Question

Copy the table. $$ \begin{array}{|l|c|c|c|c|c|c|c|c|} \hline \text { Power } & 9^{1} & 9^{2} & 9^{3} & 9^{4} & 9^{5} & 9^{6} & 9^{7} & 9^{8} \\ \hline \text { Evaluate } & ? & ? & ? & ? & ? & ? & ? & ? \\ \hline \end{array} $$ Evaluate the powers of 9 in the table. What pattern do you see for the last digit of each product?

Short answer

The evaluation of the powers of 9 from \(9^{1}\) to \(9^{8}\) yields 9, 81, 729, 6561, 59049, 531441, 4782969, and 43046721 respectively. The pattern in the last digit for each product is that it alternates between 1 and 9.

Step-by-step solution

Calculation of the powers of 9

Calculate the value of each given power:\n- \(9^{1} = 9\),- \(9^{2} = 81\),- \(9^{3} = 729\),- \(9^{4} = 6561\),- \(9^{5} = 59049\),- \(9^{6} = 531441\),- \(9^{7} = 4782969\),- \(9^{8} = 43046721\).

Observing the last digit in each product

The last digit in each of the products are as follows:- \(9^{1} = 9\),- \(9^{2} = 81\),- \(9^{3} = 729\),- \(9^{4} = 6561\),- \(9^{5} = 59049\),- \(9^{6} = 531441\),- \(9^{7} = 4782969\),- \(9^{8} = 43046721\).Thus, for the power of 1, the last digit is 9. For powers of 2 and onwards, the digit in the ones place follows a pattern of alternating between 1 and 9.