For loop with step and downTo in Kotlin - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to print numbers from 1 to 5.

Kotlin

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

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

A4

B6

C5

D10

Attempts:

3 left

💡 Hint

Common Mistakes

Using 6 instead of 5 includes an extra number.

Using 4 stops before 5.

The range 1..5 includes numbers from 1 to 5.

2fill in blank

medium

Complete the code to print even numbers from 2 to 10 using step.

Kotlin

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

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

A3

B2

C1

D5

Attempts:

3 left

💡 Hint

Common Mistakes

Using step 1 prints all numbers.

Using step 3 skips some even numbers.

The step keyword controls the increment. For even numbers, step should be 2.

3fill in blank

hard

Fix the error in the code to print numbers from 5 down to 1.

Kotlin

for (i in 5 [1] 1) {
    println(i)
}

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

Astep

Buntil

C..

DdownTo

Attempts:

3 left

💡 Hint

Common Mistakes

Using '..' creates an increasing range, causing no output.

Using 'until' is for increasing ranges and excludes the end.

The downTo keyword creates a decreasing range from 5 to 1.

4fill in blank

hard

Fill both blanks to print odd numbers from 9 down to 1.

Kotlin

for (i in 9 [1] 1 [2] 2) {
    println(i)
}

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

AdownTo

Bstep

C..

Duntil

Attempts:

3 left

💡 Hint

Common Mistakes

Using '..' instead of 'downTo' counts up, not down.

Forgetting step prints all numbers.

Use downTo for counting down and step 2 to skip even numbers.

5fill in blank

hard

Fill all three blanks to print numbers from 10 down to 2, stepping by 2.

Kotlin

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

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

A10

Bstep 2

CdownTo 2

Duntil 2

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'until 2' counts up and excludes 2.

Placing step before downTo causes syntax errors.

Start at 10, count down to 2 using downTo 2, and step by 2 with step 2.

Practice

1. What does the downTo keyword do in a Kotlin for loop?

easy

A. It makes the loop count forwards from a lower number to a higher number.

B. It makes the loop count backwards from a higher number to a lower number.

C. It skips numbers in the loop by a fixed step.

D. It stops the loop immediately.

Solution

1. Step 1: Understand the purpose of downTo

   The downTo keyword in Kotlin is used to create a range that counts backwards, from a higher number down to a lower number.

2. Step 2: Compare with other options

   It makes the loop count forwards from a lower number to a higher number. describes counting forwards, It skips numbers in the loop by a fixed step. skipping numbers by step, and It stops the loop immediately. stopping the loop immediately, none of which downTo does.

3. Final Answer:

   It makes the loop count backwards from a higher number to a lower number. -> Option B

4. Quick Check:

   downTo = count backwards [OK]

Hint: Remember: downTo means counting downwards [OK]

Common Mistakes:

• Confusing downTo with counting forwards
• Thinking downTo skips numbers
• Assuming downTo stops the loop

2. Which of the following is the correct syntax to loop from 10 down to 1, stepping by 2 in Kotlin?

easy

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

B. for (i in 1 downTo 10 step 2) { println(i) }

C. for (i in 10 downTo 1 step 2) { println(i) }

D. for (i in 10 downTo 1 by 2) { println(i) }

Solution

1. Step 1: Identify correct downTo and step syntax

   The correct syntax to count down with steps is for (i in start downTo end step stepValue). for (i in 10 downTo 1 step 2) { println(i) } matches this.

2. Step 2: Check other options for errors

   for (i in 1 downTo 10 step 2) { println(i) } has start < end, resulting in empty range. for (i in 10..1 step 2) { println(i) } uses .. which doesn't support down with step correctly. for (i in 10 downTo 1 by 2) { println(i) } uses by instead of step, which is incorrect.

3. Final Answer:

   for (i in 10 downTo 1 step 2) { println(i) } -> Option C

4. Quick Check:

   Correct syntax = for (i in 10 downTo 1 step 2) { println(i) } [OK]

Hint: Use downTo with step, not by [OK]

Common Mistakes:

• Using by instead of step
• Counting down with .. range
• Reversing start and end values

3. What is the output of the following Kotlin code?

for (i in 8 downTo 2 step 3) {
    print("$i ")
}

medium

A. 8 5 2

B. 8 6 4 2

C. 8 5 3

D. 8 5 2 1

Solution

1. Step 1: Determine the loop values

   The loop starts at 8, counts down to 2, stepping by 3. So values are 8, then 8-3=5, then 5-3=2.

2. Step 2: Confirm the loop stops at 2

   Since 2 is the end value and included, the loop prints 8, 5, and 2.

3. Final Answer:

   8 5 2 -> Option A

4. Quick Check:

   8, 5, 2 with step 3 downTo 2 = 8 5 2 [OK]

Hint: Subtract step each time, stop at end value [OK]

Common Mistakes:

• Including numbers less than end value
• Using wrong step size
• Confusing downTo with ..

4. Identify the error in this Kotlin loop:

for (i in 5 downTo 1 step 0) {
    println(i)
}

medium

A. The downTo keyword is misspelled.

B. There is no error; the code runs fine.

C. The loop should use .. instead of downTo.

D. Step value cannot be zero; it causes a runtime error.

Solution

1. Step 1: Analyze the step value

   The step value is 0, which is invalid and causes a runtime error (IllegalArgumentException).

2. Step 2: Check other options

   downTo is spelled correctly, and using .. would count upwards, not downwards. So options A, C, and D are incorrect.

3. Final Answer:

   Step value cannot be zero; it causes a runtime error. -> Option D

4. Quick Check:

   Step must be positive and non-zero [OK]

Hint: Step cannot be zero; it must be positive [OK]

Common Mistakes:

• Using zero or negative step incorrectly
• Confusing downTo with ..
• Ignoring infinite loop risk

5. You want to print all odd numbers from 15 down to 1 using a Kotlin for loop with downTo and step, without using an if statement. Which code snippet correctly does this?

hard

A. for (i in 15 downTo 1 step 2) { println(i) }

B. for (i in 15 downTo 1 step 1) { if (i % 2 == 1) println(i) }

C. for (i in 15 downTo 1 step 3) { println(i) }

D. for (i in 14 downTo 1 step 2) { println(i) }

Solution

1. Step 1: Understand the goal

   We want to print odd numbers from 15 down to 1. Starting at 15 (which is odd) and stepping by 2 will print all odd numbers downwards.

2. Step 2: Evaluate options

   for (i in 15 downTo 1 step 2) { println(i) } starts at 15 and steps by 2 down to 1, printing 15, 13, 11, ..., 1. for (i in 15 downTo 1 step 1) { if (i % 2 == 1) println(i) } uses step 1 with an if filter (disallowed). for (i in 15 downTo 1 step 3) { println(i) } steps by 3, skipping some odds. for (i in 14 downTo 1 step 2) { println(i) } starts at 14 (even), printing evens.

3. Final Answer:

   for (i in 15 downTo 1 step 2) { println(i) } -> Option A

4. Quick Check:

   Start odd, step 2 downTo 1 = odd numbers [OK]

Hint: Start at odd number, step 2 downTo 1 for odd numbers [OK]

Common Mistakes:

• Starting at even number when printing odds
• Using wrong step size
• Filtering inside loop instead of stepping