What is the last (rightmost) digit of

$\displaystyle 23^{23^{23^{23}}}$?

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- Jul 10th 2009, 06:30 PMAlephZeroLarge Number
What is the last (rightmost) digit of

$\displaystyle 23^{23^{23^{23}}}$? - Jul 10th 2009, 07:53 PMTwistedOne151
The ones digits of powers of 23 cycle as do the ones digits of powers of 3, and as 3x3=9, 3x9=27, 3x7=21, 3x1=3, we see that the ones digits cycle through the values 3, 9, 7, 1. Thus, to find the ones digit of $\displaystyle 23^{23^{23^{23}}}$, we need only find $\displaystyle 23^{23^{23}}$ mod 4. Now, 23 mod 4 = 3, and so we see that

$\displaystyle 23^2$ mod 4 = 1, and thus $\displaystyle 23^n$ mod 4 is 3 when n is odd and 1 when n is even. And we see that since 23 is odd, $\displaystyle 23^{23}$ is odd, and so $\displaystyle 23^{23^{23}}$ mod 4 = 3, and thus the ones digit of $\displaystyle 23^{23^{23^{23}}}$ is the same as that of $\displaystyle 3^3$, which is 7.

--Kevin C. - Jul 10th 2009, 08:02 PMAlephZero
Lovely.