Fibonacci

Number Theory Level 5

$\large F_n = x! + y!$

The above equation holds true for some positive integers $n$ , $x$ and $y$ . Find the largest $n<100$ satisfying this condition, and submit your answer as $n+x+y$ .

Notations :

$F_n$ denote the $n^\text{th}$ Fibonacci number , where $F_0 = 0, F_1 = 1$ and $F_n = F_{n-1} + F_{n-2}$ for $n=2,3,4,\ldots$ .

Notation : $!$ denotes the factorial notation. For example, $8! = 1\times2\times3\times\cdots\times8$ .

The answer is 21.

2 solutions

Mark Hennings
Sep 8, 2016

This paper proves that in fact $12$ is the largest possible value of $n$ such that $F_n = x! + y!$ for positive integers $x,y$ (and not just the largest in the range $n<99$ ). We have $F_{12} = 144 = 5! + 4!$ , so the solution is $\boxed{21}$ .

Hi Mark, thanks for the mention of the paper. I liked it and will get a copy.

Giorgio Coniglio - 4 years, 9 months ago

Michael Mendrin
Jul 30, 2016

Actually, there's these that work

$1!+1!=2$
$1!+2!=3$
$2!+3!=8$
$4!+5!=144$

all of which are Fibonacci numbers. Probably the question should be reworded.

Yes, you are right and I am rewording the problem. Thanks!

Giorgio Coniglio - 4 years, 10 months ago

Thanks. I've updated the problem statement.

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Brilliant Mathematics Staff - 4 years, 10 months ago

Can you please post a solution sir?

Harsh Shrivastava - 4 years, 10 months ago

Fibonacci

The answer is 21.

2 solutions

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