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Carlos Fernández-Tejeda García
4,051 PointsWhy I can not log out the even numbers from 2 to 24?
Hi all. I have to log out the even numbers from 2 to 24. I am trying the following code:
//assign i as loop counter, var i = 0; //* ej is used to add 2 to the initial value and add 2 to the set value everytime that the loop runs // var ej= 0; // until i gets the value 12, the variable sum is added + 2 every time that the loop runs and i increases one *// while (i < 12){ var sum = ej + 2; console.log(sum); i++ } But bummer says that I don´t log out the even numbers. What I am doing wrong?
Thanks
var i = 0;
var ej= 2;
while (i < 12){
var sum = ej + 2;
console.log(sum);
i++
}
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<title>JavaScript Loops</title>
</head>
<body>
<script src="script.js"></script>
</body>
</html>
3 Answers
Patrik Horváth
11,110 Points";" missing after i++ and you dont have compiler in JS so u cant see syntax error i recommended use PHP STORM or WEBSTORM its good for JavaScript witch have biuld in syntax check
AND your code will output only 4 then 4 then 4 then 4 look closer on it :) i mean line "var sum = ej + 2;"
and PS you dont need put META charset UTF-8 why ? cause its default value :)
Carlos Fernández-Tejeda García
4,051 PointsThanks Patrik.
But the solution was easier than this, after having a thought I´ve got that
var i = 0; var ej= 0;
while (i < 12){ ej +=2; i+=1; console.log(ej);
} Cheers
Patrik Horváth
11,110 Points:) yup now ej increment by 2 every time :)
Tsenko Aleksiev
3,819 PointsWhy not try to use "%" - when you devide two numbers, lets say 12 % 6 it gives back wats left of the devision in this case 0, if you try 13 % 6 the result is 1 or 2x6 = 12 and leftover 1. Start counting from 2, each iteration check if the var devided by 2 gives back 0 aka it's even number, than add +1 to the var, and check till the counter becomes 24. I would do it this way