Why ranges simplify iteration in Kotlin - Test Your Understanding

Practice - 5 Tasks

1fill in blank

easy

Complete the code to print numbers from 1 to 5 using a range.

Kotlin

for (i in 1..[1]) {
    println(i)
}

Drag options to blanks, or click blank then click option'

A0

B10

C5

D6

Attempts:

3 left

💡 Hint

Common Mistakes

Using a number outside the desired range like 6 or 10.

Using 0 which is less than the start of the range.

The range 1..5 includes numbers from 1 to 5, so the loop prints these numbers.

2fill in blank

medium

Complete the code to print numbers from 5 down to 1 using a range.

Kotlin

for (i in [1] downTo 1) {
    println(i)
}

Drag options to blanks, or click blank then click option'

A1

B0

C10

D5

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1 before downTo which would create an empty range.

Using 0 or 10 which are outside the desired range.

The downTo function creates a decreasing range from 5 down to 1.

3fill in blank

hard

Fix the error in the code to print even numbers from 2 to 10 using a range with step.

Kotlin

for (i in 2..10 step [1]) {
    println(i)
}

Drag options to blanks, or click blank then click option'

A5

B2

C1

D3

Attempts:

3 left

💡 Hint

Common Mistakes

Using step 1 which prints all numbers, not just even ones.

Using step 3 or 5 which skips some even numbers.

Using step 2 makes the loop count by twos, printing even numbers.

4fill in blank

hard

Fill both blanks to create a range that prints numbers from 1 to 10 but only odd numbers.

Kotlin

for (i in [1]..[2] step 2) {
    println(i)
}

Drag options to blanks, or click blank then click option'

A1

B2

C10

D9

Attempts:

3 left

💡 Hint

Common Mistakes

Starting at 2 which would print even numbers.

Ending at 10 which includes an even number and would print 10 if stepped by 2.

Starting at 1 and stepping by 2 up to 9 prints odd numbers: 1, 3, 5, 7, 9.

5fill in blank

hard

Fill all three blanks to create a range that prints numbers from 10 down to 1, stepping by 3.

Kotlin

for (i in [1] downTo [2] step [3]) {
    println(i)
}

Drag options to blanks, or click blank then click option'

A10

B1

C3

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Using step 2 or 1 which changes the numbers printed.

Swapping start and end values.

This loop counts down from 10 to 1, stepping by 3, printing 10, 7, 4, 1.