Math in Christmas: Christmas VS New Year

Let's find the last 2 digits of $12^{31}$ first.

A simple way to evaluate the last 2 digits of $12^{31}$ is via binomial expansion follow by repeated squaring.

$\begin{aligned} 12^{31} &=& (10 + 2)^{31} \\ &=& 100(\cdots) + 2^{31} + \dbinom{31}1 10 \cdot 2^{30} \\ & \equiv & 2^{30} (2 + 31\times10) = 2^{30} \times 312 \\ & \equiv & 2^{30} \times 12 \\ &\equiv & (2^{10})^3 \times 12 \\ &\equiv & 24^3\times 12 \\ &\equiv & 12^4 \times 8 \\ &\equiv & (12^2)^2 \times 8 \\ &\equiv & 4^2 \times 8 \\ &\equiv & 88 \pmod{100} \end{aligned}$

Now, let's find the last 2 digits of $25^{12}= (5^2)^{12} = 5^{24}$ .

Claim: The last 2 digits of $5^n$ is 25 for all positive integers $n\geq 2$ .

Proof: T̶h̶e̶ ̶p̶r̶o̶o̶f̶ ̶i̶s̶ ̶l̶e̶f̶t̶ ̶a̶s̶ ̶a̶n̶ ̶e̶x̶e̶r̶c̶i̶s̶e̶ ̶f̶o̶r̶ ̶t̶h̶e̶ ̶r̶e̶a̶d̶e̶r̶s̶.̶

This can be easily shown by induction .

Base cases: When $n=2$ , the last 2 digits of $5^n$ is 25, which is true.
Inductive hypothesis: Assume it's true that for $n=k$ be a positive integer greater or equal to 2, than the last 2 digits of $5^k$ is 25.
Similarly, the last 2 digits of $5^{k+1}$ is $25\times 5 = 125 \equiv 25 \pmod{100}$ as well.
Thus, then it is also true that $n=k+1$ .

By induction, our claim is true.

Hence, the last 2 digits of $25^{12}$ is 25.

By comparing the last 2 digits of these two numbers, the answer is Fantastic Santa with $25^{12}$ .