Well That Escalated Quickly

We know that the greatest common divisor of the first two numbers $105=3\times5\times7$ and $120 = 3\times5\times 8$ is $15$ .

So inclusive of the third large number $310000000000000000000000000000005$ , the greatest common divisor must be $15$ or less. Let's hope that we can prove that this large number is also divisible by $15$ so that the greatest common divisor of these three numbers is $15$ .

By divisiblity rules of 5, the large number is divisible by 5 because of its last digit is 5.

Similarly, by divisiblity rules of 3, the large number is divisible by 3 because the sum of digits $3+1+0 + 0 + \cdots + 0 + 5 = 9$ is divisible by 3.

Hence this number is indeed divisible by $15$ , and our answer is indeed $\boxed{15}$ .

The factorization of each number gives $105 = 3 \times 5 \times 7$ and $120 = 3 \times 5 \times 8$ , and $31000000000000000000000000000005$ is divisible by $5$ and 9 (since $3 + 1 + 5 = 9$ ). Because the common factors of the numbers are the greatest common divisor of the three numbers is 15.