A Golden Fraction II

Number Theory Level 4

$0.1123101\ldots$

Consider a decimal such that the $n^{\text{th}}$ digit to the right of the decimal place is the $n^{\text{th}}$ term of the Fibonacci Sequence $\mod 4$ , as shown above.

The exact value of this decimal can be expressed in the form $\dfrac{a}{b}$ for coprime positive integers $a$ and $b$ .

What is the value of $a+b$ ?

The answer is 101119.

1 solution

Brandon Monsen
Nov 27, 2015

since our repeating decimal looks like

$0.\overline{112310}=k$

We can count that it repeats every $6$ decimal places, and so $10^{6} \times k=112310.\overline{112310}$

We can then say that

$10^{6} \times k-k=112310$ $999999k=112310$ $k=\frac{112310}{999999}$

We can see that $gcd(999999,112310)=11$ , and so our fraction is reduced to

$\frac{10210}{90909}$

Since these two numbers are relatively prime, the value of $a+b$ is

$10210+90909=\boxed{101119}$

Don't you have to prove that the Pisano Period is 6?

Pi Han Goh - 5 years, 6 months ago

it can be proved by observing some more decimal places

Dev Sharma - 5 years, 6 months ago

Once you hit $1,0$ , the period is over as the next two places have to be $1,1$ , and thus starting the pattern over. I'm guessing that's what you wanted me to say?

Brandon Monsen - 5 years, 6 months ago

Yeah. That should be more explicit. haha

Pi Han Goh - 5 years, 6 months ago

This might help. Fibonacci numbers have a certain property that if a positive integer m is divisible by n, then the mth Fibonacci number is divisible by the nth Fibonacci number. Since 8 is the 6th Fibonacci number, every (6k)th Fibonacci number is divisible by 8 for all positive integer value of k.

Rindell Mabunga - 5 years, 4 months ago

A Golden Fraction II

The answer is 101119.

1 solution

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