<s>[INST] How to find whether the sum of a geometric series is prime or divisible by 3, 13 or 125? Let $A = 1+5+5^2+\dots+5^{99}$, then $A$ is:


*

*A prime number

*not divisible by 3

*divisible by 13

*divisible by 125


I know this is a sum of a Geometric Progression, so $ A = (5^{100}-1)/4$ but I cannot find $5^{100}$ So I thought of finding a pattern between the progression starting from 1. I get 1,6,31,131,... but cannot seem to be finding any pattern. May someone Help?
 [/INST] I will assume that you mean $1+5+5^2+\cdots +5^{99}$. Let's deal with the choices one at a time.
Of course 4.) is false, all the terms except one are multiples of $5$. So the sum cannot be a multiple of $5$. 
1.) is clearly false, we have an even number of odd numbers. The sum is therefore even, and thus not prime.
2.) Group in groups of $2$. Note that $1+5$ is a multiple of $3$, and so is $5^2+5^3$, which is $5^2(1+5)$. The same is true of $5^4+5^5=5^4(1+5)$, and so on. So our whole sum is a multiple of $3$.
3.) Note that $1+5+5^2+5^3$ is a multiple of $13$, and therefore so is $5^4+5^5+5^6+5^7$ and so on. (We are grouping in groups of $4$.)
</s>