Find the last digit of the sum `19^(81)+4^(9k),KinN`.

Options

A. 4
B. 9
C. 3
D. Cannot be determined

Answer

To find the last digit of the sum \( 19^{81} + 4^{9k} \) where \( k \) is a natural number, we can break down the problem into two parts: calculating the last digit of \( 19^{81} \) and \( 4^{9k} \).

### Step 1: Find the last digit of \( 19^{81} \)

The last digit of a number is determined by its unit digit. The unit digit of \( 19 \) is \( 9 \).

Now, we need to find the last digit of \( 9^{81} \). The last digits of powers of \( 9 \) follow a pattern:
- \( 9^1 = 9 \) (last digit is \( 9 \))
- \( 9^2 = 81 \) (last digit is \( 1 \))
- \( 9^3 = 729 \) (last digit is \( 9 \))
- \( 9^4 = 6561 \) (last digit is \( 1 \))

The pattern repeats every two powers: \( 9, 1, 9, 1, \ldots \)

Since \( 81 \) is odd, the last digit of \( 9^{81} \) is \( 9 \).

### Step 2: Find the last digit of \( 4^{9k} \)

Next, we look at \( 4^{9k} \). The last digits of powers of \( 4 \) also follow a pattern:
- \( 4^1 = 4 \) (last digit is \( 4 \))
- \( 4^2 = 16 \) (last digit is \( 6 \))
- \( 4^3 = 64 \) (last digit is \( 4 \))
- \( 4^4 = 256 \) (last digit is \( 6 \))

The pattern repeats every two powers: \( 4, 6, 4, 6, \ldots \)

Now, since \( 9k \) is always even (as \( 9 \) is odd and \( k \) is a natural number), the last digit of \( 4^{9k} \) will be \( 6 \).

### Step 3: Add the last digits

Now we can add the last digits we found:
- Last digit of \( 19^{81} \) is \( 9 \)
- Last digit of \( 4^{9k} \) is \( 6 \)

Thus, we calculate:
\[ 9 + 6 = 15 \]

The last digit of \( 15 \) is \( 5 \).

### Conclusion

The last digit of the sum \( 19^{81} + 4^{9k} \) is \( 5 \).

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