# Need Help with a Simple Javascript Problem Involving Even and Odd Checking

• Hi Everyone,

I was wondering if someone can help me with creating a Javascript function for the following problem :

• There are x number of bananas
• There are are 3 monkeys
• If the number of bananas is EVEN then monkey one and two get ALL the bananas.
• If the number of bananas is ODD, monkey three gets the extra banana.

Thank you.

• Hi there.

When you think of checking for even or odd think the mod (%) operator.

Here is one way to do it for all values greater than 0 :

``````function bananas(x){
let monkey_1 = 0;
let monkey_2 = 0;
let monkey_3 = 0;
if(x == 1){
monkey_3 = 1;
}
else if(x % 2 == 0){
// x is EVEN
monkey_1 = x / 2;
monkey_2 = x / 2;
} else {
//X is ODD
monkey_1 = (x-1) / 2;
monkey_2 = (x-1) / 2;
monkey_3 = 1;
}
console.log("For " + x + " Bananas");
console.log("Monkey 1 gets: " + monkey_1 + " Bananas");
console.log("Monkey 2 gets: " + monkey_2 + " Bananas");
console.log("Monkey 3 gets: " + monkey_3 + " Bananas");
console.log("-------------------------------");
}
``````

Please let me know if this worked for you by selecting it as the correct answer. Thank you.

• Hi there.

When you think of checking for even or odd think the mod (%) operator.

Here is one way to do it for all values greater than 0 :

``````function bananas(x){
let monkey_1 = 0;
let monkey_2 = 0;
let monkey_3 = 0;
if(x == 1){
monkey_3 = 1;
}
else if(x % 2 == 0){
// x is EVEN
monkey_1 = x / 2;
monkey_2 = x / 2;
} else {
//X is ODD
monkey_1 = (x-1) / 2;
monkey_2 = (x-1) / 2;
monkey_3 = 1;
}
console.log("For " + x + " Bananas");
console.log("Monkey 1 gets: " + monkey_1 + " Bananas");
console.log("Monkey 2 gets: " + monkey_2 + " Bananas");
console.log("Monkey 3 gets: " + monkey_3 + " Bananas");
console.log("-------------------------------");
}
``````

Please let me know if this worked for you by selecting it as the correct answer. Thank you.

• It is amazing that Avan finds the time to reply to guys like us! Kudos!