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loop that count to 100 and replace with some text any number that divide by 3 or conatins number 3

function checker (num){
for(i = 1; i <= num; ++i){
if(i % 3 === 0 ||
i === 13 ||
i === 23 ||
i === 33 ||
i === 43 ||
i === 53 ||
i === 63 ||
i === 73 ||
i === 83 ||
i === 93){
console.log('boom')
}
else{
console.log(i)
}
}
}
checker(100)
so i made this , but thats looks ugly , how to do it correct ? like any number that contains 3 replace it with some text. thank you.

You could check the string, too.
if (i % 3 === 0 || i.toString().includes('3')) ...

If your loop only counts to 100, you only need to check the tens and ones place to determine if it's a 3. You can use the JavaScript modulo operator % and the Math.floor() static method:
if (Math.floor(i / 10) === 3 || i % 10 === 3 || i % 3 === 0) // ...
The Math.floor method will return true if dividing the number by 10 and rounding down returns 3 (true for the numbers 30 to 39 inclusive), and the % modulo operator will return true when dividing i by 10 has a remainder of 3 (true for 3, 13, 23, 33, etc.) i % 3 === 0 is from your original function.

Fundamental exercise to check divisibility of a number

I got stuck on this exercise. I don't understand why they need to have the first condition. Can I ignore that part?
The exercise is from https://www.w3resource.com/javascript-exercises/javascript-basic-exercise-46.php:
Write a JavaScript program to check two given non-negative integers that whether one of the number (not both) is multiple of 7 or 11.
function valCheck(a, b) {
if (!((a % 7 == 0 || a % 11 == 0) && (b % 7 == 0 || b % 11 == 0))) {
return ((a % 7 == 0 || a % 11 == 0) || (b % 7 == 0 || b % 11 == 0));
} else
return false;
}
console.log(valCheck(14, 21));
console.log(valCheck(14, 20));
console.log(valCheck(16, 20));

TL;DR – No, it's not redundant!
Write a JavaScript program to check whether exactly one of the two given non-negative integer numbers is a multiple of 7 or 11.
To explain why it's not redundant, let's break this solution function down.
function valCheck(a, b) {
if (!((a % 7 == 0 || a % 11 == 0) && (b % 7 == 0 || b % 11 == 0))) {
return ((a % 7 == 0 || a % 11 == 0) || (b % 7 == 0 || b % 11 == 0));
} else
return false;
}
First Conditional
The first condition is !((a % 7 == 0 || a % 11 == 0) && (b % 7 == 0 || b % 11 == 0)). The x % n == 0 expressions are checking whether the given x is divisible by (a multiple of) either 7 or 11, but the expression is written in an overly verbose way that makes it a little difficult to understand.
Using De Morgan's law, we can rewrite it:
!((a % 7 == 0 || a % 11 == 0) && (b % 7 == 0 || b % 11 == 0))
⇔
!(a % 7 == 0 || a % 11 == 0) || !(b % 7 == 0 || b % 11 == 0)
⇔
(a % 7 != 0 && a % 11 != 0) || (b % 7 != 0 && b % 11 != 0)
Now, we can tell that this condition is checking for the case where either a or b isn't divisible by 7 nor 11.
When this condition is true, we return the result of a second condition. If it is false, then we know neither input is a multiple of 7 or 11, so we return false outright.
Second Conditional
The second condition is:
(a % 7 == 0 || a % 11 == 0) || (b % 7 == 0 || b % 11 == 0)
This one is much less opaque than the first, and checks whether any of the 4 sub-conditions are true (the parenthesis are irrelevant here, since they're all "or" || operators).
Since we already know from the first condition that at least one of the inputs isn't a multiple of 7 or 11, we can now simply return whether at least of the inputs is. Put another way, if we know that this second condition is true, then we can be certain that exactly one of the numbers is a multiple of 7 or 11.
Are the Conditions Redundant?
So to answer your question, no, the first conditional isn't redundant, since it checks for whether at least one of the inputs isn't a multiple or 7 or 11. We need both of the conditional expressions to properly check that exactly one of the inputs is a proper multiple or 7 or 11.
If the "but not both" clause wasn't there, then we could use just the second conditional expression, but since it is, we need both pieces of information.
Wanted to note, I lied a little in saying we need both expressions, since it is indeed possible to solve this exercise with a single expression!
function valCheck(a, b) {
return (a % 7 == 0 || a % 11 == 0) != (b % 7 == 0 || b % 11 == 0);
}
console.log(valCheck(14, 21)); // false
console.log(valCheck(14, 20)); // true
console.log(valCheck(16, 20)); // false
I'll leave it to the reader to figure out why this works!

