Greatest common divisors in a sequence

We first evaluate $a_n$ and $a_{n+1}$ . $a_n=n!+100$ , and $a_{n+1}=(n+1)!+100$

$gcd(a_n, a_{n+1})$ $=gcd(n!+100, (n+1)!+100)$ $=gcd(n!+100, ((n+1)!+100)-(n+1)(n!+100))$ $=gcd(n!+100, -100n)=gcd(n!+100, 100n)$ .

Lemma $1$ : $gcd(n!+100, 100n)$ is not divisible by any other prime aside from $2$ and $5$ .

Suppose that $100n$ is divisible by any prime other than $2$ and $5$ . That implies that $n$ is divisible by that prime, because $100$ would be relatively prime to that prime. Thus $n!$ would also be divisible by that prime, and $n!+100$ would not be divisible by it.

Hence it is not possible for both $n!+100$ and $100n$ to be divisible by a prime other than $2$ or $5$ .

Lemma $2$ : $gcd(n!+100, 100n)$ is not divisible by $2^3=8$ .

We first consider $n \in {1, 2, 3, 4, 5}$ .

If $n \in {1, 3, 5}$ , then $100n$ will not be divisible by $8$ , because the highest power of $2$ divisible by $100n$ will be $4$ .

If $n=2$ , then $n!+100=2+100=102$ , which is not divisible by $8$ .

If $n=4$ , then $n!+100=24+100=124$ , which is not divisible by $8$ .

Now consider the cases where $n \ge 6$ .

Then $n!$ will be divisible by $8$ . However, $100$ is not, so $n!+100$ will not be divisible by $8$ .

$n!+100$ and $100n$ could not be both divisible by $8$ .

Thus for all $n$ , $gcd(n!+100, 100n)$ is not divisible by $8$ .

Lemma $3$ : $gcd(n!+100, 100n)$ is not divisible by $5^3=125$

We first consider $n \in \{1, 2, ..., 14\} \setminus \{5, 10\}$ , $100n$ is not divisible by $125$ , because the highest power of $5$ divisible by $100n$ will be $25$ .

If $n=5$ , then $5!+100=120+100=220$ , which is not divisible by $125$ .

If $n=10$ , then $10!+100=3628800+100=3629900$ , which is not divisible by $125$ .

If $n \ge 15$ , then $n!$ will be divisible by $125$ . However, $100$ is not, so $n!+100$ will not be divisible by $125$ .

$n!+100$ and $100n$ could not be both divisible by $125$ .

Thus for all $n$ , $gcd(n!+100, 100n)$ is not divisible by $125$ .

Hence the maximum possible GCD is $2^2 \cdot 5^2=100$ .

We would show that this is achievable.

If we let $n=10$ , then $a_n=10!+100$ , which is divisible by $100$ as both $10!$ and $100$ are divisible by $100$ . Also, $a_{n+1}=11!+100$ , which is also divisible by $100$ as both $11!$ and $100$ are divisible by $100$

Hence $gcd(a_n, a_{n+1})$ is divisible by $100$ .

Thus, the maximum possible value of $gcd(a_n, a_{n+1})$ is $\boxed{100}$ .

We can see that for the starting terms of the sequence, the gcd of two adjacent integers is 2,4 etc. We can infer that the maximum value of gcd must occur in the higher terms of the sequence.

$10!$ and all other factorials after that contain a minimum of two zeroes at the end. Because $10!$ contains the product : $10X5X2$ and so do all the higher factorials. Now, when you add 100 to a factorial containing a minimum of two zeroes, the result contains exactly two zeroes.

For example, $40500000+100=40500100$

And so, you can check for numbers containing any number of zeroes. Now, it is clear that the later terms of the sequence will have only two zeroes at their end. So, the Greatest common divisor of two numbers ending with 2 zeroes only is 100 .