>.
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Writing out the Fibonacci numbers, the first numbers that come after <math>1536</math> are <math>1597</math>, <math>2584</math>, <math>4181</math>, and <math>6765</math>.  That means the desired three numbers are <math>1597</math>, <math>4181</math>, and <math>6765</math>.  The sum of the three numbers is <math>12543</math>, and the remainder after dividing by <math>2007</math> is <math>\boxed{501}</math>.
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==See Also==
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{{iTest box|year=2007|num-b=52|num-a=54}}

Latest revision as of 01:05, 25 June 2018

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
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