---
title: Sum All Odd Fibonacci Numbers
---
 Remember to use **`Read-Search-Ask`** if you get stuck. Try to pair program  and write your own code 
###  Problem Explanation:
You will need to gather all the **Fibonacci** numbers and then check for the odd ones. Once you get the odd ones then you will add them all. The last number should be the number given as a parameter if it actually happens to be an off Fibonacci number.
#### Relevant Links
* Fibonacci number
##  Hint: 1
To get the next number of the series, you need to add the current one to the previous and that will give you the next one.
> _try to solve the problem now_
##  Hint: 2
To check if a number is even all you have to check is if `number % 2 == 0`.
> _try to solve the problem now_
##  Hint: 3
As you get the next odd one, don't forget to add it to a global variable that can be returned at the end. `result += currNumber;` will do the trick.
> _try to solve the problem now_
## Spoiler Alert!
**Solution ahead!**
##  Basic Code Solution:
function sumFibs(num) {
var prevNumber = 0;
var currNumber = 1;
var result = 0;
while (currNumber <= num) {
if (currNumber % 2 !== 0) {
result += currNumber;
}
currNumber += prevNumber;
prevNumber = currNumber - prevNumber;
}
return result;
}
// test here
sumFibs(4);
 Run Code
### Code Explanation:
* Create a variable to keep record of the current and previous numbers along with the result that will be returned.
* Use a while loop to make sure we do not go over the number given as parameter.
* We use the modulo operand to check if the current number is odd or even. If it is even, add it to the result.
* Complete the Fibonacci circle by rotating getting the next number and swapping values after.
* Return the result.
#### Relevant Links
* JS while Loop
##  Intermediate Code Solution:
function sumFibs(num) {
// Perform checks for the validity of the input
if (num < 0) return -1;
if (num === 0 || num === 1) return 1;
// Create an array of fib numbers till num
const arrFib = [1, 1];
let nextFib = 0;
// We put the new Fibonacci numbers to the front so we
// don't need to calculate the length of the array on each
// iteration
while((nextFib = arrFib[0] + arrFib[1]) <= num) {
arrFib.unshift(nextFib);
}
// Sum only the odd numbers and return the value
// First, reverse the array to avoid starting acc with the first/greater number when it's even
return arrFib.reverse().reduce((acc, curr) => {
return acc + curr * (curr % 2);
});
}
// test here
sumFibs(4);
 Run Code
### Code Explanation:
* Create an array of fibonacci numbers till **num**.
* Use `reduce()` method to find the sum of odd members of array.
* Return the sum.
#### Relevant Links
* JS Array Prototype Push
* JS For Loops Explained
* JS Array Prototype Reduce
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