Finding the Unit Digit of 124ⁿ + 124⁽ⁿ⁺¹⁾

The question asks for the unit digit of the expression $124^n + 124^{(n+1)}$, where $n$ is any whole number. We only need to consider the unit digit of the base, which is 4.

Unit Digit Pattern for Powers of 4

Let's examine the unit digits of powers of 4:

• $4^1 = 4$
• $4^2 = 16 \implies$ Unit digit is 6
• $4^3 = 64 \implies$ Unit digit is 4
• $4^4 = 256 \implies$ Unit digit is 6

The pattern for the unit digit of $4^n$ (for $n \ge 1$) alternates between 4 and 6. Specifically:

• If the exponent $n$ is odd, the unit digit is 4.
• If the exponent $n$ is even, the unit digit is 6.

Calculating the Unit Digit of the Sum

The unit digit of $124^n + 124^{(n+1)}$ is determined by the sum of the unit digits of $124^n$ and $124^{(n+1)}$. Let $U(\text{number})$ denote the unit digit.

We need to find $U(4^n) + U(4^{(n+1)})$.

Case 1: $n$ is odd

If $n$ is odd, then $n+1$ is even.

• $U(124^n) = U(4^n) = 4$
• $U(124^{(n+1)}) = U(4^{(n+1)}) = 6$
• The unit digit of the sum is $U(4 + 6) = U(10) = 0$.

Case 2: $n$ is even

If $n$ is even, then $n+1$ is odd.

• $U(124^n) = U(4^n) = 6$
• $U(124^{(n+1)}) = U(4^{(n+1)}) = 4$
• The unit digit of the sum is $U(6 + 4) = U(10) = 0$.

Conclusion

In both cases (when $n$ is odd and when $n$ is even), the unit digit of $124^n + 124^{(n+1)}$ is 0.

Therefore, the digit in the unit's place is 0.