AP CSA 2.7 Exercise 1: Digit Peeler

Unit 2: Selection and Iteration · Lesson 2.7 · Exercise 1

Digit Peeler

while Loops. Write real Java, run it, and submit it to be graded against hidden test cases.

Back to the Lesson 2.7 page All Unit 2 lessons

Why this one is worth doing

A while loop is the right shape when you do not know how many times you are going round. Peeling the digits off a number is the classic case: the number itself decides when to stop.

What to write

Read one non-negative integer n.
Using a while loop, print each digit of n on its own line, from the LAST digit to the first.
After the loop, print how many digits you printed.
Print the sum of those digits on the final line.
Treat 0 as having one digit, which is 0.

Your program reads

A single non-negative integer.

Your program prints

One line per digit from last to first, then the digit count, then the digit sum.

Worked examples

These are the cases you can see. There are more you cannot, and they use different values, so an answer that prints these numbers as constants will fail.

Example 1 input

Example 1 output

Example 2 input

Example 2 output

9
1
9

Your answer

Main.java

import java.util.Scanner;

public class Main {
  public static void main(String[] args) {
    Scanner input = new Scanner(System.in);
    int n = input.nextInt();

    // n % 10 is the last digit. n / 10 removes it.
  }
}

Input for the Run button

Run sends whatever is in the input box below. Submitting runs your program against every test case, including hidden ones with different input.

Stuck?

Hint 1

n % 10 gives the last digit and n / 10 throws it away. Doing both every pass is the whole loop.

Hint 2

If you never change n inside the loop, the condition never becomes false and the run times out.

Hint 3

Zero needs its own line before the loop, because while (n > 0) never runs for it.

Where to go next

Rework the Lesson 2.7 page Pick another Unit 2 lesson

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