Step 1: Understanding the Concept:
Finding the last two digits is equivalent to finding the remainder when the number is divided by 100. Binomial expansion is the best tool for this. Step 3: Detailed Explanation:
1. Rewrite 9 as \((10 - 1)\):
\[ 9^{50} = (10 - 1)^{50} \]
2. Expand using the Binomial Theorem:
\[ (10 - 1)^{50} = \binom{50}{0}10^{50} - \binom{50}{1}10^{49} + \dots - \binom{50}{49}10^{1} + \binom{50}{50}10^{0} \]
3. Since we need the remainder modulo 100, all terms containing \(10^{k}\) for \(k \geq 2\) are divisible by 100 and can be ignored.
4. Consider only the last two terms:
\[ 9^{50} \equiv \left[ -\binom{50}{49}10 + \binom{50}{50} \right] \pmod{100} \]
5. Calculate the values:
\[ -50 \times 10 + 1 = -500 + 1 = -499 \]
6. Reduce \(-499\) modulo 100:
\[ -499 \equiv 1 \pmod{100} \]
Thus, the last two digits are 01. Step 4: Final Answer:
The last two digits are 01.