Fibonacci 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

(The foregoing is given without proof regarding the properties of the Pisano period function, usually denoted as $\pi \left( n \right)$ , where $n$ is the base)

The sequence of the last digit of Fibonacci numbers repeats with a period of $60$ , so in base $10$ , we have $\pi \left( 10 \right) =60$ . In other bases $n=1,2,3,4,5,6,7,8,9,10,…$ we have

$1,3,8,6,20,24,16,12,24,60,…$

One property of the Pisano period function is

$\pi \left( { p }^{ a } \right) ={ p }^{ a-1 }\pi \left( p \right)$

where $p$ is a prime number, and $a$ is any integer

Another property of same is, for primes ${ p }_{ 1 },{ p }_{ 2 },{ p }_{ 3 },...$

$\pi \left( { { { { p }_{ 1 } }^{ a }{ { p }_{ 2 } }^{ b }{ { p }_{ 3 } }^{ c }... } } \right) =$

$LCM\left( \pi \left( { { p }_{ 1 } }^{ a } \right) ,\pi \left( { { p }_{ 2 } }^{ b } \right) ,\pi \left( { { p }_{ 3 } }^{ c } \right) ,...\quad \right)$

Hence, for $n={ 10 }^{ 7 }={ 2 }^{ 7 }{ 5 }^{ 7 }$ , we have

$\pi \left( { 10 }^{ 7 } \right) =LCM\left( { 2 }^{ 6 }\pi \left( 2 \right) ,{ 5 }^{ 6 }\pi \left( 5 \right) \right) =LCM\left( { 2 }^{ 6 }3,{ 5 }^{ 6 }20 \right) =15000000$

Now, consider the Fibonacci series from ${ F }_{ -1 }$ to ${ F }_{ 30 }$

$1,0,1,1,2,3,5,8,13,2,34, 55, 89, 144, 233, 377, 610, 987, 1597,$
$2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025,$
$121393, 196418, 317811, 514229, 832040$

after which the terms are greater than $6$ digits in length. This sequence will repeat at

${ F }_{ 15000000-1 }={ F }_{ 14999999 }$

so that the answer is $14999999$

Note that in order for this sequence of $32$ numbers to repeat exactly, the $7th$ digit (and others not shown) must be $0$ , making for a sequence of $32$ consecutive $0s$ in the $7th$ digit.

Also note that the Pisano period function, with its properties, applies to many other additive number sequences, such as Lucas numbers, so it's an useful function to be familiar with, one with interesting properties.