What are the last 2 digits of 7 7 7 7 . . . 7 ?
The tower has 7 sevens.
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Surely try my problem: The Game of 7's
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I just tried your problem, nice but it wasn't easy.
Thank you for the solution. Very nice and elaborate one.
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I have provided a better explanation for the first solution.
@Chew-Seong Cheong you could use Carmiachel Function for faster covergence
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Thanks. But I am unsure of the method actually. If you can show.
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We note that 7 7 ≡ 7 4 7 3 ≡ 4 9 2 7 3 ≡ ( 5 0 − 1 ) 2 7 3 ≡ 7 3 ≡ 3 4 3 ≡ 4 3 (mod 100) ; ⟹ 7 7 ≡ 1 0 0 n + 4 3 , where n is an integer.
Now consider 7 7 7 ≡ 7 1 0 0 n + 4 3 ≡ 4 9 5 0 n + 2 0 7 3 ≡ ( 5 0 − 1 ) 5 0 n + 2 0 7 3 ≡ 7 3 ≡ 3 4 3 ≡ 4 3 (mod 100) .
Similarly, 7 7 7 ⋅ ⋅ ⋅ ≡ 4 3 (mod 100) .
Alternative solution
7 7 7 7 7 7 7 ≡ 7 7 7 7 7 7 7 mod ϕ ( 1 0 0 ) (mod 100) ≡ 7 7 7 7 7 7 7 mod 4 0 (mod 100) ≡ 7 7 7 7 7 7 7 mod 2 mod 4 mod 8 mod 1 6 mod 4 0 (mod 100) ≡ 7 7 7 7 7 1 mod 4 mod 8 mod 1 6 mod 4 0 (mod 100) ≡ 7 7 7 7 3 mod 8 mod 1 6 mod 4 0 (mod 100) ≡ 7 7 7 7 mod 1 6 mod 4 0 (mod 100) ≡ 7 7 7 mod 4 0 (mod 100) ≡ 7 2 3 (mod 100) ≡ 4 3 (mod 100) As g cd ( 7 , 1 0 0 ) = 1 , Euler’s theorem applies. Euler’s totient function ϕ ( 1 0 0 ) = 4 0 Repeated use of Euler’s theorem. 7 ≡ 3 (mod 4) 7 3 ≡ 7 ⋅ 4 9 ≡ 7 ( 4 8 + 1 ) ≡ 7 (mod 8) 7 7 ≡ 7 ⋅ 4 9 3 ≡ 7 ( 4 8 + 1 ) 3 ≡ 7 (mod 16) See note: 7 7 ≡ 2 3 (mod 16) See note
Notes:
7 7 ≡ 7 ⋅ 4 9 3 ≡ 7 ( 4 0 + 9 ) 3 ≡ 7 ⋅ 9 3 ≡ 7 ⋅ 8 1 ⋅ 9 ≡ 7 ⋅ 9 ≡ 6 3 ≡ 2 3 (mod 40)
7 2 3 ≡ 7 ⋅ ( 4 9 ) 1 1 ≡ 7 ⋅ 4 9 ( 5 0 − 1 ) 1 0 ≡ 7 ⋅ 4 9 ≡ 3 4 3 ≡ 4 3 (mod 100)