Using the powers of 9, it is found that the last digit of the product will always be either 1 or 9.
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The multiplication of as many 9's as you want is equivalent to the powers of 9, which are:
[tex]9^2 = 9 \times 9 = 81[/tex]
[tex]9^3 = 9 \times 9 \times 9 = 729[/tex]
[tex]9^4 = 9 \times 9 \times 9 \times 9 = 6561[/tex]
[tex]9^5 = 9 \times 9 \times 9 \times 9 \times 9= 59049[/tex]
....
[tex]9^{10} = 3486784401[/tex]
Thus, the last digit of the products will always be either 1 or 9.
A similar problem is given at https://brainly.com/question/20692373