Prime Exponentials

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Find the sum of the last two digits of 2 9 2 3 1 9 . . . 7 5 3 2 29^{23^{19^{...^{7^{5^{3^{2}}}}}}}


The answer is 9.

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1 solution

Pi Han Goh
Dec 20, 2013

I assume the number above 19 19 is 17 17 as you have listed the first 10 10 prime numbers in ascending order.

The trick is to apply Euler's Totient Function repeatedly

φ ( 100 ) = 40 , φ ( 40 ) = 16 , φ ( 100 ) = 8 \varphi(100) = 40, \varphi(40) = 16, \varphi(100) = 8

And gcd ( 29 , 100 ) = gcd ( 23 , 40 ) = gcd ( 19 , 16 ) = gcd ( 17 , 8 ) = 1 \text{gcd} (29,100) = \text{gcd} (23,40) = \text{gcd} (19,16) = \text{gcd} (17,8) = 1

2 9 2 3 1 9 1 7 . . 7 5 3 2 2 9 2 3 1 9 1 7 . . 7 5 3 2 ( m o d 100 ) 2 9 2 3 1 9 1 7 . . 7 5 3 2 ( m o d 40 ) ( m o d 100 ) 2 9 2 3 1 9 1 7 . . 7 5 3 2 ( m o d 16 ) ( m o d 40 ) ( m o d 100 ) 2 9 2 3 1 9 1 7 . . 7 5 3 2 ( m o d 8 ) ( m o d 16 ) ( m o d 40 ) ( m o d 100 ) 2 9 2 3 1 9 1 ( m o d 16 ) ( m o d 40 ) ( m o d 100 ) 2 9 2 3 3 ( m o d 40 ) ( m o d 100 ) 2 9 7 ( m o d 100 ) 9 \begin{aligned} \LARGE 29^{23^{19^{17^{.^{.^{7^{5^{3^2}}} }}}}} & \equiv & 29^{23^{19^{17^{.^{.^{7^{5^{3^2}}} }}}}} \pmod {100} \\ & \equiv & 29^{23^{19^{17^{.^{.^{7^{5^{3^2}}} }}}} \pmod {40} } \pmod {100} \\ & \equiv & 29^{23^{19^{17^{.^{.^{7^{5^{3^2}}} }}} \pmod {16} } \pmod {40} } \pmod {100} \\ & \equiv & 29^{23^{19^{17^{.^{.^{7^{5^{3^2}}} }} \pmod {8} } \pmod {16} } \pmod {40} } \pmod {100} \\ & \equiv & 29^{23^{19^{1 } \pmod {16} } \pmod {40} } \pmod {100} \\ & \equiv & 29^{23^3 \pmod {40} } \pmod {100} \\ & \equiv & 29^7 \pmod {100} \\ & \equiv & \boxed{9} \\ \end{aligned}

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