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The last two digits of $9^{9^9}$
How to find the last two digits of $9^{9^{9^9}}$ (a power tower of 4 $9$'s) ?
Is there any special approach to these kind of problems?
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Hint: the last digit is the number mod $10$,$81 mod 10=1$ – Belgi Jul 07 '12 at 15:04
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3You can follow the approach in this problem. Is this sufficiently similar to be a duplicate? – Zev Chonoles Jul 07 '12 at 15:04
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2@Zev, methinks yes. – J. M. ain't a mathematician Jul 07 '12 at 15:11
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Yes, enough @ZevChonoles – Hyperbola Jul 07 '12 at 15:22
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1Also some discussion at http://math.stackexchange.com/questions/166083/last-few-digits-of-nnn-cdot-cdot-cdotn/166133#166133 – Ross Millikan Jul 07 '12 at 15:22
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Related: How do I compute $a^b,\bmod c$ by hand? – Bart Michels Aug 20 '15 at 11:45