Anonymous at Sun, 5 May 2024 12:59:44 UTC No. 16160593

the last eight digits of z are between 0 and 9.

Anonymous at Sun, 5 May 2024 13:02:32 UTC No. 16160599

>>16160590
The last 8 digits are divisible by 5^8.

Anonymous at Sun, 5 May 2024 13:40:02 UTC No. 16160649

08203125

Anonymous at Sun, 5 May 2024 14:08:24 UTC No. 16160679

>>16160649
Correct!

Would you mind share your solving method?

Anonymous at Sun, 5 May 2024 14:12:00 UTC No. 16160686

>>16160679
wolframalpha

Anonymous at Sun, 5 May 2024 14:35:04 UTC No. 16160722

>>16160679
z mod 10^8 = ((z mod 5^8)(2^-8 mod 5^8)2^8 + (z mod 2^8)(5^-8 mod 2^8)5^8) mod 10^8

z mod 5^8 = 0 since z is a big power of 5.
z mod 2^8 = 117 which can be easily calculated with Euler's theorem.
5^-8 mod 2^8 = 33

117 * 33 * 5^8 mod 10^8 = 1508203125 mod 10^8 = 8203125

Anonymous at Sun, 5 May 2024 14:45:23 UTC No. 16160729

You're calculating 5^5^5.... 26 times mod 10^8, this is basically 5*5*5*... a billion times in sets of 5^5

Because its all multiplying by 5 you can just do it until you have a value larger than 8 digits, this occurs on 5^5^5 which is already a massive number

5^3125 mod 10^8 = 08203125

You can break this into smaller pieces, using exponential rules but I can't be assed