There are many repeating functions, let's refactor for clarity:
function isMultipleOf7Or11(num) {
return num % 7 == 0 || num % 11 == 0;
}
function valCheck(a, b) {
var isAVarMultiple = isMultipleOf7Or11(a);
var isBVarMultiple = isMultipleOf7Or11(b);
if(!(isAVarMultiple && isBVarMultiple) {
return isAVarMultiple || isBVarMultiple;
else
return false;
}
}
Now to simplify the condition:
!(isAVarMultiple && isBVarMultiple) is also equal to !isAVarMultiple || !isBVarMultiple according to De Morgan's negation of a conjuction, which returns true if at least one of them is not a multiple of 7 or 11
and in the return value the condition isAVarMultiple || isBVarMultiple returns true if at least one of them is a multiple of 7 or 11
We can conclude that it is a XOR operation on two values, which would return true only if one and only one of the variables is a multiple of 7 or 11
Outputs:
console.log(valCheck(14, 21));
returns false
console.log(valCheck(14, 20));
returns true
console.log(valCheck(16, 20));
returns false

I recommend you look at the code by the debugger, not for this example only, but when you will be stuck in difficult situations, you can split your code in a small and readable pieces and see what is happening in there.
For example this can be transformed to this
Not Sure If I can make this more readable :)
function valCheck (a, b) {
debugger;
const canADividedTo7 = a % 7 == 0;
const canADividedTo11 = a % 11 == 0;
const canADividedTo7Or11 = canADividedTo7 || canADividedTo11;
const canBDividedTo7 = b % 7 == 0;
const canBDividedTo11 = b % 11 == 0;
const canBDividedTo7Or11 = canBDividedTo7 || canBDividedTo11;
// true when both both side of expression are true
const canAAndBDividedTo7Or11 = canADividedTo7Or11 && canBDividedTo7Or11;
// true when at least one of the expression is true
const canAOrBDividedTo7Or11 = canADividedTo7Or11 || canBDividedTo7Or11;
//
if (!canAAndBDividedTo7Or11) {
return canAOrBDividedTo7Or11;
} else {
return false;
}
}
console.log(valCheck(14, 21));
console.log(valCheck(14, 20));
console.log(valCheck(16, 20));

Who would write anything that inane and complicated?
Let's take this mess:
function valCheck(a, b) {
if (!((a % 7 == 0 || a % 11 == 0) && (b % 7 == 0 || b % 11 == 0))) {
return ((a % 7 == 0 || a % 11 == 0) || (b % 7 == 0 || b % 11 == 0));
} else
return false;
}
and refactor it into something readable:
function valChec( a, b ) {
const aIsDivisible = a % 7 === 0 || a % 11 === 0 ;
const bIsDivisible = b % 7 === 0 || b % 11 === 0 ;
return aIsDivisible ? !bIsDivisible : bIsDivisible;
}
You could even make it a one-liner with a ternary expression (at the expense of a few more division ops... but given the original code, that's not an issue):
const isMagic = x => x % 7 === 0 || x % 11 === 0;
const valCheck = (a,b) => isMagic(a) ? !isMagic(b) : isMagic(b);

Yes, you're right in finding the answer to that question redundant; also, I believe the ! is incorrect and makes the code always return false.
A more reasonable solution to the problem would be to just return the boolean expression:
function valCheck(a, b) {
return ((a % 7 == 0 || a % 11 == 0) || (b % 7 == 0 || b % 11 == 0));
}
Alternatively, if you really wanted to use an if/else construct, it would be better to just return true in the first case:
function valCheck(a, b) {
if ((a % 7 == 0 || a % 11 == 0) && (b % 7 == 0 || b % 11 == 0))
return true;
else
return false;
}
Hope this helps with your understanding of JavaScript! But as suggested in the comments, you might want to learn JavaScript from other sources.

