Number literal formats (underscore, hex, binary) in Kotlin - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Kotlin number literals with underscores

What is the output of this Kotlin code snippet?

Kotlin

fun main() {
    val number = 1_000_000
    println(number)
}

A 1000000

B 1000_000

C 1_000_000

D 100000

💡 Hint

Underscores in number literals are ignored by the compiler and only improve readability.

In Kotlin, underscores in numeric literals are just for readability and do not affect the actual value. So 1_000_000 is the same as 1000000.

❓ Predict Output

intermediate

Output of Kotlin hexadecimal literal

What will this Kotlin code print?

Kotlin

fun main() {
    val hexNumber = 0x1A
    println(hexNumber)
}

A 0x1A

B 1A

C 26

D 31

💡 Hint

Hexadecimal literals start with 0x and represent base 16 numbers.

0x1A in hexadecimal equals 26 in decimal, so the program prints 26.

❓ Predict Output

advanced

Output of Kotlin binary literal with underscores

What is the output of this Kotlin program?

Kotlin

fun main() {
    val binaryNumber = 0b1010_1101
    println(binaryNumber)
}

A 173

B 10101101

C 0b10101101

D 165

💡 Hint

Binary literals start with 0b and represent base 2 numbers. Underscores are ignored.

0b10101101 is binary for decimal 173, so the program prints 173.

❓ Predict Output

advanced

Sum of mixed number literals in Kotlin

What is the output of this Kotlin code?

Kotlin

fun main() {
    val a = 0xF
    val b = 0b10_10
    val c = 10_4
    println(a + b + c)
}

A 135

B 129

C 140

D 150

💡 Hint

Convert each literal to decimal before adding: 0xF = 15, 0b10_10 = 10, 10_4 = 104.

0xF is 15 decimal, 0b10_10 is binary 1010 which is 10 decimal, 10_4 is 104 decimal. Sum is 15 + 10 + 104 = 129.

❓ Predict Output

expert

Output of Kotlin code with invalid underscore placement

What error does this Kotlin code produce?

Kotlin

fun main() {
    val invalidNumber = 0x_1F
    println(invalidNumber)
}

A No error, prints 31

B NumberFormatException

C IllegalArgumentException

D SyntaxError

💡 Hint

Underscores cannot be placed right after the 0x prefix in hexadecimal literals.

Kotlin does not allow underscores immediately after the 0x prefix, so this code causes a syntax error.

Practice

1. Which of the following is a valid way to write the number one million in Kotlin using underscores for readability?

easy

A. 1__000000

B. 1000_000_

C. 10__00_000

D. 1_000_000

Solution

1. Step 1: Understand underscore usage in Kotlin numbers

   Kotlin allows underscores between digits to improve readability, but they must not be consecutive, leading, or trailing.

2. Step 2: Check each option for correct underscore placement

   1_000_000 uses single underscores separating groups of three digits correctly. A has consecutive underscores (invalid). B has trailing underscore (invalid). C has consecutive underscores (invalid).

3. Final Answer:

   1_000_000 -> Option D

4. Quick Check:

   Underscores separate digits properly = 1_000_000 [OK]

Hint: Use single underscores between digit groups only [OK]

Common Mistakes:

• Using double underscores
• Placing underscores at start or end
• Trailing underscores

2. Which of the following Kotlin number literals is NOT valid syntax?

easy

A. 123_456_

B. 0b1010_1101

C. 0x1A3F

D. 987_654

Solution

1. Step 1: Review Kotlin rules for underscores in number literals

   Underscores can be used between digits but not at the end or start of a number.

2. Step 2: Check each option for trailing underscores

   123_456_ ends with an underscore, which is invalid syntax. Options A, B, and D are valid hexadecimal, binary, and decimal literals respectively.

3. Final Answer:

   123_456_ -> Option A

4. Quick Check:

   Trailing underscore invalid = 123_456_ [OK]

Hint: Never put underscore at the end of a number [OK]

Common Mistakes:

• Trailing underscores in numbers
• Confusing hex prefix 0x with decimal
• Misplacing underscores in binary literals

3. What is the output of this Kotlin code?

val x = 0xF_A
val y = 0b1010_1010
println(x + y)

medium

A. 270

B. 250

C. 420

D. 280

Solution

1. Step 1: Convert hexadecimal and binary literals to decimal

   0xF_A means hex FA: F=15, A=10, so 15*16 + 10 = 250. 0b1010_1010 is binary 10101010 which equals 128+32+8+2=170.

2. Step 2: Add the decimal values

   250 + 170 = 420.

3. Final Answer:

   420 -> Option C

4. Quick Check:

   (0xFA + 0b10101010) = 420 [OK]

Hint: Convert hex and binary to decimal before adding [OK]

Common Mistakes:

• Misreading hex digits
• Incorrect binary to decimal conversion
• Ignoring underscores in literals

4. Identify the error in this Kotlin code snippet:

val num = 0b1021_0010

medium

A. Underscores are not allowed in binary literals

B. Binary literals cannot contain digits other than 0 or 1

C. Binary literals must start with 0x

D. The number is too large for binary format

Solution

1. Step 1: Understand valid digits in binary literals

   Binary literals can only contain digits 0 and 1, optionally separated by underscores.

2. Step 2: Check the given literal for invalid digits

   The literal contains '2', which is invalid in binary. Underscores are allowed, and binary literals start with 0b, not 0x.

3. Final Answer:

   Binary literals cannot contain digits other than 0 or 1 -> Option B

4. Quick Check:

   Binary digits must be 0 or 1 only [OK]

Hint: Binary digits only 0 or 1, no other digits allowed [OK]

Common Mistakes:

• Using digits other than 0 or 1 in binary
• Confusing hex prefix 0x with binary 0b
• Thinking underscores are disallowed in binary

5. You want to represent the decimal number 255 in Kotlin using a binary literal with underscores for readability. Which of the following is the best way to write it?

hard

A. 0b1111_1111

B. 0b11111111

C. 0b1111_1110

D. 0b1111_11112

Solution

1. Step 1: Convert decimal 255 to binary

   255 in binary is 11111111 (eight ones).

2. Step 2: Add underscores for readability

   Grouping bits in fours is common, so 1111_1111 is clearer and valid.

3. Step 3: Check options for correctness

   0b1111_1111 correctly uses underscores and matches 255. 0b11111111 is correct but less readable. 0b1111_1110 is 254, and 0b1111_11112 has invalid digit '2'.

4. Final Answer:

   0b1111_1111 -> Option A

5. Quick Check:

   Binary 1111_1111 = 255 decimal [OK]

Hint: Group binary digits in fours with underscores [OK]

Common Mistakes:

• Using wrong binary digits
• Adding invalid digits like 2
• Not grouping bits for readability