# AP calculus

Consider the sequence t0 = 3, t1 = 3^3, t2 = 3^3^3 , t3 = 3^3^3^3,….. , defined by t0 = 3 and tn+1 = 3tn for n ≥ 0. What are the last 2 digits of t3 = 3^3^3^3 ? Show that the last 10 digits of tk are the same for all k ≥ 10

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1. (3^3)^3 which would be 27^3 = 19683
or
2. 3^(3^3) ? which would be 3^27 = 7.625... x 10^12

e.g. (9^9)^9 = (387,420,489)^9 or 1.966.. x 10^77 , quite large , but ...
9^(9^9) = 9^(387,420,489) , which is unimaginably larger.

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posted by Reiny
2. It's the 1.
3^3^3 = 19683

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posted by xikhun

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