Find the last digit, last two digits and last three digits of the number `(81)^(25)`

Answer

To find the last digit, last two digits, and last three digits of the number \( (81)^{25} \), we can follow these steps:

### Step 1: Rewrite the base
We start by rewriting \( 81 \) in terms of \( 10 \):
\[ 81 = 9^2 \]
Thus,
\[ (81)^{25} = (9^2)^{25} = 9^{50} \]

### Step 2: Use the Binomial Theorem
Next, we can express \( 9 \) as \( 10 - 1 \):
\[ 9^{50} = (10 - 1)^{50} \]
Now we apply the Binomial Theorem:
\[ (10 - 1)^{50} = \sum_{k=0}^{50} \binom{50}{k} 10^k (-1)^{50-k} \]
This expands to:
\[ = \binom{50}{0} 10^{50} - \binom{50}{1} 10^{49} + \binom{50}{2} 10^{48} - \ldots + (-1)^{50} \binom{50}{50} \]

### Step 3: Identify significant terms
To find the last digit, last two digits, and last three digits, we only need to consider the terms that contribute to these digits:
- The terms with \( k = 0, 1, 2 \) will be significant for the last three digits.
- All higher powers of \( 10 \) will contribute \( 0 \) to the last three digits.

The relevant terms are:
1. \( \binom{50}{0} 10^{50} = 1 \cdot 10^{50} \) (not relevant for last three digits)
2. \( -\binom{50}{1} 10^{49} = -50 \cdot 10^{49} \) (not relevant for last three digits)
3. \( \binom{50}{2} 10^{48} = 1225 \cdot 10^{48} \) (not relevant for last three digits)
4. \( -\binom{50}{3} 10^{47} = -19600 \cdot 10^{47} \) (not relevant for last three digits)
5. \( \ldots \)
6. \( + \binom{50}{50} = 1 \)

### Step 4: Calculate the last three digits
We will focus on the last few terms:
- The last significant term for our calculation is:
\[ (-1)^{50} \cdot \binom{50}{50} = 1 \]
Thus, the last three digits of \( (81)^{25} \) can be calculated as:
\[ 1 \text{ (from the last term)} \]

### Step 5: Conclusion
From the above calculations:
- The last digit of \( (81)^{25} \) is \( 1 \).
- The last two digits of \( (81)^{25} \) are \( 01 \).
- The last three digits of \( (81)^{25} \) are \( 001 \).

### Final Answer:
- Last digit: \( 1 \)
- Last two digits: \( 01 \)
- Last three digits: \( 001 \)