Exponential factorials

Number Theory Level 4

Find the remainder when 3 ! 4 ! 5 ! . . . . 2013 ! 3!^{4!^{5!^{.{.{.{.^{2013!}}}}}}} is divided by 11.


The answer is 5.

Rohit Udaiwal by Rohit Udaiwal , 20, India

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2 solutions

Otto Bretscher
Nov 24, 2015

We will be "working backwards", using Fermat twice, for 5 and 11: 5 ! 6 ! . . . ≡ 0 ( m o d 4 ) , 4 ! 5 ! . . ≡ 1 ( m o d 5 ) , 4 ! 5 ! . . ≡ 6 ( m o d 10 ) , 3 ! 4 ! . . ≡ 6 6 ≡ 3 3 ≡ 5 ( m o d 11 ) 5!^{6!...}\equiv 0 \pmod4, \hspace{6mm} 4!^{5!..}\equiv1 \pmod5, \hspace{6mm} 4!^{5!..}\equiv 6 \pmod{10}, \hspace{6mm} 3!^{4!..}\equiv6^6\equiv 3^3\equiv \boxed{5} \pmod{11}

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Which theorem of Fermat's are you talking about? And how would you apply that theorem?

Bloons Qoth - 4 years, 8 months ago
Ramiel To-ong
Dec 12, 2015

Fermat's theorem

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