What are the last two digit of the number `9^(200)` ?

Options

A. A)19
B. B)21
C. C)41
D. D)1

Answer

To find the last two digits of \( 9^{200} \), we need to calculate \( 9^{200} \mod 100 \). Here’s how we can do it step by step:

### Step 1: Use the Binomial Theorem
We can express \( 9^{200} \) as \( (10 - 1)^{200} \). This allows us to use the Binomial Theorem.

### Step 2: Apply the Binomial Theorem
According to the Binomial Theorem:
\[ (a - b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} (-b)^k \]
In our case, \( a = 10 \), \( b = 1 \), and \( n = 200 \):
\[ (10 - 1)^{200} = \sum_{k=0}^{200} \binom{200}{k} 10^{200-k} (-1)^k \]

### Step 3: Identify Relevant Terms
We are interested in the last two digits, which means we only need to consider terms where \( 200-k \leq 2 \) (i.e., \( k \geq 198 \)). The relevant terms are:
- For \( k = 198 \):
\[ \binom{200}{198} 10^{2} (-1)^{198} = \binom{200}{2} 10^{2} = 19900 \]
- For \( k = 199 \):
\[ \binom{200}{199} 10^{1} (-1)^{199} = \binom{200}{1} 10^{1} (-1) = -2000 \]
- For \( k = 200 \):
\[ \binom{200}{200} 10^{0} (-1)^{200} = 1 \]

### Step 4: Combine the Terms
Now we combine these terms:
\[ 19900 - 2000 + 1 = 17901 \]

### Step 5: Find the Last Two Digits
Now, we need to find \( 17901 \mod 100 \):
\[ 17901 \mod 100 = 01 \]

### Conclusion
Thus, the last two digits of \( 9^{200} \) are **01**.

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