?
Nekkamath 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.
I'm not quite sure what you mean by this...? (Note: It is generally more helpful to
show your work, rather than just mention that you did some work. We can't correct steps we cannot see!)
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.