What's wrong with my solution?

Question about the task

Under discussion

public class Solution {
    public static void main(String[] args) {
        checkInterval(60);
        checkInterval(112);
        checkInterval(10);
    }

    public static void checkInterval(int a) {

        if(a < 50 || a > 100){
            System.out.println("The number a is not in the interval.");

        }
        else{
            System.out.println("The number a is in the interval.");
        }
    }
}

I don't get it, why is the last bit failing? This happened on the same question before, I got the output right..?

Write the checkInterval method. The method should check whether an integer is in the range from 50 to 100 and display the result on the screen as follows: "The number a is not in the interval." or "The number a is in the interval.", where a is the method argument.

Example for 112:
The number 112 is not in the interval.

Example for 60:
The number 60 is in the interval.

Requirements:

The program should display text on the screen.
The main method should not call System.out.println or System.out.print().
The main method must call the checkInterval method.
The checkInterval method must be static void.
The checkInterval method should display text on the screen in accordance with the task conditions.

package com.codegym.task.task04.task0410; /* Come on, lucky seven! */ public class Solution { public static void main(String[] args) { checkInterval(60); checkInterval(112); checkInterval(10); } public static void checkInterval(int a) { if(a < 50 || a > 100){ System.out.println("The number a is not in the interval."); } else{ System.out.println("The number a is in the interval."); } } }

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Odhran

Level 18 , Sheffield, United Kingdom

12 August 2021, 10:01

The logic needs to be within the specified range, so '>50 || <100' is within this. This way if its equal to or outside of the range then the else value result will be valid.

wasnt me

Level 4 , United Kingdom

12 August 2021, 10:01

right so they wanted the "a" to be the integer in display. nvm.

Pavel

Level 20 , Kyiv, Ukraine

17 August 2021, 10:36

Maybe you should concatenate strings with variable, like this: "The number" + a + "is not in the interval."