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I want to filter only odd negative numbers from an array. Why does n % 2 === 1 not work, but n % 2 !== 0 does?

Write a function that returns only negative odd numbers from an array.
const arr = [4, -7, -6]
I first tried:
let negativeOdd = arr.filter(n => n % 2 === 1 && n < 0);
return negativeOdd;
result was an empty array. []. The answer should be [-5].
But when I replaced n % 2 === 1 with n % 2 !== 0, it workded. I am new to JS and was hopeing somenone could help me understand why this is happening. Thank you.

The modulo % operator in Javascript divides a number by a divisor (in this case 2), and returns the remainder.
-5 divided by 2 is -2.5, or -2 with a remainder of -1. 2 * -2 + -1 = -5
5 divided by 2 is 2.5, or 2 with a remainder of 1. 2 * 2 + 1 = 5
console.log(-5 % 2);
console.log(5 % 2);

A negative odd number gives
n % 2 === -1.
Just try it out. In the console, type
-5 % 2

There are different possible definitions of the modulus with respect to negative numbers. For some implementations, it's always true that 0 <= |a % b| < |b|; this is tantamount to always using the floor() function to find the integer quotient before computing the remainder. There are many languages with this interpretation, including Common Lisp and Python. You can get a negative value out of the % operator in these languages, but only if the divisor is negative; the sign of the dividend doesn't change the sign of the result.
In other languages, -|b| < |a % b| < |b|, and the sign of the result does depend on the sign of the dividend. This is equivalent to obtaining the integer quotient by rounding toward zero before taking the remainder; C is in this category (since C99; previous editions of the spec left it up to the implementation). So is Javascript, which is why -1 % 2 is not 1, but -1.
The first definition is often more useful, but easy to define in terms of the second:
const mod = (a,b) => (a % b + b) % b // mod(-1,2) is 1

reduction of complexity numbers procedure

I'm looking for a solution to my problem
if I have numbers
var first = 14:1
var next = 13:8
therefore, the console should give a result
console.log(first_result) // 141
console.log(next_result) // 141
and I want the numbers counted like 141 in the result to be
simply if there is an example
13:8 if both digits are the last, 3 and 8 are bigger than ten, then turn the tens and turn and insert into the previous number and leave the rest at the end
so 13:8 becomes 141

If you are starting with strings, then you just simply split the string on : to get your 2 numbers.
To get the last digit, you can simply use x % 10. Then just add the numbers and see what happens.
let value = '13:8',
nums = value.split(':').map(Number),
last_digits = nums.map(x => x % 10),
// or last_digits.reduce((x,y) => x+y, 0)
sum = last_digits[0] + last_digits[1],
result;
if (sum > 10) {
nums[0]++;
nums[1] = sum - 10; // or, probably sum % 10
}
result = nums.join('');
console.log(result);

How do i return a number excluding the last digit?

So I have a number like 5467. I want my code to return 546.
I tried taking the last number and subtracting it from the original number but I get 5460 instead of 546.

Combine / with %:
(5467 - (5467 % 10)) / 10
564

Sounds like you also need to divide my 10. You could do something like this:
var number = 5467;
number = number - (number % 10); // This will subtract off the last digit.
number = number / 10;
console.log(number); // 546
We first use the modulo operator % to get the last digit, and we subtract it from number. That reduces the number from 5467 to 5460. Now to chop off the last digit (which is guaranteed to be a 0) we divide by 10 and get 546.
Written more concisely you could do:
number = (number - ( number % 10)) / 10;

There's a few things you can do the most concise being:
Math.floor(num / 10);
Or, convert to a string, remove the last character and convert back to number.
parseInt(num.toString().slice(0, -1));

If string representation would be fine for you then one other way is
var num = 5467,
cut = (num/10).toFixed(); // <-'547'
Well... warning..! i have to say toFixed() method rounds if necessary. So in this particular example it doesn't work.

