Solution:

Divisibility by 25:

To check whether (13^100 17^100) is divisible by 25 or not, we need to check whether 25 divides the given number or not.

25 = 5 x 5

If a number is divisible by 25, then it should be divisible by both the factors 5 and 5.

Divisibility by 5:

Let's check whether the given number is divisible by 5 or not.

The last two digits of 13^100 are 29.

The last two digits of 17^100 are 01.

When we add these two numbers, we get 30.

Since 30 is divisible by 5, the given number is also divisible by 5.

Divisibility by 5:

Let's check whether the given number is divisible by 5 or not.

The last two digits of 13^100 are 29.

The last two digits of 17^100 are 01.

When we add these two numbers, we get 30.

Since 30 is divisible by 5, the given number is also divisible by 5.

Remainder when divided by 25:

Since the given number is divisible by 5, the last digit of the number should be either 0 or 5.

Let's try to find the last two digits of the number.

We can use Euler's Totient Function to find the last two digits of the number.

Euler's Totient Function:

If a and n are two coprime positive integers, then a^(phi(n)) is congruent to 1 modulo n, where phi(n) is the Euler's Totient Function.

phi(n) = n x (1 - 1/p1) x (1 - 1/p2) x ... x (1 - 1/pk), where p1, p2, ..., pk are the prime factors of n.

Let's find the value of phi(100):

phi(100) = 100 x (1 - 1/2) x (1 - 1/5) = 40

Now, let's find the last two digits of 13^40 and 17^40.

13^40 = (13^10)^4

Using binomial theorem, we can expand (13^10)^4.

(13^10)^4 = (169^5) x (10C0 + 10C1 x 169 + 10C2 x 169^2 + ... + 10C10 x 169^10)

We need to find the remainder of 10C0 + 10C1 x 169 + 10C2 x 169^2 + ... + 10C10 x 169^10 when divided by 100.

Using congruence modulo 100, we can simplify the above expression.

10C0 + 10C1 x 169 + 10C2 x 169^2 + ... + 10C10 x 169^10 = (1 + 169)^10 - (1 + 1)^10

= 170^10 - 2^10

= 048

Therefore, the last two digits of 13^40 are 48.

17^40 = (17^10)^4

Using binomial theorem, we can expand (17^10)^4.

(17^10)^