[JazzyJessie]

The method evenSquares takes a single int parameter, n, (for example, 10000), and then prints all of the (positive) even perfect squares less than n, each on a separate line. Notice that evenSquares has a void return type, since all it does is print integers to the console. Be sure to use the println method to print each entry of your output.

Example: if n = 40, your code should print

4
16
36

(Hint: your method should be built around a for loop with a test component that asks if the square of the control variable (say, k) is < n. Thus, the loop should terminate as soon as k*k equals or exceeds n.)

[Kakerergodt]

Seems simple enough, have you given it a try? Anything specific you need help with?

[JazzyJessie]

Kakerergodt, on 03 October 2012 - 09:33 AM, said:

Seems simple enough, have you given it a try? Anything specific you need help with?

i got this:


public void evenSquares(int n){
for(k = 2; Math.sqrt(k) <= n; k = k + 2){
System.out.println(Math.sqrt(k));
}
}

but it doesn't work

[pbl]

System.out.println(Math.sqrt(k));

will print the square root of all the k specified
you want the exact square root and you want the input parameter to be even
I would rather go the other way:
- take all numbers
- raise them to of 2
- break out of the loop if you execeed the limits
- found if the result is even