Reverse Engineering

Transformation
Solved by: ret2basic
Challenge
I wonder what this really is... enc ''.join([chr((ord(flag[i]) << 8) + ord(flag[i + 1])) for i in range(0, len(flag), 2)])
Analysis
The encoding scheme acts on two characters from the flag each time. The first character is used as higher 8 bits, while the second character is used as lower 8 bits. Together, the 2-character pair is transformed into a 16-bit binary number and this binary number is converted to ASCII character.
Implementation
1
#!/usr/bin/env python3
2
​
3
with open("enc", "r") as f:
4
    encoded = f.read()
5
​
6
    flag = ""
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    for ch in encoded:
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        binary = "{0:016b}".format(ord(ch))
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        first_half, second_half = binary[:8], binary[8:]
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        flag += chr(int(first_half, 2))
11
        flag += chr(int(second_half, 2))
12
​
13
    print(flag)
Copied!
Flag
1
picoCTF{16_bits_inst34d_of_8_04c0760d}
Copied!
keygenme-py
Solved by: ret2basic
Challenge
​keygenme-trial.py​
Source Code
1
def check_key(key, username_trial):
2
​
3
    global key_full_template_trial
4
​
5
    if len(key) != len(key_full_template_trial):
6
        return False
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    else:
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        # Check static base key part --v
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        i = 0
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        for c in key_part_static1_trial:
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            if key[i] != c:
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                return False
13
​
14
            i += 1
15
​
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        # TODO : test performance on toolbox container
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        # Check dynamic part --v
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        if key[i] != hashlib.sha256(username_trial).hexdigest()[4]:
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            return False
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        else:
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            i += 1
22
​
23
        if key[i] != hashlib.sha256(username_trial).hexdigest()[5]:
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            return False
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        else:
26
            i += 1
27
​
28
        if key[i] != hashlib.sha256(username_trial).hexdigest()[3]:
29
            return False
30
        else:
31
            i += 1
32
​
33
        if key[i] != hashlib.sha256(username_trial).hexdigest()[6]:
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            return False
35
        else:
36
            i += 1
37
​
38
        if key[i] != hashlib.sha256(username_trial).hexdigest()[2]:
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            return False
40
        else:
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            i += 1
42
​
43
        if key[i] != hashlib.sha256(username_trial).hexdigest()[7]:
44
            return False
45
        else:
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            i += 1
47
​
48
        if key[i] != hashlib.sha256(username_trial).hexdigest()[1]:
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            return False
50
        else:
51
            i += 1
52
​
53
        if key[i] != hashlib.sha256(username_trial).hexdigest()[8]:
54
            return False
55
​
56
​
57
​
58
        return True
Copied!
Analysis
According to the global variables declared at the beginning of the source code, the flag is picoCTF{1n_7h3_|<3y_of_xxxxxxxx} where x stands for a "dynamic" character. Our objective is to reverse engineer the check_key function and pass the check.
Note that check_key is not fully implemented and there are eight if statements checking if a certain character is valid. We could simply write a script and recover each character.
Implementation
1
#!/usr/bin/env python3
2
import hashlib
3
​
4
username_trial = b"PRITCHARD"
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digest = hashlib.sha256(username_trial).hexdigest()
6
​
7
print(f"index 4: {digest[4]}")
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print(f"index 5: {digest[5]}")
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print(f"index 3: {digest[3]}")
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print(f"index 6: {digest[6]}")
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print(f"index 2: {digest[2]}")
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print(f"index 7: {digest[7]}")
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print(f"index 1: {digest[1]}")
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print(f"index 8: {digest[8]}")
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​
16
flag_part_1 = "picoCTF{1n_7h3_|<3y_of_"
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flag_part_2 = ""
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flag_part_2 += digest[4] + digest[5]
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flag_part_2 += digest[3] + digest[6]
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flag_part_2 += digest[2] + digest[7]
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flag_part_2 += digest[1] + digest[8]
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flag_part_3 = "}"
23
​
24
print(f"flag: {flag_part_1 + flag_part_2 + flag_part_3}")
Copied!
