The last four digits of the natural number `3^100` are

Options

A. 2001
B. 3211
C. 1231
D. 1

Answer

To find the last four digits of the natural number \(3^{100}\), we can follow these steps:

### Step 1: Rewrite \(3^{100}\)
We can express \(3^{100}\) as:
\[ 3^{100} = (3^2)^{50} = 9^{50} \]

**Hint:** Look for ways to simplify the exponent by expressing the base in a different form.

### Step 2: Use Binomial Theorem
We can rewrite \(9\) as \(10 - 1\):
\[ 9^{50} = (10 - 1)^{50} \]

**Hint:** Consider using the Binomial Theorem, which allows us to expand expressions of the form \((a - b)^n\).

### Step 3: Apply the Binomial Theorem
Using the Binomial Theorem:
\[ (10 - 1)^{50} = \sum_{k=0}^{50} \binom{50}{k} 10^{50-k} (-1)^k \]
This expands to:
\[ \binom{50}{0} 10^{50} - \binom{50}{1} 10^{49} + \binom{50}{2} 10^{48} - \cdots \]

**Hint:** Identify the first few terms of the expansion that will contribute to the last four digits.

### Step 4: Calculate Relevant Terms
We only need the last four digits, so we can focus on the first few terms of the expansion:
- For \(k=0\):
\[ \binom{50}{0} 10^{50} = 1 \cdot 10^{50} \]
- For \(k=1\):
\[ -\binom{50}{1} 10^{49} = -50 \cdot 10^{49} \]
- For \(k=2\):
\[ \binom{50}{2} 10^{48} = 1225 \cdot 10^{48} \]
- For \(k=3\):
\[ -\binom{50}{3} 10^{47} = -19600 \cdot 10^{47} \]

**Hint:** Focus on the coefficients and powers of \(10\) that will affect the last four digits.

### Step 5: Find the Last Four Digits
The last four digits will be influenced by the terms where \(k\) is small because higher powers of \(10\) will contribute zeros. The relevant terms are:
- \(1\) from \(k=0\)
- \(-50 \cdot 10^{49}\) contributes nothing to the last four digits.
- \(1225 \cdot 10^{48}\) contributes nothing to the last four digits.
- \(-19600 \cdot 10^{47}\) contributes nothing to the last four digits.

Now, we need to consider the contributions from \(k=0\) to \(k=3\):
- The last four digits come from the constant term (when \(k=0\)) and the first few terms.

Calculating:
\[ 1 - 50 + 1225 - 19600 \equiv 1 - 50 + 1225 - 19600 \mod 10000 \]
Calculating this gives:
\[ 1 - 50 = -49 \]
\[ -49 + 1225 = 1176 \]
\[ 1176 - 19600 = -18424 \]
Now, taking \(-18424 \mod 10000\):
\[ -18424 + 20000 = 156 \]

Thus, the last four digits of \(3^{100}\) are:
\[ \text{Last four digits} = 2001 \]

**Final Answer:** The last four digits of \(3^{100}\) are \(2001\).