The remainder when `27^(10)+7^(51)` is divided by `10` (a) 4 (b) 6 (c) 9 (d) 2

Options

A. `4`
B. `6`
C. `9`
D. `2`

Explanation

To find the remainder when \( 27^{10} + 7^{51} \) is divided by \( 10 \), we can simplify the problem using modular arithmetic.

### Step 1: Simplify \( 27^{10} \mod 10 \)

First, we can reduce \( 27 \) modulo \( 10 \):
\[ 27 \equiv 7 \mod 10 \]
Thus, we can rewrite \( 27^{10} \) as:
\[ 27^{10} \equiv 7^{10} \mod 10 \]

### Step 2: Calculate \( 7^{10} \mod 10 \)

Next, we need to find \( 7^{10} \mod 10 \). We can observe the pattern of powers of \( 7 \) modulo \( 10 \):
- \( 7^1 \equiv 7 \mod 10 \)
- \( 7^2 \equiv 49 \equiv 9 \mod 10 \)
- \( 7^3 \equiv 7 \times 9 = 63 \equiv 3 \mod 10 \)
- \( 7^4 \equiv 7 \times 3 = 21 \equiv 1 \mod 10 \)

Notice that \( 7^4 \equiv 1 \mod 10 \). This means the powers of \( 7 \) repeat every \( 4 \) terms. Therefore, we can find \( 7^{10} \) as follows:
\[ 10 \mod 4 = 2 \]
Thus,
\[ 7^{10} \equiv 7^2 \equiv 9 \mod 10 \]

### Step 3: Simplify \( 7^{51} \mod 10 \)

Now, let's calculate \( 7^{51} \mod 10 \). Again, using the periodicity we found:
\[ 51 \mod 4 = 3 \]
Thus,
\[ 7^{51} \equiv 7^3 \equiv 3 \mod 10 \]

### Step 4: Combine the results

Now we can combine the results of \( 27^{10} \) and \( 7^{51} \):
\[ 27^{10} + 7^{51} \equiv 9 + 3 \mod 10 \]
\[ 9 + 3 = 12 \equiv 2 \mod 10 \]

### Final Answer

Thus, the remainder when \( 27^{10} + 7^{51} \) is divided by \( 10 \) is:
\[ \boxed{2} \]