When `7^(21)+7^(22)+7^(23)+7^(24)` is divided by 25 the remainder is :

Options

A. 5
B. 17
C. 7
D. 0

Answer

To solve the problem of finding the remainder when \( 7^{21} + 7^{22} + 7^{23} + 7^{24} \) is divided by 25, we can follow these steps:

### Step 1: Factor out the common term
We notice that \( 7^{21} \) is a common factor in all the terms:
\[ 7^{21} + 7^{22} + 7^{23} + 7^{24} = 7^{21}(1 + 7 + 7^2 + 7^3) \]

### Step 2: Calculate the sum inside the parentheses
Now, we need to calculate \( 1 + 7 + 7^2 + 7^3 \):
\[ 7^2 = 49 \quad \text{and} \quad 7^3 = 343 \]
Thus,
\[ 1 + 7 + 49 + 343 = 1 + 7 + 49 + 343 = 400 \]

### Step 3: Substitute back into the expression
Now, we can substitute back into our expression:
\[ 7^{21}(400) \]

### Step 4: Find \( 7^{21} \mod 25 \)
Next, we need to find \( 7^{21} \mod 25 \). To do this, we can use Euler's theorem. First, we find \( \phi(25) \):
\[ \phi(25) = 25 \left(1 - \frac{1}{5}\right) = 20 \]
According to Euler's theorem, since \( 7 \) and \( 25 \) are coprime:
\[ 7^{20} \equiv 1 \mod 25 \]
Thus,
\[ 7^{21} = 7^{20} \cdot 7 \equiv 1 \cdot 7 \equiv 7 \mod 25 \]

### Step 5: Calculate \( 400 \mod 25 \)
Now, we need to find \( 400 \mod 25 \):
\[ 400 \div 25 = 16 \quad \text{(exactly, so the remainder is 0)} \]
Thus,
\[ 400 \equiv 0 \mod 25 \]

### Step 6: Combine results
Now we combine our results:
\[ 7^{21} \cdot 400 \mod 25 \equiv 7 \cdot 0 \equiv 0 \mod 25 \]

### Conclusion
Therefore, the remainder when \( 7^{21} + 7^{22} + 7^{23} + 7^{24} \) is divided by 25 is:
\[ \boxed{0} \]