Finding the Unit Digit of Large Powers and Products

The problem asks us to find the unit digit of the expression $(123)^{123} \times (347)^{347} \times (568)^{568}$. The key idea is that the unit digit of a product is determined solely by the unit digits of the numbers being multiplied. Similarly, the unit digit of a number raised to a power depends only on the unit digit of the base and the exponent.

We need to find the unit digit of each part of the product separately and then multiply their unit digits.

Unit Digit of $(123)^{123}$

To find the unit digit of $(123)^{123}$, we only need to consider the unit digit of the base, which is 3. We look at the pattern of the unit digits of powers of 3:

• $3^1 = 3$
• $3^2 = 9$
• $3^3 = 27 \rightarrow$ unit digit is 7
• $3^4 = 81 \rightarrow$ unit digit is 1
• $3^5 = 243 \rightarrow$ unit digit is 3

The cycle of unit digits for powers of 3 is (3, 9, 7, 1). This cycle has a length of 4.

To find the unit digit of $3^{123}$, we need to find the position in this cycle, which is determined by the remainder of the exponent (123) when divided by the cycle length (4).

Calculate the remainder of $123 \div 4$: $123 = 4 \times 30 + 3$. The remainder is 3.

The unit digit corresponds to the 3rd element in the cycle (3, 9, 7, 1), which is 7.

Therefore, the unit digit of $(123)^{123}$ is 7.

Unit Digit of $(347)^{347}$

For $(347)^{347}$, we focus on the unit digit of the base, which is 7. Let's examine the cycle of unit digits for powers of 7:

• $7^1 = 7$
• $7^2 = 49 \rightarrow$ unit digit is 9
• $7^3 = 343 \rightarrow$ unit digit is 3
• $7^4 = 2401 \rightarrow$ unit digit is 1
• $7^5 = 16807 \rightarrow$ unit digit is 7

The cycle of unit digits for powers of 7 is (7, 9, 3, 1). This cycle also has a length of 4.

We find the unit digit of $7^{347}$ by looking at the remainder of the exponent (347) divided by the cycle length (4).

Calculate the remainder of $347 \div 4$: $347 = 4 \times 86 + 3$. The remainder is 3.

The unit digit is the 3rd element in the cycle (7, 9, 3, 1), which is 3.

Therefore, the unit digit of $(347)^{347}$ is 3.

Unit Digit of $(568)^{568}$

For $(568)^{568}$, we consider the unit digit of the base, which is 8. Let's find the pattern for powers of 8:

• $8^1 = 8$
• $8^2 = 64 \rightarrow$ unit digit is 4
• $8^3 = 512 \rightarrow$ unit digit is 2
• $8^4 = 4096 \rightarrow$ unit digit is 6
• $8^5 = 32768 \rightarrow$ unit digit is 8

The cycle of unit digits for powers of 8 is (8, 4, 2, 6). This cycle has a length of 4.

To determine the unit digit of $8^{568}$, we need the remainder of the exponent (568) when divided by the cycle length (4).

Calculate the remainder of $568 \div 4$: $568 = 4 \times 142 + 0$. The remainder is 0.

When the remainder is 0, it means the exponent is a multiple of the cycle length. In this case, the unit digit is the same as the last element of the cycle (the 4th element). The 4th element in the cycle (8, 4, 2, 6) is 6.

Therefore, the unit digit of $(568)^{568}$ is 6.

Calculating the Final Unit Digit

We found the unit digits of the three parts:

• Unit digit of $(123)^{123}$ is 7.
• Unit digit of $(347)^{347}$ is 3.
• Unit digit of $(568)^{568}$ is 6.

To find the unit digit of the product $(123)^{123} \times (347)^{347} \times (568)^{568}$, we multiply their unit digits:

Unit digit = Unit digit of ($7 \times 3 \times 6$)

First, multiply the first two unit digits: $7 \times 3 = 21$. The unit digit is 1.

Now, multiply this result's unit digit by the third unit digit: $1 \times 6 = 6$. The unit digit is 6.

Conclusion

The unit digit of the expression $(123)^{123} \times (347)^{347} \times (568)^{568}$ is 6.