I dont mind some of the other answers, but i feel that this maybe too fixed on it being a number.
Which it is, but you want to remove the last digit/char, regardless of the number, so why not substr?
http://www.w3schools.com/jsref/jsref_substr.asp
var s = 5467;
s = s.toString().substr(0, s.toString().length - 1);
console.log(s)
or even easier:
var s = (5467).toString();
s = s.substr(0, s.length - 1);
console.log(s)
These dont take into account single digit numbers, so passing in 1 would return blank. To answer that you could simply do a check like:
var s = (1).toString();
if(s.length > 1)
s = s.substr(0, s.length - 1);
console.log(s)
Also, similar question to:
Remove last digits from an int
Remove the last digits of a number (not string)
Removing the last digits in string

To truncate digits from the right hand side until the number is less than 30, keep dividing by 10 and rounding down until a suitable value is reached:
var n = 12341235;
while (n > 30) n = n/10|0;
document.write(n);
The greater than and division operations will coerce n to a number, so it can be a number or string. If ToNumber(n) results in NaN (e.g. n = 'foo'), then the value of n is not modified.

You can simply divide the number by 10 and parseInt()
var num = 5467;
num = parseInt(num/10);
Update :
To repeat the process until the answer is less than 30, use while loop as
var num = 5467;
while(num >= 30) {
num = parseInt(num/10);
}
document.write(num);

How to check the sum of a divison with javascript?

I want to check For EXAMPLE:
12 - 13 - 14 - 15
12 / 3 = 4 -> OK.
13 / 3 = 4.33 -> NOT OK.
14 / 3 = 4.67 -> NOT OK.
15 / 3 = 5 -> OK.
I want to create a while loop where i use the number i want to divide to 3 as "x".
so:
var x = 0
while (x<20) {
//SOMETHING HERE
x++;
}
Something like that but i dont know the command to check if the sum is a decimal number or not.
Than i want to see the "OK. numbers" in my web browser with a documetn.write( OK numbers )

You can use the modulo operator to check if a number is divisible by another.
Demo
var x = 0;
while (x < 20) {
if (x % 3 === 0) {
document.write(x + ' ');
}
x++;
}
Also, the W3C recommends against using document.write now. Instead it is better to use document.createElement to create an element and insert it that way, like this:
Demo
var span = document.createElement('span');
span.innerHTML = text;
document.body.appendChild(span);

Need an explanation of this javascript

I have a question about this script I found and used. It works but I don't get why. The exercise was to make a list with random numbers from -50 to 50. The function below uses Math.floor(Math.random() * (the part i dont understand).
If I put this calculation on google I got as answer 151 and Math.random()*151 does not do from -50 to 50.
Can someone give me a clear explanation about this function below because I am sure that I am missing something.
this script works but I only want a clear explanation how
for (i = 0; i <= 100; i++)
{
Rnumber[i] = randomFromTo(-50,50);
}
function randomFromTo(from, to)
{
return Math.floor(Math.random() * (to - from + 1) + from);
}

to - from + 1 = 50 - (-50) + 1 = 101
Math.random() * 101 = number in range [0,101[
Math.floor([0,101[) = integer in range [0,100]
[0,100] + from = [0,100] + (-50) = integer in range [-50,50]
Which is exactly what is asked for.

https://developer.mozilla.org/en/JavaScript/Reference/Global_Objects/Math/random
Math.random returns a floating-point, pseudo-random number in the
range [0, 1) that is, from 0 (inclusive) up to but not including 1
(exclusive), which you can then scale to your desired range.
which when multiplied with a number > 1 and floored gives you an integer

Math.random() - get only value between 0 and 1.
Math.floor( number ) get integer down rounded value from number.
You should:
function randomFromTo(from, to)
{
// you can use doubled bitwise NOT operator which also as Math.floor get integer value from number but is much faster.
// ~1 == -2 , ~-2 == 1 and ~1.5 == -2 :)
return ~~( --from + ( Math.random() * ( ++to - from )) )
}