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![Hina K.](https://secure.gravatar.com/avatar/4c25215ab300f04b5fde84f5ad49e132?s=96&d=https%3A%2F%2Fecs-static.teamtreehouse.com%2Fassets%2Fcontent%2Fdefault_avatar-445fbbabfc8dc9188fb5967fe43322ee0c3e0dd1e10f378bf8343784af5a13eb.webp&r=pg)
Hina K.
4,806 PointsMy solution (different from the tutorial). Does this work? I used the '&&' operator.
const random = (lowNum, highNum) => {
if (!isNaN(lowNum) && !isNaN(highNum)) {
return Math.floor((Math.random() * (highNum - lowNum + 1) + lowNum));
} else {
console.log("Enter two numbers and try again.");
}
}
** Call functions I used **
console.log(random(10, 30));
console.log(random(20, 40));
console.log(random('a', 40));
console.log(random(0, 2));
1 Answer
![Steven Parker](https://uploads.teamtreehouse.com/production/profile-photos/969492/micro_tool_chess_set_2.jpg)
Steven Parker
230,688 PointsInverting the logic to test for a good condition instead of a bad one is perfectly legitimate from a functional perspective, and a good example of how there's rarely just one solution when programming. Good job!