Find last three digits in `19^(100)`

Answer

To find the last three digits of \( 19^{100} \), we can follow these steps:

### Step 1: Rewrite the expression
We can express \( 19^{100} \) as:
\[ 19^{100} = (19^2)^{50} \]
Calculating \( 19^2 \):
\[ 19^2 = 361 \]
Thus, we have:
\[ 19^{100} = 361^{50} \]

### Step 2: Use the Binomial Theorem
We can rewrite \( 361 \) as \( 360 + 1 \):
\[ 361^{50} = (360 + 1)^{50} \]
Using the Binomial Theorem, we expand this:
\[ (360 + 1)^{50} = \sum_{r=0}^{50} \binom{50}{r} \cdot 360^{50-r} \cdot 1^r \]

### Step 3: Identify the relevant terms
We need only the last three digits of this expansion. The last three digits are influenced primarily by the terms where \( 360^{50-r} \) is small enough to not contribute significantly to the last three digits. The significant terms are \( r = 0, 1, 2 \).

Calculating the first few terms:
1. For \( r = 0 \):
\[ \binom{50}{0} \cdot 360^{50} \cdot 1^0 = 1 \cdot 360^{50} \]
The last three digits of \( 360^{50} \) are \( 000 \).

2. For \( r = 1 \):
\[ \binom{50}{1} \cdot 360^{49} \cdot 1^1 = 50 \cdot 360^{49} \]
The last three digits of \( 360^{49} \) are also \( 000 \).

3. For \( r = 2 \):
\[ \binom{50}{2} \cdot 360^{48} \cdot 1^2 = \frac{50 \cdot 49}{2} \cdot 360^{48} = 1225 \cdot 360^{48} \]
The last three digits of \( 360^{48} \) are still \( 000 \).

4. For \( r = 3 \):
\[ \binom{50}{3} \cdot 360^{47} \cdot 1^3 = \frac{50 \cdot 49 \cdot 48}{6} \cdot 360^{47} \]
The last three digits of \( 360^{47} \) are still \( 000 \).

5. For \( r = 50 \):
\[ \binom{50}{50} \cdot 360^{0} \cdot 1^{50} = 1 \]

### Step 4: Combine the results
The last three digits of \( (360 + 1)^{50} \) are determined by the constant term:
\[ 1 \]

### Final Answer
Thus, the last three digits of \( 19^{100} \) are:
\[ \boxed{001} \]