Flag
1
picoCTF{1n_7h3_|<3y_of_54ef6292}
Copied!
crackme-py
Solved by: ret2basic
Challenge
​crackme.py​
Source Code
1
# Hiding this really important number in an obscure piece of code is brilliant!
2
# AND it's encrypted!
3
# We want our biggest client to know his information is safe with us.
4
bezos_cc_secret = "A:[email protected]%uL`M-^M0c0AbcM-MFE055a4ce`eN"
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​
6
# Reference alphabet
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alphabet = "!\"#$%&'()*+,-./0123456789:;<=>[email protected]"+ \
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            "[\\]^_`abcdefghijklmnopqrstuvwxyz{|}~"
9
​
10
​
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​
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def decode_secret(secret):
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    """ROT47 decode
14
​
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    NOTE: encode and decode are the same operation in the ROT cipher family.
16
    """
17
​
18
    # Encryption key
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    rotate_const = 47
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​
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    # Storage for decoded secret
22
    decoded = ""
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​
24
    # decode loop
25
    for c in secret:
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        index = alphabet.find(c)
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        original_index = (index + rotate_const) % len(alphabet)
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        decoded = decoded + alphabet[original_index]
29
​
30
    print(decoded)
31
​
32
​
33
​
34
def choose_greatest():
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    """Echo the largest of the two numbers given by the user to the program
36
​
37
    Warning: this function was written quickly and needs proper error handling
38
    """
39
​
40
    user_value_1 = input("What's your first number? ")
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    user_value_2 = input("What's your second number? ")
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    greatest_value = user_value_1 # need a value to return if 1 & 2 are equal
43
​
44
    if user_value_1 > user_value_2:
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        greatest_value = user_value_1
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    elif user_value_1 < user_value_2:
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        greatest_value = user_value_2
48
​
49
    print( "The number with largest positive magnitude is "
50
        + str(greatest_value) )
51
​
52
​
53
​
54
choose_greatest()
Copied!
Analysis
Modify the source code to call the decode_secret function.
Implementation
1
# choose_greatest()
2
print(decode_secret(bezos_cc_secret))
Copied!
Flag
1
picoCTF{1|\/|_4_p34|\|ut_dd2c4616}
Copied!
ARMssembly 0
Solved by: ret2basic
Challenge
What integer does this program print with arguments 1765227561 and 1830628817? File: chall.S Flag format: picoCTF{XXXXXXXX} -> (hex, lowercase, no 0x, and 32 bits. ex. 5614267 would be picoCTF{0055aabb})
Assembly with Comments
1
    .arch armv8-a
2
    .file    "chall.c"
3
    .text
4
    .align    2
5
    .global    func1
6
    .type    func1, %function
7
func1:
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    sub    sp, sp, #16 @ allocate 16 bytes on the stack
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    str    w0, [sp, 12] @ store w0 into [sp+12]
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    str    w1, [sp, 8] @ store w1 into [sp+8]
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    ldr    w1, [sp, 12] @ load [sp+12] into w1
12
    ldr    w0, [sp, 8] @ load [sp+8] into w0
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    cmp    w1, w0 @ compare w1 and w0 => compare [sp+12] and [sp+8]
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    bls    .L2 @ branch to .L2 if ls
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    ldr    w0, [sp, 12] @ load [sp+12] into w0
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    b    .L3 @ branch to .L3
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.L2:
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    ldr    w0, [sp, 8]
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.L3:
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    add    sp, sp, 16
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    ret
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    .size    func1, .-func1
23
    .section    .rodata
24
    .align    3
25
.LC0:
26
    .string    "Result: %ld\n"
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    .text
28
    .align    2
29
    .global    main
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    .type    main, %function
31
main:
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    stp    x29, x30, [sp, -48]!
