Display n even numbers (not upto n)

Question

guys i know this is a dumb question from me
so could anyone tell me how to do that? and plz consider this one too
i have a loop which runs like upto 100000 even numbers but i want it to stop at the number specified by the user
for example: i am trying to print binary even numbers under 100000 so consider user input is 2 so only 2 binary numbers shd be displayed rather than going all way to 100000...
thanks in advance

Answer 1

for(int a=2,b=0;b<n;b++) System.out.println(a+=2);

Answer 2

i thought u want even numbers after seeing one of the answers here 😅

Answer 3

I think it should be like this :
I used JS.
var n = prompt ("enter a number");
for (i=0;i<100000;i++){
if(i==n){
break ;
}
if(i%2==0){
console.log(i);
}
}

Answer 4

gaurav i didnt get you?
well here is my code to make things more clearer
https://code.sololearn.com/cMZPX3dLGhOY/?ref=app

Answer 5

lol that was just an internal part but if u see the code i would understand my actual problem haha
the problem is "a number shd be even as well as it shd contain 2 active bits i.e no of 1's" so i tried to get all the satisfying numbers below 'm' so instead of printing all I want to print only 'n' numbers which is specified by the user ..if n=2 then first two of the satisfying Condition numbers shd be displayed

Answer 6

lol i always do silly mistakes and make it a big mess out of it.. haha
well i got it gaurav ..and guys thanks for replying :)
I'll update my correct code :)