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Jeremy
2026-02-20 23:49:03 -05:00
parent 0193445394
commit bbc9ef7828
3 changed files with 28 additions and 20 deletions
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37
static/posts/crackmes-the-matrix/post.typ
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@@ -23,24 +23,25 @@ Notice again this part iterates through 10 characters of input. In this loop the
Here's the math on how I found a password that reached 984:
`a=password2=qqqqqqqqqt`\
`b=password3=?`\
`∑4((a[i] + b[i]) >> 3) == 984`\
`∑((a[i] + b[i]) >> 3) == 246`\
`9((113 + b1) >> 3) + ((116 + b2) >> 3) == 246`\
\# Note `b1, b2 ∈ [32, 126] = usable ascii characters`\
\# Since the sum of 10 characters is 246, a good b1 value will cause the character function to return 25.\
`(113 + b1) >> 3 = 25`\
`113 + b1 = 200`\
`b1 = 87 = "W"`\
\# Sub b1 back into the whole equation\
`9((113 + 87) >> 3) + ((116 + b2) >> 3) = 246`\
`9(200 >> 3) + ((116 + b2) >> 3) = 246`\
`9(25) + ((116 + b2) >> 3) = 246`\
`225 + ((116 + b2) >> 3) = 246`\
`(116 + b2) >> 3 = 21`\
`116 + b2 = 168`\
`b2 = 52 = "4"`\
```
a=password2=qqqqqqqqqt
b=password3=?
Σ4((a[i] + b[i]) >> 3) == 984
Σ((a[i] + b[i]) >> 3) == 246
9((113 + b1) >> 3) + ((116 + b2) >> 3) == 246
# Note `b1, b2 ∈ [32, 126] = usable ascii characters
# Since the sum of 10 characters is 246, a good b1 value will cause the character function to return 25.
(113 + b1) >> 3 = 25
113 + b1 = 200
b1 = 87 = "W"
# Sub b1 back into the whole equation
9((113 + 87) >> 3) + ((116 + b2) >> 3) = 246
9(200 >> 3) + ((116 + b2) >> 3) = 246
9(25) + ((116 + b2) >> 3) = 246
225 + ((116 + b2) >> 3) = 246
(116 + b2) >> 3 = 21
116 + b2 = 168
b2 = 52 = "4"```
Therefore a solution for password3 is `WWWWWWWWW4`