๐งต Aproximation
Anonymous at Sun, 5 May 2024 12:56:22 UTC No. 16160590
A good friend of mine send me this but I cant figure out a way to aproximate the result. Is there a way?
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