Find element that appears odd number of times

Question

I'm trying to solve this exercise of finding the number that appears an odd number of times in an array. I have this so far but the output ends up being an integer that appears an even number of times. For example, the number 2 appears 3 times and the number 4 appears 6 times, but the output is 4 because it counts it as appearing 5 times. How can it be that it returns the first set that it finds as odd? Any help is appreciated!

         function oddInt(array) {
         var count = 0;
         var element = 0;
         for(var i = 0; i < array.length; i++) {
           var tempInt = array[i];
           var tempCount = 0;
             for(var j = 0; j <array.length; j++) {
                if(array[j]===tempInt) {
                tempCount++;
                  if(tempCount % 2 !== 0 && tempCount > count) {
                  count = tempCount; 
                  element = array[j];
                }
               }
              }
             }
           return element;
           }
           oddInt([1,2,2,2,4,4,4,4,4,4,5,5]);

Answer 1

function findOdd(numbers) {
  var count = 0;
  for(var i = 0; i<numbers.length; i++){
    for(var j = 0; j<numbers.length; j++){
      if(numbers[i] == numbers[j]){
        count++;
      }
    }
    if(count % 2 != 0 ){
      return numbers[i];
    }
  }
};

console.log(findOdd([20,1,-1,2,-2,3,3,5,5,1,2,4,20,4,-1,-2,5])); //5
console.log(findOdd([1,1,1,1,1,1,10,1,1,1,1])); //10