# The greatest common divisor of $\sum_\limits{k=0}^{5n-2}2^k$

I am not sure how the greatest common divisor of one number is defined. It can be $1$, or first greater number than $1$, or that number, or first smaller number from that given number.

If $n=1\Rightarrow 5n-2=3\Rightarrow\sum_\limits{k=0}^{5n-2}=15$.

The first number greater than $1$ that divides $15$ is $3$, and the first smaller than $15$ is $5$.

If $n=2\Rightarrow 5n-2=8\Rightarrow\sum_\limits{k=0}^{5n-2}=511$.

The first number greater than $1$ that divides $511$ is $7$, and the first smaller than $511$ is $73$.

Is it possible to find a property for greatest common divisor of $\sum_\limits{k=0}^{5n-2}2^k$ in both cases (first greater than $1$, and the second, smaller than that number)?

## 1 Answer

I imagine what your problem is asking is this:

What is the greatest common divisor of all the numbers of the form $$\sum_{k=0}^{5n-2} 2^k$$ for positive integer $n$.

This pretty much means:

Evaluate $$\gcd(\sum_{k=0}^3 2^k,\sum_{k=0}^{8} 2^k,\sum_{k=0}^{13} 2^k,...)$$

• Number of the form $\sum_\limits{k=0}^{8}2^k=511$ can't be divided by $3,5,15$. Is it correct that $gcd(...)=1$? – user300045 Nov 18 '16 at 20:59