If `A_(n)=7^(n)-1`, what is the units digit of `A_(33)`?

Answer

To find the units digit of \( A_{33} = 7^{33} - 1 \), we will first determine the units digit of \( 7^{33} \).

### Step 1: Identify the pattern in the units digits of powers of 7
Let's calculate the units digits for the first few powers of 7:

- \( 7^1 = 7 \) (units digit is **7**)
- \( 7^2 = 49 \) (units digit is **9**)
- \( 7^3 = 343 \) (units digit is **3**)
- \( 7^4 = 2401 \) (units digit is **1**)

### Step 2: Recognize the repeating cycle
From the calculations above, we see that the units digits repeat every 4 powers:
- \( 7, 9, 3, 1 \)

### Step 3: Determine the position of \( 7^{33} \) in the cycle
To find the units digit of \( 7^{33} \), we need to find the remainder when 33 is divided by 4 (the length of the cycle):
\[ 33 \div 4 = 8 \quad \text{(remainder } 1\text{)} \]
This means \( 33 \equiv 1 \mod 4 \).

### Step 4: Find the corresponding units digit
Since \( 33 \equiv 1 \mod 4 \), the units digit of \( 7^{33} \) corresponds to the units digit of \( 7^1 \), which is **7**.

### Step 5: Calculate \( A_{33} \)
Now we need to find the units digit of \( A_{33} = 7^{33} - 1 \):
\[ \text{Units digit of } A_{33} = \text{Units digit of } (7 - 1) = 6 \]

### Conclusion
Thus, the units digit of \( A_{33} \) is **6**.

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