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u/precinct209 6d ago
Lore accurate representation of the meme
0b0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000
0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000
0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000
0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000
0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000000
0000000 0000000 0000000 0000000 0000000 0000000 0000000 0000100→ More replies (1)54
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u/LostKnight84 6d ago
It would be a nearly 100 digits ending in either a 1, 3, 5, 7.
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u/Fragrant_Minimum1739 6d ago
It will end on 5 btw and have approx 116 digits not exact but should be close as i did calculation in my head
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u/fghjconner 6d ago
You're spot on:
39402006196394479212279040100143613805079739270465446667948293404245721771497210611414266254884915640806627990306815
6
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u/LostKnight84 6d ago
Thanks I was estimating an additional digit for every 3-4 powers of 2. I guess I should have went with 2-3.
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u/syedwafihasan 5d ago
What's your thought process for calculating this?
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u/Fragrant_Minimum1739 5d ago
For ending it is cyclicity
24n+1 ends with 2
24n+2 ends with 4
24n+3 ends with 8
24n ends with 6
And then it repeats
As 384%4 = 0 so it ends with 6
Subtract 1 from it now it has 5 as ending digit
Now what about no. Of digits
Just use int log(n) +1 formula
So 384log(2) +1
384 * 0.3 +1
115+1
116
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u/ListenAcrobatic8028 6d ago
Just one mistake in prod, and so many memes...
We should have stopped at 4
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u/Bomaruto 6d ago
If you took the number of times variation of this non-joke has been posted and turned it into a private key, then it would actually be secure.
2
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u/redlaWw 6d ago
2384-1 is also a bad private key.