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    add    x29, sp, 0
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    str    x19, [sp, 16]
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    str    w0, [x29, 44]
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    str    x1, [x29, 32]
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    ldr    x0, [x29, 32]
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    add    x0, x0, 8
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    ldr    x0, [x0]
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    bl    atoi
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    mov    w19, w0
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    ldr    x0, [x29, 32]
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    add    x0, x0, 16
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    ldr    x0, [x0]
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    bl    atoi
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    mov    w1, w0
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    mov    w0, w19
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    bl    func1
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    mov    w1, w0
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    adrp    x0, .LC0
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    add    x0, x0, :lo12:.LC0
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    bl    printf
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    mov    w0, 0
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    ldr    x19, [sp, 16]
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    ldp    x29, x30, [sp], 48
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    ret
57
    .size    main, .-main
58
    .ident    "GCC: (Ubuntu/Linaro 7.5.0-3ubuntu1~18.04) 7.5.0"
59
    .section    .note.GNU-stack,"",@progbits
Copied!
Solution
bls: Branch on Lower than or Same.
The main function calls atoi twice to convert the arguments to integers and then calls func1 to compare those two integers. Since 1765227561≤18306288171765227561 \leq 18306288171765227561≤1830628817
, the control flow goes to .L2. Now w0 = [sp+8] = 1830628817 = x0.
Eventually, when printf is called, the value stored in x0 will be printed (calling convention). Hence this program prints 1830628817, which is 0x6d1d2dd1 in hex.
Flag
1
picoCTF{6d1d2dd1}
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speeds and feeds
Solved by: ret2basic
Challenge
There is something on my shop network running at mercury.picoctf.net:53740, but I can't tell what it is. Can you?
Analysis
Save the G-code to a file:
1
$ nc mercury.picoctf.net 53740 > G-code.txt
Copied!
Use NC Viewer to plot the flag.
Flag
1
picoCTF{num3r1cal_c0ntr0l_775375c7}
Copied!
Shop
Solved by: ret2basic
Challenge
Best Stuff - Cheap Stuff, Buy Buy Buy... Store Instance: source. The shop is open for business at nc mercury.picoctf.net 34938.
Analysis
Buy -10 "Quiet Quiches" and then buy 1 flag. Convert ASCII numbers to text.
Implementation
1
#!/usr/bin/env python3
2
​
3
characters = [112, 105, 99, 111, 67, 84, 70, 123, 98, 52, 100, 95, 98, 114, 111, 103, 114, 97, 109, 109, 101, 114, 95, 98, 97, 54, 98, 56, 99, 100, 102, 125]
4
​
5
flag = ""
6
for character in characters:
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    flag += chr(character)
8
​
9
print(flag)
Copied!
Flag
1
picoCTF{b4d_brogrammer_ba6b8cdf}
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ARMssembly 1
Solved by: ret2basic
Challenge
For what argument does this program print win with variables 85, 6 and 3? File: chall_1.S Flag format: picoCTF{XXXXXXXX} -> (hex, lowercase, no 0x, and 32 bits. ex. 5614267 would be picoCTF{0055aabb})
Assembly with Comments
1
    .arch armv8-a
2
    .file    "chall_1.c"
3
    .text
4
    .align    2
5
    .global    func
6
    .type    func, %function
7
func:
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    sub    sp, sp, #32 @ allocate 32 bytes on the stack
9
    str    w0, [sp, 12] @ [sp+12] = w0 = [x29+44] = arg
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    mov    w0, 85 @ w0 = 85
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    str    w0, [sp, 16] @ [sp+16] = w0 = 85
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    mov    w0, 6 @ w0 = 6
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    str    w0, [sp, 20] @ [sp+20] = w0 = 6
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    mov    w0, 3 @ w0 = 3
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    str    w0, [sp, 24] @ [sp+24] = w0 = 3
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    ldr    w0, [sp, 20] @ w0 = [sp+20] = 6
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    ldr    w1, [sp, 16] @ w1 = [sp+16] = 85
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    lsl    w0, w1, w0 @ w0 = w1 * (2**w0) = 85 * (2**6) = 85 * 64 = 5440
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    str    w0, [sp, 28] @ [sp+28] = w0 = 5440
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    ldr    w1, [sp, 28] @ w1 = [sp+28] = 5440
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    ldr    w0, [sp, 24] @ w0 = [sp+24] = 3
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    sdiv    w0, w1, w0 @ w0 = w1 // w2 = 5440 // 3 = 1813
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    str    w0, [sp, 28] @ [sp+28] = w0 = 1813
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    ldr    w1, [sp, 28] @ w1 = [sp+28] = 1813
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    ldr    w0, [sp, 12] @ w0 = [sp+12] = arg
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    sub    w0, w1, w0 @ w0 = w1 - w0 = 1813 - arg
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    str    w0, [sp, 28] @ [sp+28] = w0 = 1813 - arg
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    ldr    w0, [sp, 28] @ w0 = [sp+28] = 1813 - arg
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    add    sp, sp, 32
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    ret
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    .size    func, .-func
32
    .section    .rodata
33
    .align    3
34
.LC0:
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    .string    "You win!"
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    .align    3
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.LC1:
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    .string    "You Lose :("
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    .text
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    .align    2
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    .global    main
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    .type    main, %function
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main:
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    stp    x29, x30, [sp, -48]!
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    add    x29, sp, 0
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    str    w0, [x29, 28]
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    str    x1, [x29, 16]
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    ldr    x0, [x29, 16]
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    add    x0, x0, 8
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    ldr    x0, [x0]
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    bl    atoi
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    str    w0, [x29, 44]
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    ldr    w0, [x29, 44]
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    bl    func
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    cmp    w0, 0
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    bne    .L4
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    adrp    x0, .LC0
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    add    x0, x0, :lo12:.LC0
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    bl    puts
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    b    .L6
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.L4:
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    adrp    x0, .LC1
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    add    x0, x0, :lo12:.LC1
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    bl    puts
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.L6:
66
    nop
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    ldp    x29, x30, [sp], 48
68
    ret
69
    .size    main, .-main
70
    .ident    "GCC: (Ubuntu/Linaro 7.5.0-3ubuntu1~18.04) 7.5.0"
71
    .section    .note.GNU-stack,"",@progbits
Copied!
Solution
lsl: Logical Shift Left. It provides the value of a register multiplied by a power of two, inserting zeros into the vacated bit positions.
sdiv: Signed Divide.
Note that the x29 register is the frame pointer in ARM64. It is equivalent to the RBP register in Intel x86-64. The function func is just doing math:
85⋅263=arg\frac{85 \cdot 2^6}{3} = arg385⋅26​=arg
Hence arg = 1813, which is 0x715 in hex.
Flag
1
picoCTF{00000715}
Copied!
ARMssembly 2
Solved by: ret2basic
Challenge
What integer does this program print with argument 3848786505? File: chall_2.S Flag format: picoCTF{XXXXXXXX} -> (hex, lowercase, no 0x, and 32 bits. ex. 5614267 would be picoCTF{0055aabb})
Solution
b: Branch. This is the uncoditional branch.
bcc: Branch on Carry Clear. This is the conditional branch.
Note that the wzr register is equivalent to 0. The instruction str wzr, [sp, 24] zeros out the content of [sp+24].
Here [sp+24] is the result and [sp+28] is the counter. The loop keeps increment [sp+24] by 3 and compares the counter [sp+28] with 3848786505. In the end, the result [sp+24] is printed out. In other word, the program computes 3848786505 * 3 = 11546359515, which is 0x2b03776db in hex. Since the flag requires 32-bit hex, this hex number is truncated as 0xb03776db.
Flag
1
picoCTF{b03776db}
Copied!
Hurry up! Wait!
Solved by: ret2basic
Challenge
​svchost.exe​
Solution
Each function call prints out a character:
Pseudocode
Flag
1
picoCTF{d15a5m_ftw_dfbdc5d}
Copied!
gogo
Solved by: 
Challenge
Hmmm this is a weird file... enter_password. There is a instance of the service running at mercury.picoctf.net:35862.
Solution
Flag
1
​
Copied!
ARMssembly 3
Solved by: ret2basic
Challenge
What integer does this program print with argument 3350728462? File: chall_3.S Flag format: picoCTF{XXXXXXXX} -> (hex, lowercase, no 0x, and 32 bits. ex. 5614267 would be picoCTF{0055aabb})
Assembly with Comments
1
    .arch armv8-a
2
    .file    "chall_3.c"
3
    .text
4
    .align    2
5
    .global    func1
6
    .type    func1, %function
7
func1:
8
    stp    x29, x30, [sp, -48]! @ push x29 and x30 onto the stack
9
    add    x29, sp, 0 @ x29 = sp + 0 = sp
10
    str    w0, [x29, 28] @ [x29+28] = w0 = 3350728462
11
    str    wzr, [x29, 44] @ [x29+44] = wzr = 0
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    b    .L2 @ branch to .L2 unconditionally
13
.L4:
14
    ldr    w0, [x29, 28] @ w0 = [x29+28]
15
    and    w0, w0, 1 @ w0 &= 1 => => w0 = 1 if all bits of w0 are 1, w0 = 0 otherwise
16
    cmp    w0, 0 @ compare w0 and 0
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    beq    .L3 @ branch to .L3 if w0 == 0
18
    ldr    w0, [x29, 44] @ w0 = [x29+44]
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    bl    func2 @ branch to func2
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    str    w0, [x29, 44] @ [x29+44] = w0
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.L3:
22
    ldr    w0, [x29, 28] @ w0 = [x29+28]
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    lsr    w0, w0, 1 @ w0 >> 1
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    str    w0, [x29, 28] @ [x29+28] = w0
25
.L2:
26
    ldr    w0, [x29, 28] @ w0 = [x29+28]
27
    cmp    w0, 0 @ compare w0 and 0 => compare [x29+28] and 0
28
    bne    .L4 @ if [x29+28] != 0, branch to .L4
29
    ldr    w0, [x29, 44] @ w0 = [x29+44]
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    ldp    x29, x30, [sp], 48
31
    ret
32
    .size    func1, .-func1
33
    .align    2
34
    .global    func2
35
    .type    func2, %function
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func2:
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    sub    sp, sp, #16 @ allocate 16 bytes on the stack
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    str    w0, [sp, 12] @ [sp+12] = w0
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    ldr    w0, [sp, 12] @ w0 = [sp+12]
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    add    w0, w0, 3 @ w0 += 3 
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    add    sp, sp, 16
42
    ret
43
    .size    func2, .-func2
44
    .section    .rodata
45
    .align    3
46
.LC0:
47
    .string    "Result: %ld\n"
48
    .text
49
    .align    2
50
    .global    main
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    .type    main, %function
52
main:
53
    stp    x29, x30, [sp, -48]!
54
    add    x29, sp, 0
55
    str    w0, [x29, 28]
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    str    x1, [x29, 16]
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    ldr    x0, [x29, 16]
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    add    x0, x0, 8
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    ldr    x0, [x0]
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    bl    atoi
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    bl    func1
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    str    w0, [x29, 44]
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    adrp    x0, .LC0
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    add    x0, x0, :lo12:.LC0
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    ldr    w1, [x29, 44]
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    bl    printf
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    nop
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    ldp    x29, x30, [sp], 48
69
    ret
70
    .size    main, .-main
71
    .ident    "GCC: (Ubuntu/Linaro 7.5.0-3ubuntu1~18.04) 7.5.0"
72
    .section    .note.GNU-stack,"",@progbits
Copied!
