/etc/passwd and /etc/shadow in Linux CLI - Practice Problems & Coding Challenges

Challenge - 5 Problems

💻 Command Output

intermediate

What is the output of this command?

You run the command cut -d: -f1 /etc/passwd | head -3. What is the expected output?

Linux CLI

cut -d: -f1 /etc/passwd | head -3

A

root
sys
adm

B

root
bin
daemon

C

root
daemon
bin

D

daemon
bin
sys

💡 Hint

The first field in /etc/passwd is the username. The command extracts usernames and shows the first three.

The /etc/passwd file lists user accounts. The first three usernames are usually root, daemon, and bin in that order.

🧠 Conceptual

intermediate

What is the main difference between /etc/passwd and /etc/shadow?

Choose the correct statement about the difference between /etc/passwd and /etc/shadow files.

A /etc/passwd is readable by all users, /etc/shadow is readable only by root and stores encrypted passwords.

B /etc/passwd stores encrypted passwords, /etc/shadow stores user IDs.

C /etc/shadow stores usernames, /etc/passwd stores passwords in plain text.

D /etc/passwd and /etc/shadow both store passwords in plain text.

💡 Hint

Think about file permissions and security of password storage.

/etc/passwd contains user account info and is world-readable. /etc/shadow contains encrypted passwords and is only readable by root for security.

🔧 Debug

advanced

Why does this script fail to read /etc/shadow?

A user runs this script to read the first line of /etc/shadow:

#!/bin/bash
head -1 /etc/shadow

But it shows a permission denied error. Why?

A /etc/shadow is only readable by root, normal users cannot read it.

B head command does not work on /etc/shadow file.

C The script is missing execute permission.

D The script syntax is incorrect.

💡 Hint

Check file permissions of /etc/shadow.

/etc/shadow contains sensitive password hashes and is only readable by root. Normal users get permission denied errors.

🚀 Application

advanced

Extract usernames with locked passwords from /etc/shadow

Which command correctly lists usernames whose password field in /etc/shadow starts with an exclamation mark (!) indicating locked accounts?

A sed -n '/^!/p' /etc/shadow

B awk -F: '$2 ~ /^!/ {print $1}' /etc/shadow

C cut -d: -f1 /etc/shadow | grep '!'

D grep '^!' /etc/shadow | cut -d: -f1

💡 Hint

Look for lines where the second field starts with ! and print the first field.

The password field is the second field in /etc/shadow. Using awk with field separator ':' and regex /^!/ on $2 correctly filters locked accounts.

📝 Syntax

expert

What error does this command produce?

What error will this command produce?

cut -d: -f1 /etc/shadow | head -3

A No output, command runs silently

B File not found error

C Syntax error due to wrong delimiter

D Permission denied error because /etc/shadow is restricted

💡 Hint

Consider file permissions of /etc/shadow for normal users.

Normal users cannot read /etc/shadow, so cut command fails with permission denied error.

Practice

1. What is the primary purpose of the /etc/passwd file in Linux?

easy

A. To store encrypted user passwords securely

B. To configure network settings

C. To store user account information accessible to all users

D. To log system errors

Solution

1. Step 1: Understand the role of /etc/passwd

   The /etc/passwd file contains user account details like username, user ID, group ID, home directory, and shell. It is readable by all users.

2. Step 2: Differentiate from /etc/shadow

   The encrypted passwords are stored separately in /etc/shadow for security, not in /etc/passwd.

3. Final Answer:

   To store user account information accessible to all users -> Option C

4. Quick Check:

   /etc/passwd = user info file [OK]

Hint: Remember: passwd = public user info, shadow = private passwords [OK]

Common Mistakes:

• Confusing /etc/passwd with /etc/shadow
• Thinking passwords are stored in /etc/passwd
• Assuming /etc/passwd stores system logs

2. Which of the following lines correctly shows the format of a typical /etc/shadow entry?

easy

A. username:x:1000:1000:User Name:/home/user:/bin/bash

B. username:$6$randomsalt$encryptedpassword:18000:0:99999:7:::

C. username:password:1000:1000:User Name:/home/user:/bin/bash

D. username:encryptedpassword:1000:1000:User Name:/home/user:/bin/bash

Solution

1. Step 1: Identify /etc/shadow format

   The /etc/shadow file stores encrypted passwords and related info. The password field looks like $6$randomsalt$encryptedpassword for SHA-512 encryption.

2. Step 2: Compare options

   The line username:$6$randomsalt$encryptedpassword:18000:0:99999:7::: matches the shadow file format with encrypted password (indicated by $6$) and password aging fields separated by colons. The other options resemble /etc/passwd format or have incorrect password fields.

