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

Answer

To find the last three digits of \( 19^{100} \), we can use the Binomial Theorem. Here’s a step-by-step solution:

### Step 1: Rewrite \( 19^{100} \)
We can express \( 19 \) as \( 20 - 1 \). Therefore, we can rewrite \( 19^{100} \) as:
\[ 19^{100} = (20 - 1)^{100} \]

### Step 2: Apply the Binomial Theorem
According to the Binomial Theorem, we can expand \( (a - b)^n \) as follows:
\[ (a - b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} (-b)^k \]
In our case, \( a = 20 \), \( b = 1 \), and \( n = 100 \):
\[ (20 - 1)^{100} = \sum_{k=0}^{100} \binom{100}{k} 20^{100-k} (-1)^k \]

### Step 3: Focus on the last three digits
To find the last three digits of \( 19^{100} \), we need to compute \( (20 - 1)^{100} \mod 1000 \).

### Step 4: Calculate relevant terms
We need to consider the terms in the expansion that will contribute to the last three digits:
- The term when \( k = 0 \):
\[ \binom{100}{0} 20^{100} (-1)^0 = 20^{100} \]
- The term when \( k = 1 \):
\[ \binom{100}{1} 20^{99} (-1)^1 = -100 \cdot 20^{99} \]
- The term when \( k = 2 \):
\[ \binom{100}{2} 20^{98} (-1)^2 = \frac{100 \cdot 99}{2} \cdot 20^{98} = 4950 \cdot 20^{98} \]

### Step 5: Calculate \( 20^{100} \mod 1000 \)
Calculating \( 20^{100} \mod 1000 \):
\[ 20^{100} = (20^3)^{33} \cdot 20^1 \equiv 8000^{33} \cdot 20 \equiv 0 \mod 1000 \]

### Step 6: Calculate \( 20^{99} \mod 1000 \)
Calculating \( 20^{99} \mod 1000 \):
\[ 20^{99} = (20^3)^{33} \cdot 20^0 \equiv 8000^{33} \cdot 1 \equiv 0 \mod 1000 \]

### Step 7: Calculate \( 20^{98} \mod 1000 \)
Calculating \( 20^{98} \mod 1000 \):
\[ 20^{98} = (20^3)^{32} \cdot 20^2 \equiv 8000^{32} \cdot 400 \equiv 0 \mod 1000 \]

### Step 8: Combine the results
Since \( 20^{100} \), \( 20^{99} \), and \( 20^{98} \) are all \( 0 \mod 1000 \), we can conclude:
\[ 19^{100} \equiv 0 + 0 + 0 \equiv 0 \mod 1000 \]

### Final Result
Thus, the last three digits of \( 19^{100} \) are:
\[ \boxed{000} \]