Giant Power Divisors

The first term is odd, since any product of 5s (odd numbers) is odd. The second term is 1

The sum of two odd numbers is even, so this gigantic number is even. Its smallest prime divisor is 2.

One should note that positive integral powers of $5$ always ends in $5$ , that is, the unit digit of any positive integral power of $5$ will always be $5$ . If $1$ is added to such power, the unit digit will then become $6$ . Since the smallest prime divisor of $6$ is $2$ , then, the smallest prime divisor of $5^{8^{12^{15^{104}}}} + 1$ must be $2$ .

Any odd number raised at any power results in a odd number too. So, an odd number plus 1 results in a even number, which have 2 as its smallest prime divisor.

5 raised to any power is an odd number. Adding one to that makes it an even number, and the smallest prime divisor of any even number will always be 2. Therefore, 2.

i guess,5 to the power anything such as 5^1=5;

5^2-25;

5^3=125

always the last digit is 5.so 5+1=6,which can be divided by 2(smallest prime divisor).

so 2 is the answer

(Odd Number)^n = odd , Hence the answer is 2

We can also explain this in terms of modulo.

we see that $5 \equiv 1 \pmod{2}$ , therefore $5^n \equiv 1^n \equiv 1 \pmod{2}$ Also, $1 \equiv -1 \pmod{2}$ therefore $5^n+1 \equiv 1-1 \equiv 0\pmod{2}$ here 2 is a factor of any $5^n+1$

5^8^12^15^104 unit place is 5 +1 = 6 So smallest prime divisible is 2

No importance how many time you multiply 5 by itself, the result will always finish by 5. If you add 1, the result will be even so the smallest prime divisor will be 2

The last digit of $5^n$ is always 5 and $5^n+1$ is 6 since 6 is a even number so the answer is $\boxed{\large{2}}$

Let N be a divisor (not necessarily prime) of the gigantic number. Thus,

$5^{8^{12^{15^{104}}}} + 1 \equiv 0 \pmod{N}$

$5^{8^{12^{15^{104}}}} \equiv -1 \pmod{N}$

By the exponentiation property in modular arithmetic,

$5 \equiv -1 \pmod{N}$

Since $-1 \equiv (N - 1) \pmod{N}$ , we can say,

$5 \equiv N - 1 \pmod{N}$

Thus, $5 = N - 1$ or equivalenty $N = 6$ is a solution.

Thus, $6$ is a divisor of this gigantic number.

Since $6 = 2 * 3$ , the smallest prime divisor of the gigantic number is $2$ .

as we can see if 5 is raised to any power then unit digit is 5 so unit digit of resulting number is 5+1=6 smallest prime divisor =2. ANSWER =2

$\LARGE\color{#3D99F6}{5}^{\color{#D61F06}{\textbf{whatever}}}\normalsize\textbf{ is forever an odd number}\textbf{ and the last number (units) will be 5}$

$\large\textbf{so}$

$\large XXXXXX..5+1=XXXXXX..6 \text{ which is divisible by 2 the smallest prime}$

I am curious. 1 is a prime number and ,while leaving the number itself as an answer, is a divisor of any number. Why is 1 incorrect?

all Powers of five have unit digit 5. and the given no is adding +1 to that . it means the number comes to be even number (as ending with 6 digit ) and 2 is smallest prime divisor for all even number

Alternate : (odd)^n =odd
odd+1= even (divisible by 2)

5 raised to any positive integer will have 5 in its units place. So 5+1 = 6 in units place of the given expression. Hence smallest prime divisor will be 2.

5^∞ will always have 5 as its last digit so 5 +1 = 6 and smallest divisor of 6 is 2