GeminiHeat wrote:
If \(A_{n} = 7^n − 1\), what is the units digit of \(A_{33}\)?
A. 1
B. 3
C. 6
D. 7
E. 9
\(A_{33}\) = \(7^{33}\) - 1
7 has a repeating pattern of 4 (1, 9, 3, and 1) i.e. the units digit repeats itself every 4 powers of 7
\(7^1\) =
7\(7^2\) = 4
9\(7^3\) = 34
3\(7^4\) = 240
1\(7^5\) = 1680
7Now, to find the units digit in \(7^{33}\), we will divide 33 with 4 and find the remainder.
Here the remainder is 1, so the first digit in the repeating pattern would be the units digit of \(7^{33}\), which is 7
So, \(A_{33}\) = \(7^{33}\) - 1 = ____7 - 1 = ____6
Hence, option C