Solution
stp: Store Pair.
bl: Branch with Link.
lsr: Logical Shift Right.
The program computes and w0, w0, 1, where w0 is the argument. If the w0 & 1 = 1, it increments the counter by 3. Otherwise, it does nothing and continues. After each round of testing, the program divides w0 by 2 and repeats this test until w0 = 0.
Implementation
1
#!/usr/bin/env python3
2
​
3
target = 3350728462
4
​
5
result = 0
6
while target != 0:
7
    if target & 1 != 0:
8
        result += 3
9
    target >>= 1
10
​
11
print(hex(result))
Copied!
Flag
1
picoCTF{00000030}
Copied!
Let's get dynamic
Solved by: ret2basic
Challenge
Can you tell what this file is reading? chall.S​
Solution
Compile the assembly code:
1
$ gcc -c chall.S -o chall.o
2
$ gcc chall.o -o chall
Copied!
Disassemble the main function:
1
pwndbg> disass main
Copied!
PIE
Since PIE is enabled, all the addresses here are just offsets. We need to run the program for the correct addresses to load:
1
pwndbg> r
2
...
3
pwndbg> disass main
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main
The memcmp function compares our input with the flag. Set a breakpoint on memcmp and run the program again:
1
pwndbg> b *0x00005555555552f6
2
pwndbg> r
Copied!
The flag is located in RSI at this moment:
flag
Flag
1
picoCTF{dyn4m1c_4n4ly1s_1s_5up3r_us3ful_56e35b54}
Copied!
Easy as GDB
Solved by: ret2basic
Challenge
The flag has got to be checked somewhere... File: brute​
Solution
Take a look at the pseudocode:
strncpy()
Here unk_2008 is the encrypted flag:
Ecrypted flag
This is a nice use case for the angr template. What we have to do here is:
1.
Figure out the flag length
2.
Find the address of the "Correct!" state
3.
Find the address of the "Incorrect." state
The length of the encrypted flag is 30, so the flag in clear should be 30-byte long as well.
The call puts instruction for "Correct!":
Correct
The call puts instruction for "Incorrect.":
Incorrect
Implementation
1
#!/usr/bin/env python3
2
import angr
3
import claripy
4
​
5
FLAG_LEN = 30
6
STDIN_FD = 0
7
​
8
base_addr = 0x100000 # To match addresses to Ghidra
9
​
10
proj = angr.Project("brute", main_opts={'base_addr': base_addr}) 
11
​
12
flag_chars = [claripy.BVS('flag_%d' % i, 8) for i in range(FLAG_LEN)]
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flag = claripy.Concat( *flag_chars + [claripy.BVV(b'\n')]) # Add \n for scanf() to accept the input
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​
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state = proj.factory.full_init_state(
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        args=['brute'],
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        add_options=angr.options.unicorn,
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        stdin=flag,
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)
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​
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# Add constraints that all characters are printable
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for k in flag_chars:
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    state.solver.add(k >= ord('!'))
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    state.solver.add(k <= ord('~'))
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​
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simgr = proj.factory.simulation_manager(state)
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find_addr  = 0x100a72 #  the address of "call puts" for SUCCESS
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avoid_addr = 0x100a86 # the address of "call puts" for FAILURE
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simgr.explore(find=find_addr, avoid=avoid_addr)
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​
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if (len(simgr.found) > 0):
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    for found in simgr.found:
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        print(found.posix.dumps(STDIN_FD))
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Run this script in docker:
1
$ docker run -it angr/angr
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The script takes some time to finish. Be patient.
