Excellent Crackme

We know one can do pretty much everything in Excel spreadsheets, but this...

excel_crackme

Open the excel file:

Unzipping the xlsm file and decompiled the vba source code using pcode2code:

pcode2code xl/vbaProject.bin -o decompile

Analyse the code, only VolgaCTF function is important for finding the flag

Private Sub VolgaCTF()
Dim input As String
Dim solution As Long
Dim number As Long
Dim answer As Long
input = Range("L15")  'L15 is the flag input box cell 
For i = 1 To Len(input) 'i will loop to 1 to length of input
	solution = 0
	For j = 1 To Len(input) 'j will loop to 1 to length of input
		'Get the value of cell and assign into number
		number = CInt(Cells(99 + i, 99 + j).Value)
		'Get a character from input from index j
		c = Mid(input, j, 1)
		'Add the number times the ASCII value of c to solution
		solution = solution + number * Asc(c)
	Next j
	'Get the value of cell and assign to answer
	answer = CLng(Cells(99 + i, 99 + Len(input) + 1).Value)
	'If answer not equal solution means wrong flag
	If (answer <> solution) Then
		MsgBox "Wrong Flag!"
		Exit Sub
	End If
Next i
'Only if all answer equal to all solution then the flag is correct
MsgBox "Congratz"
End Sub

After I reversing the code, immediately found a secret number table starts at (CV,100), because of Cells(99 + i, 99 + j) and i j starts at 1

It is a 45*46 table, and I notice last column of the table is quite large compared to other number:

After more deep analysing, I found these are Simultaneous equations!

Assume abcdefgh (unknown) is the input characters
First row:
620a +340b +895c -39d +945e +321f +586g +487h... = 490493
Second row:
-19a +85b -456c +228d -127e -777f +191g +605h... = -7845
etc

Therefore, the flag length must be 45 (45 column)

I decide to use z3 solver in python to solve this:

from z3 import *
n = 45
inp = [Int('inp%i' %i) for i in range(n)]
numbers = [i.strip().split(" ") for i in open("numbers.txt",'r').read().split("\n")]
solver = Solver()
solver.add(inp[0] == ord("V"))
solver.add(inp[1] == ord("o"))
solver.add(inp[2] == ord("l"))
solver.add(inp[3] == ord("g"))
solver.add(inp[4] == ord("a"))
solver.add(inp[5] == ord("C"))
solver.add(inp[6] == ord("T"))
solver.add(inp[7] == ord("F"))
solver.add(inp[8] == ord("{"))
for i in range(9,45):
	solver.add(inp[i] >= 32)
	solver.add(inp[i] <= 126) 
solver.add(inp[44] == ord("}"))
for i in range(44):
	condition = ""
	for j in range(n):
		condition += "("+ numbers[i][j] + "* inp[%i])+" % j
	solver.add(eval(condition[:-1] + "==" + numbers[i][n]))
print(solver.check())
modl = solver.model()
print(''.join([chr(modl[inp[i]].as_long()) for i in range(n)]))

It took around 2 minutes to solve:

time python solve2.py
sat
VolgaCTF{7h3_M057_M47h_cr4ckM3_y0u_3V3R_533N}

real	1m41.494s
user	1m41.331s
sys	0m0.095s

Flag

VolgaCTF{7h3_M057_M47h_cr4ckM3_y0u_3V3R_533N}

Better Solution

After the CTF ends, I saw better solution that can solve faster: Unhappi writeups Better solution

It use matrix to solve like the picture below: