Finding the Units Digit of an Expression with Powers

To find the digit in the units place of an expression involving sums and differences of numbers raised to powers, we only need to determine the units digit of each term separately and then perform the addition and subtraction using only those units digits.

The units digit of a number raised to a power follows a cycle. We need to find the units digit of $(34)^9$, $(46)^{21}$, and $(43)^{27}$.

Units Digit of (34)⁹

The units digit of the base number 34 is 4.

Let's look at the units digits of powers of 4:

• $4^1$ ends in 4
• $4^2 = 16$ ends in 6
• $4^3 = 64$ ends in 4
• $4^4 = 256$ ends in 6

The units digits of powers of 4 follow a cycle of 2: (4, 6). The units digit is 4 for odd powers and 6 for even powers.

Since the power is 9 (which is odd), the units digit of $(34)^9$ is 4.

Units Digit of (46)²¹

The units digit of the base number 46 is 6.

Let's look at the units digits of powers of 6:

• $6^1$ ends in 6
• $6^2 = 36$ ends in 6
• $6^3 = 216$ ends in 6

The units digit of any positive integer power of 6 is always 6.

So, the units digit of $(46)^{21}$ is 6.

Units Digit of (43)²⁷

The units digit of the base number 43 is 3.

Let's look at the units digits of powers of 3:

• $3^1$ ends in 3
• $3^2 = 9$ ends in 9
• $3^3 = 27$ ends in 7
• $3^4 = 81$ ends in 1
• $3^5 = 243$ ends in 3

The units digits of powers of 3 follow a cycle of 4: (3, 9, 7, 1).

To find the units digit of $3^{27}$, we divide the power (27) by the cycle length (4) and find the remainder.

$\frac{27}{4} = 6$ with a remainder of $3$.

The units digit of $3^{27}$ is the same as the units digit of $3^3$, which is 7.

So, the units digit of $(43)^{27}$ is 7.

Calculating the Final Units Digit

We need to find the units digit of $(34)^9 + (46)^{21} - (43)^{27}$.

This is equivalent to finding the units digit of (units digit of $(34)^9$) + (units digit of $(46)^{21}$) - (units digit of $(43)^{27}$).

The units digit of the expression is the units digit of $4 + 6 - 7$.

First, perform the addition: $4 + 6 = 10$. The units digit of this sum is 0.

Now, perform the subtraction using the units digits: (units digit of $10$) - (units digit of $7$). This is $0 - 7$.

When subtracting, if the units digit of the first number is smaller than the units digit of the second number, we effectively borrow 10 from the tens place in the units column. So, the units digit is $10 + 0 - 7 = 3$.

Alternatively, the expression's units digit is the units digit of $10 - 7$, which is 3.

Conclusion

The units digit of the expression $(34)^9 + (46)^{21} - (43)^{27}$ is 3.