3. Final Answer:

   username:$6$randomsalt$encryptedpassword:18000:0:99999:7::: -> Option B

4. Quick Check:

   /etc/shadow entries have encrypted passwords with $ signs [OK]

Hint: Look for $ signs and many colon-separated fields in shadow entries [OK]

Common Mistakes:

• Confusing /etc/passwd format with /etc/shadow
• Expecting plain text passwords in /etc/shadow
• Missing fields after encrypted password

3. Given the following command:

grep '^root:' /etc/passwd | cut -d':' -f1,3,7

What will be the output format?

medium

A. root:/bin/bash

B. root:x:0:0:root:/root:/bin/bash

C. root:0:0

D. root:0:/bin/bash

Solution

1. Step 1: Understand the command parts

   grep '^root:' /etc/passwd finds the line starting with 'root:'. cut -d':' -f1,3,7 extracts fields 1 (username), 3 (UID), and 7 (shell) separated by colons.

2. Step 2: Identify root line fields

   Typical root line: root:x:0:0:root:/root:/bin/bash. Fields: 1=root, 2=x, 3=0, 4=0, 5=root, 6=/root, 7=/bin/bash.

3. Step 3: Extract fields 1,3,7

   Fields 1=root, 3=0, 7=/bin/bash. Joined by colons: root:0:/bin/bash.

4. Final Answer:

   root:0:/bin/bash -> Option D

5. Quick Check:

   Fields 1,3,7 from /etc/passwd line = root:0:/bin/bash [OK]

Hint: Count fields carefully; cut -f1,3,7 picks username, UID, shell [OK]

Common Mistakes:

• Mixing up field numbers in cut command
• Expecting full /etc/passwd line output
• Confusing shadow file format with passwd

4. You run this command to check password expiry info:

sudo chage -l username

But it shows an error: chage: user 'username' does not exist. What is the most likely cause?

medium

A. The username is missing in /etc/passwd

B. The /etc/shadow file is empty

C. The chage command is not installed

D. The user has no password set

Solution

1. Step 1: Understand chage user check

   The chage command checks user password expiry info by looking up the username in /etc/passwd.

2. Step 2: Analyze error cause

   Error 'user does not exist' means the username is not found in /etc/passwd. If the user was missing there, chage cannot proceed.

3. Final Answer:

   The username is missing in /etc/passwd -> Option A

4. Quick Check:

   User must exist in /etc/passwd for chage to work [OK]

Hint: Check /etc/passwd for username before using chage [OK]

Common Mistakes:

• Assuming /etc/shadow emptiness causes this error
• Thinking chage command missing causes user error
• Believing no password set triggers 'user does not exist'

5. You want to write a script that lists all users who have a valid login shell (not /usr/sbin/nologin or /bin/false) by reading /etc/passwd. Which command snippet correctly filters these users?

hard

A. awk -F':' '$7 != "/usr/sbin/nologin" && $7 != "/bin/false" {print $1}' /etc/passwd

B. grep -v '/usr/sbin/nologin' /etc/shadow | cut -d':' -f1

C. cat /etc/passwd | grep '/bin/bash' | awk '{print $1}'

D. sed '/nologin/d' /etc/passwd | cut -d':' -f7

Solution

1. Step 1: Identify shell field in /etc/passwd

   The 7th field in /etc/passwd is the user's login shell. We want to exclude /usr/sbin/nologin and /bin/false.

2. Step 2: Analyze each option

   awk -F':' '$7 != "/usr/sbin/nologin" && $7 != "/bin/false" {print $1}' /etc/passwd uses awk with field separator ':' and filters out unwanted shells, printing usernames (field 1). grep -v '/usr/sbin/nologin' /etc/shadow | cut -d':' -f1 wrongly reads /etc/shadow and uses grep -v incorrectly. cat /etc/passwd | grep '/bin/bash' | awk '{print $1}' filters only /bin/bash but misses other valid shells. sed '/nologin/d' /etc/passwd | cut -d':' -f7 deletes lines with 'nologin' but then prints shell field, not usernames.

3. Final Answer:

   awk -F':' '$7 != "/usr/sbin/nologin" && $7 != "/bin/false" {print $1}' /etc/passwd -> Option A

4. Quick Check:

   Use awk on /etc/passwd to filter shell field and print usernames [OK]

Hint: Use awk with FS=':' and check field 7 for valid shells [OK]

Common Mistakes:

• Using /etc/shadow instead of /etc/passwd for shell info
• Printing wrong fields (shell instead of username)
• Using grep without proper field filtering