To find the unit place digit of $(1768)^{1293}$, we only need to consider the unit digit of the base number, which is 8. The problem simplifies to finding the unit digit of $(8)^{1293}$.

Identifying the Cyclicity Pattern

Let's examine the pattern of the unit digits of the powers of 8:

• $8^1 = 8$ (Unit digit is 8)
• $8^2 = 64$ (Unit digit is 4)
• $8^3 = 8^2 \times 8 = 64 \times 8 = 512$ (Unit digit is 2)
• $8^4 = 8^3 \times 8 = 512 \times 8 = 4096$ (Unit digit is 6)
• $8^5 = 8^4 \times 8 = 4096 \times 8 = 32768$ (Unit digit is 8)

The pattern of the unit digits of powers of 8 is (8, 4, 2, 6). This pattern repeats every 4 powers. Therefore, the cyclicity of the unit digit of 8 is 4.

Calculating the Position in the Cycle

To find the unit digit of $(8)^{1293}$, we need to determine where in this cycle the 1293rd power falls. We can do this by finding the remainder when the exponent (1293) is divided by the cycle length (4).

We perform the division:

$$ 1293 = (4 \times 323) + 1 $$

The remainder is 1.

Determining the Final Unit Digit

Since the remainder is 1, the unit digit of $(8)^{1293}$ will be the same as the first unit digit in the cycle (8, 4, 2, 6). The first unit digit in the cycle is 8.

Therefore, the unit place of $(1768)^{1293}$ is 8.