I see a pattern of ones for the even powered, such as 9 to the 2nd, 4th, 6th powers. And I see nines in the ones place when powering 9 to the 3rd, 5th, 7 powers. So I say the answer is 9.

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I see a pattern of ones for the even powered, such as 9 to the 2nd, 4th, 6th powers. And I see nines in the ones place when powering 9 to the 3rd, 5th, 7 powers. So I say the answer is 9.

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I'm not quite sure what you mean by this...? (Note: It is generally more helpful toNekkamath said:I see a pattern of ones for the even powered, such as 9 to the 2nd, 4th, 6th powers. And I see nines in the ones place when powering 9 to the 3rd, 5th, 7 powers.

Try to find a pattern or cycle in the ones digits:

. . . . .9[sup:3jcm1vx1]0[/sup:3jcm1vx1] = 1

. . . . .9[sup:3jcm1vx1]1[/sup:3jcm1vx1] = 9

. . . . .9[sup:3jcm1vx1]2[/sup:3jcm1vx1] = 81

. . . . .9[sup:3jcm1vx1]3[/sup:3jcm1vx1] = 729

. . . . .9[sup:3jcm1vx1]4[/sup:3jcm1vx1] = 6561

. . . . .9[sup:3jcm1vx1]5[/sup:3jcm1vx1] = 59049

. . . . .9[sup:3jcm1vx1]6[/sup:3jcm1vx1] = 531441

Clearly, this pattern must continue, due simply to the nature of multiplication. Every even power has a ones digit of "1"; every odd power has a ones digit of "9".

Since fifty-five is odd, the ones digit has to be "9".

Eliz.

\(\displaystyle \text{What is the ones-digit when }9^{55}\text{ is expressed as an integer?}\)

\(\displaystyle \text{I see a pattern of 1s for the even-powered, such as }9^2,\;9^4,\;9^6 \text{powers.}\) . . . . Good!

\(\displaystyle \text{And I see }9s\text{ in the ones-place for }9^3,\;9^5,\,9^7\text{ powers.}\) . . . . Yes!

\(\displaystyle \text{ So I say the answer is 9.}\) . . . . Correct!

That is probably the way they expected you to solve it.