Output the square root just for the whole numbers

Question

I need to Output the results of square root just for the whole numbers.

Example:

1  - 1
4  - 2
9  - 3
16 - 4...

up to 961 - 31... Which is the last square root before 1000.

Until now I have this... But, this is showing square roots for all numbers until 1000, and I want the square root for WHOLE NUMBERS ONLY.

What should I do?

public class HelloWorld {
    public static void main(String args[]) {
        double i = 1;
        do {
            System.out.printf("%.0f\t%.0f\n", i, Math.sqrt(i));
            i++;
        } while (i <= 1000);
    }
}

The output that I want should look like this:

Consider looping `i` from 1 to `Math.sqrt(1000)` and printing `i` and `i*i` — M.S., Mar 19 '16 at 16:50

score 1 · Answer 1 · edited Jun 20 '20 at 09:12

1

The trick that can solve your request is in the Math class too,

Math.floor(...)

that method will round a double to the lower integer, so you need to check if the root and the lower floor are the same..

Example:

for (int j = 0; j < 1000; j++) {
            final double res = Math.sqrt(j);
            if (res == Math.floor(res)) {
                System.out.printf("%d\t%.0f\n", j, res);
            }
        }

edited Jun 20 '20 at 09:12

Community

1
1

answered Mar 19 '16 at 17:01

ΦXocę 웃 Пepeúpa ツ

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Thanks, I'll try that – Philip Santana Dias Mar 19 '16 at 17:04
sure, give a try and let e know – ΦXocę 웃 Пepeúpa ツ Mar 19 '16 at 17:31

score 1 · Answer 2 · answered Mar 19 '16 at 17:03

1

It is not efficient to loop through all 1000 values. Instead, consider the following code:

for(int i = 1; i <= Math.sqrt(1000); i++) {
    System.out.println(i*i + " - " + i);
}

answered Mar 19 '16 at 17:03

M.S.

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That's clever! Prints the perfect squares from 1 to 31. – BryanT Mar 19 '16 at 17:47
Happy it helped! If this answered your question then please remember to accept this answer. – M.S. Mar 20 '16 at 02:46

score 0 · Answer 3 · answered Mar 19 '16 at 17:14

0

Try this solution

for (int i = 1; i <= 1000; i++) {
    double sqrt = Math.sqrt(i);
    if (sqrt % 1 == 0) {
        System.out.println(i + " " + (int) sqrt);
    }
}

For int values sqrt % 1 will be equal to zero.

answered Mar 19 '16 at 17:14

Johny

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you are looping 1000 times but you only need 31 loops – Mixone Mar 19 '16 at 20:16

score 0 · Answer 4 · edited Aug 20 '16 at 10:19

Your method of solving this problem is slightly flawed.

You can know the biggest number immediately, it is the floor of the square root of the limit you have. So for 1000 => sqrt(1000) = 31.6.... therefore the closest you will get with a whole number is 31 squared, anything bigger will be greater than 1000
Also, you have a function, f(x) = x^2 , but you are looping through f(x) values instead of x values, this will make for a much larger amount of tests.

I would just do:

for(int i = 0; i < Math.sqrt(1000); i++) {
    System.out.println((i+1)^2 + " - " + i);
}

Output the square root just for the whole numbers

4 Answers4

Example: