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easyBasicsSometimes asked

Multiples of a Number

Basics · 6 min

Given a number k and a count n, return the first n positive multiples of k as a list.

In plain English

Multiply k by 1..n, or keep adding k.

Examples

Example 1

Input: k = 3, n = 4
Output: [3, 6, 9, 12]
Note: 3·1..3·4.

Example 2

Input: k = 5, n = 2
Output: [5, 10]
Note: First two.

Example 3

Input: k = 1, n = 3
Output: [1, 2, 3]
Note: Multiples of 1.

Constraints

1 ≤ n ≤ 10^4
-10^4 ≤ k ≤ 10^4

Clarifying questions1First question: First n multiples, starting at k?Explore questions

Q1.First n multiples, starting at k?
Yes — k·1 through k·n.

Quick interview revision

Understand

30-second explanation

Build the interview-ready mental model before comparing implementations.

Loop i from 1 to n, appending `k * i`.

Interview contextWhat is testedWhere candidates slipHow to stand outExplore interview context

What's being tested

Whether you generate by index-times-value correctly.

Where candidates slip

Off-by-one emitting 0 or n+1 terms.

How to stand out

Mention the add-k variant to avoid multiplication.

Choose

Compare the available approaches

Use constraints and trade-offs to select the route you can defend in an interview.

Learn & trace

Build each solution step by step

Move from the core idea to a visual trace, implementation, proof, and edge cases.

03.1

Approach 1 of 1 · Optimal

Index times k

TimeO(n)SpaceO(n)

This is the standard, expected solution — lead with it.

Stuck? Progressive hintsReveal one at a time

1Hint 1Explore hint

What's the i-th multiple of k in terms of i?

2Hint 2Explore hint

Multiply the index by k (or add k repeatedly).

3Hint 3Explore hint

Start at i = 1 so you don't emit 0.

A · Core idea

Produce the first n positive multiples of k. The i-th multiple is simply k·i, so a single index-driven loop generates them in order.

Why index × k. Multiples of k are k·1, k·2, …, k·n. Multiplying the loop index by k gives each directly; alternatively, keep a running sum and add k each step (no multiplication).

Why start i at 1. The 'first' multiple is k·1 = k, not k·0 = 0. Starting at 0 would emit a spurious 0.

Overflow. For large k and i, k·i can exceed int range — accumulate or produce into a long when the values can be big.

B · Visual dry run

Interactive dry run

Select a frame to connect the decision with its visual state.

k = 3, n = 4

i=1

3·1

State3

Final result[3, 6, 9, 12]

C · Code with teaching comments

solution.javaJava

Index times k — optimal

1
public java.util.List<Long> multiples(int k, int n) {
2
    java.util.List<Long> out = new java.util.ArrayList<>();
3
    for (int i = 1; i <= n; i++) {
4
        out.add((long) k * i);
5
    }
6
    return out;
7
}

D · Why it works and complexityO(n) timeO(n) spaceOpen the invariant and proofExplore the proof

ComplexityO(n) time, O(n) space — n iterations, one multiply (or add) each.

E · Edge cases and pitfalls4k = 5, n = 2 — [5, 10].Explore edge cases

Edge cases handled

k = 5, n = 2[5, 10].
k = 1, n = 3[1, 2, 3].

Pitfalls

Starting at i = 0 (gives a 0 multiple).
Producing the wrong count.

Practice

Try the solution yourself

Edit the optimal implementation, test another input, and observe the result.

Editable code previewpublic java.util.List<Long> multiples(int k, int n) { java.util.List<Long> out = new java.util.ArrayList<>(); for (int i = 1; i <= n; i++) { out.add((long) k * i); }Explore the playground

Try it — Java· run a complete program; method snippets need a driver

Avoid mistakes

Recognize the interview traps

Understand why common attempts fail before reviewing the corrected implementation.

Starting at i = 0 (gives a 0 multiple).
Producing the wrong count.

Revise

Carry the pattern into the interview

Quick interview revision

Pattern to remember

Index × k

Code pattern to memorize

for i in 1..n: out.append(k*i)

Rules to remember

1i from 1 to n.
2Multiple = k·i.
3Or add k each step.

Use this pattern when

Generating arithmetic sequences

Interview takeaway

The i-th multiple is k·i — index-driven generation.

Continue

Transfer the pattern to another problem

Use a direct variation or pattern sibling to prove the idea is reusable.

Transferable patternA core fresher pattern: Index × k. Master it on small inputs and it transfers to harder problems.

Practice queueDirect variations and pattern siblings

Direct follow-ups

Print Even Numbers up to N
Multiples of 2, bounded by n.

easyBasicsSometimes asked

Multiples of a Number

Basics · 6 min

Given a number k and a count n, return the first n positive multiples of k as a list.

In plain English

Multiply k by 1..n, or keep adding k.

Examples

Example 1

Input: k = 3, n = 4
Output: [3, 6, 9, 12]
Note: 3·1..3·4.

Example 2

Input: k = 5, n = 2
Output: [5, 10]
Note: First two.

Example 3

Input: k = 1, n = 3
Output: [1, 2, 3]
Note: Multiples of 1.

Constraints

1 ≤ n ≤ 10^4
-10^4 ≤ k ≤ 10^4

Clarifying questions1First question: First n multiples, starting at k?Explore questions

Q1.First n multiples, starting at k?
Yes — k·1 through k·n.

Quick interview revision

Understand

30-second explanation

Build the interview-ready mental model before comparing implementations.

Loop i from 1 to n, appending `k * i`.

Interview contextWhat is testedWhere candidates slipHow to stand outExplore interview context

What's being tested

Whether you generate by index-times-value correctly.

Where candidates slip

Off-by-one emitting 0 or n+1 terms.

How to stand out

Mention the add-k variant to avoid multiplication.

Choose

Compare the available approaches

Use constraints and trade-offs to select the route you can defend in an interview.

Learn & trace

Build each solution step by step

Move from the core idea to a visual trace, implementation, proof, and edge cases.

03.1

Approach 1 of 1 · Optimal

Index times k

TimeO(n)SpaceO(n)

This is the standard, expected solution — lead with it.

Stuck? Progressive hintsReveal one at a time

1Hint 1Explore hint

What's the i-th multiple of k in terms of i?

2Hint 2Explore hint

Multiply the index by k (or add k repeatedly).

3Hint 3Explore hint

Start at i = 1 so you don't emit 0.

A · Core idea

Produce the first n positive multiples of k. The i-th multiple is simply k·i, so a single index-driven loop generates them in order.

Why index × k. Multiples of k are k·1, k·2, …, k·n. Multiplying the loop index by k gives each directly; alternatively, keep a running sum and add k each step (no multiplication).

Why start i at 1. The 'first' multiple is k·1 = k, not k·0 = 0. Starting at 0 would emit a spurious 0.

Overflow. For large k and i, k·i can exceed int range — accumulate or produce into a long when the values can be big.

B · Visual dry run

Interactive dry run

Select a frame to connect the decision with its visual state.

k = 3, n = 4

i=1

3·1

State3

Final result[3, 6, 9, 12]

C · Code with teaching comments

solution.javaJava

Index times k — optimal

1
public java.util.List<Long> multiples(int k, int n) {
2
    java.util.List<Long> out = new java.util.ArrayList<>();
3
    for (int i = 1; i <= n; i++) {
4
        out.add((long) k * i);
5
    }
6
    return out;
7
}

D · Why it works and complexityO(n) timeO(n) spaceOpen the invariant and proofExplore the proof

ComplexityO(n) time, O(n) space — n iterations, one multiply (or add) each.

E · Edge cases and pitfalls4k = 5, n = 2 — [5, 10].Explore edge cases

Edge cases handled

k = 5, n = 2[5, 10].
k = 1, n = 3[1, 2, 3].

Pitfalls

Starting at i = 0 (gives a 0 multiple).
Producing the wrong count.

Practice

Try the solution yourself

Edit the optimal implementation, test another input, and observe the result.

Try it — Java· run a complete program; method snippets need a driver

Avoid mistakes

Recognize the interview traps

Understand why common attempts fail before reviewing the corrected implementation.

Starting at i = 0 (gives a 0 multiple).
Producing the wrong count.

Revise

Carry the pattern into the interview

Quick interview revision

Pattern to remember

Index × k

Code pattern to memorize

for i in 1..n: out.append(k*i)

Rules to remember

1i from 1 to n.
2Multiple = k·i.
3Or add k each step.

Use this pattern when

Generating arithmetic sequences

Interview takeaway

The i-th multiple is k·i — index-driven generation.

Continue

Transfer the pattern to another problem

Use a direct variation or pattern sibling to prove the idea is reusable.

Transferable patternA core fresher pattern: Index × k. Master it on small inputs and it transfers to harder problems.

Practice queueDirect variations and pattern siblings

Direct follow-ups

Print Even Numbers up to N
Multiples of 2, bounded by n.