Find the last digit of `2^(35)`.

Answer

To find the last digit of \(2^{35}\), we can observe the pattern of the last digits of the powers of 2. Let's calculate the last digits of the first few powers of 2:

1. \(2^1 = 2\) (last digit is 2)
2. \(2^2 = 4\) (last digit is 4)
3. \(2^3 = 8\) (last digit is 8)
4. \(2^4 = 16\) (last digit is 6)
5. \(2^5 = 32\) (last digit is 2)
6. \(2^6 = 64\) (last digit is 4)
7. \(2^7 = 128\) (last digit is 8)
8. \(2^8 = 256\) (last digit is 6)

From the above calculations, we can see that the last digits of the powers of 2 repeat every 4 terms:

- Last digit of \(2^1\) is 2
- Last digit of \(2^2\) is 4
- Last digit of \(2^3\) is 8
- Last digit of \(2^4\) is 6
- Then it repeats: \(2^5\) is 2, \(2^6\) is 4, \(2^7\) is 8, \(2^8\) is 6, and so on.

Now, to find the last digit of \(2^{35}\), we need to determine where \(35\) falls in this cycle of 4. We can do this by finding the remainder when \(35\) is divided by \(4\):

\[ 35 \div 4 = 8 \quad \text{remainder } 3 \]

This means that \(35 \mod 4 = 3\). According to our pattern:

- If the remainder is \(1\), the last digit is \(2\).
- If the remainder is \(2\), the last digit is \(4\).
- If the remainder is \(3\), the last digit is \(8\).
- If the remainder is \(0\), the last digit is \(6\).

Since \(35 \mod 4 = 3\), the last digit of \(2^{35}\) is \(8\).

Thus, the last digit of \(2^{35}\) is **8**.