Flag
1
picoCTF{I_5D3_A11DA7_e5458cbf}
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ARMssembly 4
Solved by: ret2basic
Challenge
What integer does this program print with argument 3964545182? File: chall_4.S Flag format: picoCTF{XXXXXXXX} -> (hex, lowercase, no 0x, and 32 bits. ex. 5614267 would be picoCTF{0055aabb})
Assembly with Comments
1
    .arch armv8-a
2
    .file    "chall_4.c"
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    .text
4
    .align    2
5
    .global    func1
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    .type    func1, %function
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func1:
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    stp    x29, x30, [sp, -32]!
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    add    x29, sp, 0
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    str    w0, [x29, 28]
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    ldr    w0, [x29, 28]
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    cmp    w0, 100
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    bls    .L2 @ if w0 <= 100, branch to .L2
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    ldr    w0, [x29, 28] @ w0 = [x29+28]
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    add    w0, w0, 100 @ w0 += 100
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    bl    func2 @ branch to fun2
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    b    .L3 @ branch to .L3 unconditionally
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.L2:
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    ldr    w0, [x29, 28]
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    bl    func3 @ branch to func3
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.L3:
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    ldp    x29, x30, [sp], 32
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    ret
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    .size    func1, .-func1
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    .align    2
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    .global    func2
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    .type    func2, %function
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func2:
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    stp    x29, x30, [sp, -32]!
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    add    x29, sp, 0
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    str    w0, [x29, 28]
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    ldr    w0, [x29, 28]
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    cmp    w0, 499
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    bhi    .L5 @ branch to .L5 if w0 > 499
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    ldr    w0, [x29, 28]
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    sub    w0, w0, #86
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    bl    func4 @ branch to func4
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    b    .L6 @ branch to .L6 unconditionally
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.L5:
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    ldr    w0, [x29, 28]
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    add    w0, w0, 13 @ w0 += 13
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    bl    func5 @ branch to func5
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.L6:
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    ldp    x29, x30, [sp], 32
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    ret
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    .size    func2, .-func2
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    .align    2
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    .global    func3
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    .type    func3, %function
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func3:
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    stp    x29, x30, [sp, -32]!
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    add    x29, sp, 0
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    str    w0, [x29, 28]
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    ldr    w0, [x29, 28]
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    bl    func7 @ branch to func7
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    ldp    x29, x30, [sp], 32
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    ret
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    .size    func3, .-func3
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    .align    2
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    .global    func4
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    .type    func4, %function
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func4:
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    stp    x29, x30, [sp, -48]!
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    add    x29, sp, 0
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    str    w0, [x29, 28]
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    mov    w0, 17 @ w0 = 17
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    str    w0, [x29, 44]
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    ldr    w0, [x29, 44]
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    bl    func1 @ branch to func1
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    str    w0, [x29, 44]
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    ldr    w0, [x29, 28]
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    ldp    x29, x30, [sp], 48
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    ret
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    .size    func4, .-func4
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    .align    2
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    .global    func5
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    .type    func5, %function
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func5:
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    stp    x29, x30, [sp, -32]!
