Last two digits

What are the last 2 digits of 7 7 7 7 . . . 7 ? \Large7^{7^{7^{7^{.^{.^{.^{7}}}}}}}?

The tower has 7 sevens.

30 33 34 43 31

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1 solution

Chew-Seong Cheong
Feb 20, 2017

We note that 7 7 7 4 7 3 4 9 2 7 3 ( 50 1 ) 2 7 3 7 3 343 43 (mod 100) 7^7 \equiv 7^47^3 \equiv 49^2 7^3 \equiv (50-1)^27^3 \equiv 7^3 \equiv 343 \equiv 43 \text{ (mod 100)} ; 7 7 100 n + 43 \implies 7^7 \equiv 100n + 43 , where n n is an integer.

Now consider 7 7 7 7 100 n + 43 4 9 50 n + 20 7 3 ( 50 1 ) 50 n + 20 7 3 7 3 343 43 (mod 100) 7^{7^7} \equiv 7^{100n+43} \equiv 49^{50n+20}7^3 \equiv (50-1)^{50n+20}7^3 \equiv 7^3 \equiv 343 \equiv 43 \text{ (mod 100)} .

Similarly, 7 7 7 43 (mod 100) \large 7^{7^{7^{\cdot ^{\cdot ^\cdot}}}} \equiv \boxed{43} \text{ (mod 100)} .


Alternative solution

7 7 7 7 7 7 7 7 7 7 7 7 7 7 mod ϕ ( 100 ) (mod 100) As gcd ( 7 , 100 ) = 1 , Euler’s theorem applies. 7 7 7 7 7 7 7 mod 40 (mod 100) Euler’s totient function ϕ ( 100 ) = 40 7 7 7 7 7 7 7 mod 2 mod 4 mod 8 mod 16 mod 40 (mod 100) Repeated use of Euler’s theorem. 7 7 7 7 7 1 mod 4 mod 8 mod 16 mod 40 (mod 100) 7 3 (mod 4) 7 7 7 7 3 mod 8 mod 16 mod 40 (mod 100) 7 3 7 49 7 ( 48 + 1 ) 7 (mod 8) 7 7 7 7 mod 16 mod 40 (mod 100) 7 7 7 4 9 3 7 ( 48 + 1 ) 3 7 (mod 16) 7 7 7 mod 40 (mod 100) See note: 7 7 23 (mod 16) 7 23 (mod 100) 43 (mod 100) See note \large \begin{aligned} 7^{7^{7^{7^{7^{7^7}}}}} & \equiv 7^{7^{7^{7^{7^{7^7}}}} \color{#3D99F6} \text{ mod }\phi(100)} \text{ (mod 100)} & \small \color{#3D99F6} \text{As }\gcd(7,100)=1 \text{, Euler's theorem applies.} \\ & \equiv 7^{7^{7^{7^{7^{7^7}}}} \color{#3D99F6} \text{ mod }40} \text{ (mod 100)} & \small \color{#3D99F6} \text{Euler's totient function }\phi(100) = 40 \\ & \equiv 7^{7^{7^{7^{7^{7^7\text{ mod }2}\text{ mod }4}\text{ mod }8}\text{ mod }16} \text{ mod }40} \text{ (mod 100)} & \small \color{#3D99F6} \text{Repeated use of Euler's theorem.} \\ & \equiv 7^{7^{7^{7^{\color{#3D99F6}7^1 \text{ mod }4} \text{ mod }8} \text{ mod }16} \text{ mod }40} \text{ (mod 100)} & \small \color{#3D99F6} 7 \equiv 3 \text{ (mod 4)} \\ & \equiv 7^{7^{7^{\color{#3D99F6}7^3 \text{ mod }8} \text{ mod }16} \text{ mod }40} \text{ (mod 100)} & \small \color{#3D99F6} 7^3 \equiv 7\cdot 49 \equiv 7(48+1) \equiv 7 \text{ (mod 8)} \\ & \equiv 7^{7^{\color{#3D99F6}7^7 \text{ mod }16} \text{ mod }40} \text{ (mod 100)} & \small \color{#3D99F6} 7^7 \equiv 7\cdot 49^3 \equiv 7(48+1)^3 \equiv 7 \text{ (mod 16)} \\ & \equiv 7^{\color{#3D99F6}7^7 \text{ mod }40} \text{ (mod 100)} & \small \color{#3D99F6} \text{See note: } 7^7 \equiv 23 \text{ (mod 16)} \\ & \equiv 7^{23} \text{ (mod 100)} \\ & \equiv \boxed{43} \text{ (mod 100)} & \small \color{#3D99F6} \text{See note} \end{aligned}


Notes:

7 7 7 4 9 3 7 ( 40 + 9 ) 3 7 9 3 7 81 9 7 9 63 23 (mod 40) 7^7 \equiv 7\cdot 49^3 \equiv 7(40+9)^3 \equiv 7\cdot 9^3 \equiv 7\cdot 81 \cdot 9 \equiv 7\cdot 9 \equiv 63 \equiv 23 \text{ (mod 40)}

7 23 7 ( 49 ) 11 7 49 ( 50 1 ) 10 7 49 343 43 (mod 100) 7^{23} \equiv 7\cdot (49)^{11} \equiv 7 \cdot 49(50-1)^{10} \equiv 7\cdot 49 \equiv 343 \equiv 43 \text{ (mod 100)}

Surely try my problem: The Game of 7's

Mrigank Shekhar Pathak - 4 years, 3 months ago

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I just tried your problem, nice but it wasn't easy.

Hana Wehbi - 4 years, 3 months ago

Thank you for the solution. Very nice and elaborate one.

Hana Wehbi - 4 years, 3 months ago

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I have provided a better explanation for the first solution.

Chew-Seong Cheong - 4 years, 3 months ago

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Thank you. I really appreciate it.

Hana Wehbi - 4 years, 3 months ago

@Chew-Seong Cheong you could use Carmiachel Function for faster covergence

Anirudh Sreekumar - 4 years, 3 months ago

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Thanks. But I am unsure of the method actually. If you can show.

Chew-Seong Cheong - 4 years, 3 months ago

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