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    add    x29, sp, 0
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    str    w0, [x29, 28]
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    ldr    w0, [x29, 28]
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    bl    func8 @ branch to func8
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    str    w0, [x29, 28]
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    ldr    w0, [x29, 28]
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    ldp    x29, x30, [sp], 32
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    ret
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    .size    func5, .-func5
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    .align    2
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    .global    func6
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    .type    func6, %function
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func6:
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    sub    sp, sp, #32
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    str    w0, [sp, 12]
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    mov    w0, 314 @ w0 = 314
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    str    w0, [sp, 24] @ [sp+24] = w0 = 314
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    mov    w0, 1932 @ w0 = 1932
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    str    w0, [sp, 28] @ [sp+28] = w0
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    str    wzr, [sp, 20] @ [sp+20] = 0
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    str    wzr, [sp, 20]
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    b    .L14 @ branch to .L14
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.L15:
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    ldr    w1, [sp, 28] @ w1 = [sp+28]
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    mov    w0, 800 @ w0 = 800
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    mul    w0, w1, w0 @ w0 *= w1
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    ldr    w1, [sp, 24] @ w1 = [sp+24]
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    udiv    w2, w0, w1 @ w2 = w0 // w1
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    ldr    w1, [sp, 24] @ w1 = [sp+24]
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    mul    w1, w2, w1 @ w1 *= w2
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    sub    w0, w0, w1 @ w0 -= w1
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    str    w0, [sp, 12] @ [sp+12] = w0
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    ldr    w0, [sp, 20] @ w0 = [sp+20]
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    add    w0, w0, 1 @ w0 += 1
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    str    w0, [sp, 20] @ [sp+20] = w0
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.L14:
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    ldr    w0, [sp, 20]
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    cmp    w0, 899
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    bls    .L15 @ branch to .L15 if w0 <= 899
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    ldr    w0, [sp, 12]
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    add    sp, sp, 32
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    ret
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    .size    func6, .-func6
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    .align    2
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    .global    func7
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    .type    func7, %function
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func7:
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    sub    sp, sp, #16
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    str    w0, [sp, 12]
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    ldr    w0, [sp, 12]
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    cmp    w0, 100
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    bls    .L18 @ branch to .L18 if w0 <= 100
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    ldr    w0, [sp, 12]
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    b    .L19 @ branch to .L19 unconditionally
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.L18:
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    mov    w0, 7 @ w0 = 7
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.L19:
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    add    sp, sp, 16
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    ret
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    .size    func7, .-func7
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    .align    2
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    .global    func8
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    .type    func8, %function
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func8:
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    sub    sp, sp, #16
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    str    w0, [sp, 12]
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    ldr    w0, [sp, 12]
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    add    w0, w0, 2 @ w0 += 2
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    add    sp, sp, 16
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    ret
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    .size    func8, .-func8
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    .section    .rodata
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    .align    3
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.LC0:
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    .string    "Result: %ld\n"
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    .text
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    .align    2
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    .global    main
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    .type    main, %function
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main:
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    stp    x29, x30, [sp, -48]!
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    add    x29, sp, 0
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    str    w0, [x29, 28]
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    str    x1, [x29, 16]
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    ldr    x0, [x29, 16]
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    add    x0, x0, 8
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    ldr    x0, [x0]
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    bl    atoi
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    str    w0, [x29, 44]
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    ldr    w0, [x29, 44]
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    bl    func1
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    mov    w1, w0
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    adrp    x0, .LC0
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    add    x0, x0, :lo12:.LC0
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    bl    printf
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    nop
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    ldp    x29, x30, [sp], 48
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    ret
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    .size    main, .-main
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    .ident    "GCC: (Ubuntu/Linaro 7.5.0-3ubuntu1~18.04) 7.5.0"
180
    .section    .note.GNU-stack,"",@progbits
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Solution
bhi: Branch on Higher than.
Here is my note:
1
arg = 3964545182
2
​
3
1. func1: 3964545182 + 100 = 3964545282
4
2. func2: 3964545282 + 13 = 3964545295
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3. func5: do nothing
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4. func8: 3964545295 + 2 = 3964545297
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5. func1: do nothing
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6. func3: do nothing
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7. func7: do nothing
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​
11
result = 3964545297 = 0xec4e2911
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Flag
1
picoCTF{ec4e2911}
Copied!
Powershelly
Solved by:
Challenge
It's not a bad idea to learn to read Powershell. We give you the output, but do you think you can find the input? rev_PS.ps1 output.txt​
Solution
Todo!
Flag
1
​
Copied!
Rolling My Own
Solved by:
Challenge
I don't trust password checkers made by other people, so I wrote my own. It doesn't even need to store the password! If you can crack it I'll give you a flag. remote nc mercury.picoctf.net 17615
Solution
Todo!
Flag
1
​
Copied!
Checkpass
Solved by:
Challenge
What is the password? File: checkpass Flag format: picoCTF{...}
Solution
Todo!
Flag
1
​
Copied!