Conditions with assigned variables vs inline condition check

Can't find a better Title for this question so edit suggestions will be much appreciated.
I'm wondering if there's a difference between checking condition with assigned variables and inline condition.
for example:
Option 1:
// inline conditions check
function isSomething(){
return (1 > 2 || 'a' == 'a' || 2 < 4) ||
(55 == 1 || (32 > 4 || 'a' == 'a') && 6 > 2) ||
('abc' != 'bca' && 3 == 3);
}
Option 2:
// pre assigned variables condition check
function isSomething(){
const conditionA = 1 > 2 || 'a' == 'a' || 2 < 4; // some complex condition
const conditionB = 55 == 1 || (32 > 4 || 'a' == 'a') && 6 > 2; // some complex condition
const conditionC = 'abc' != 'bca' && 3 == 3 // some complex condition
const result = conditionA || conditionB || conditionC;
return result;
}
It seems that in option 2 it must check all 3 conditions but in option 1 theoretically it can return after the first check if it's true.
Obviously option 2 is my choice as it's much more readable, though i was wondering is there a difference in behavior or performance?
Is there a way to test performance between the two options?

If you want to combine the benefits of short-circuited evaluation with readability and named variables, then
function isSomething(){
const conditionA = () => 1 > 2 || 'a' == 'a' || 2 < 4;
const conditionB = () => 55 == 1 || (32 > 4 || 'a' == 'a') && 6 > 2;
const conditionC = () => 'abc' != 'bca' && 3 == 3;
const result = conditionA() || conditionB() || conditionC();
return result;
}

As far as checking the performance I would look at jsperf.
Also please take a look at console.time(), console.profile(), and performance.now() if you haven't already.
In option 2 you are creating 3 new objects and assigning them to variables, creating object and allocating them in memory, the effect that this will have on performance would tend to be insignificant.
In option 1 if the first value is true the second option will not be evaluated as || is a short circuit operator whilst in the second option all three conditions will be evaluated regardless of the result that they return.
If performance is an issue as this method is being used multiple times I would always suggest that the performance test simulates the real world application as much as possible.

Logical operator || returns number instead of boolean value

I attached an example with two if conditions. The first if condition works as expected. The second if condition returns 11, but why? I know that the second if condition is wrong, but I would like to understand why Javascript returns in that case 11.
function exception(number) {
// if(number === 10 || number === 11) { // Working as expected
if(number === 10 || 11) { // Why 11?
console.log(number);
}
}
function loop(f) {
for (i = 0; i <= 100; i++) {
f(i);
}
}
loop(exception);

from this question.
(expr1 || expr2)
"Returns expr1 if it can be converted to true; otherwise, returns
expr2."
source
So when expr1 is (or evaluates to) one of these 0,"",false,null,undefined,NaN, then expr2 is returned, otherwise expr1 is returned

Some Information about what you where trying to achieve:
number === 10 || number === 11 is the same as (number === 10) || (number === 11)
number === 10 || 11 is the same as (number === 10) || (11) it does not compare 11 to number here
Now let's have a closer look atnumber === 10 || 11 :
number === 10 will be true if number is of type number and equal to 10
if the first was false, it will evaluate the boolean value of the next statement: 11 (wich is accepted as true, for beeing a number not equal to 0)

because Boolean(11) is true (try on your console)
so even if first condition is not true (if the number is not 10), then second condition will be true always

A quick way to test equality of more than 2 values at once?

I was wondering if there was a quick way to test the equality of more than two values in js. Something similar to (= 6 6 6).
In the console, I tried things like...
1 == 1 == 1 == 1
true
2 == 2 == 2 == 2
false
0 == 0 == 0
false
0 == 0 == 0 == 0
true
...which was amusing, but also puzzling.
Is there a quick way of doing this in js?
Thanks.

The reason you got unexpected behavior is because we need to adjust your expectations in js a bit ;) 2 == 2 == 2 == 2 does 3 comparisons, all from left to right. The first comparison is the leftmost 2 == 2, which evaluates to true. After that we get the result of the first comparison being compared to (what is in this case) the 3rd 2. Ie, true === 2, which is false. And finally, we get false === 2, which is also false.
It might help to visualize it as such:
(((2 == 2) == 2) == 2)
I think in general a === b && b === c might be what you're looking for.
EDIT: Ah, and sorry I keep switching out the == for ===. It's just habit. And it's a habit I'd recommend. the === operator doesn't do type casting, so it evaluates the value proper, not a casted version of the value.

It's because true == 1 but true != 2
You can try:
function isEquals() {
var flag = true;
for(var i=1; i<arguments.length; i++) flag = flag && (arguments[i] == arguments[0]);
return flag;
}
isEquals(2,2,2); // true
or:
function isEquals() {
var ar = arguments;
return Array.prototype.every.call(arguments, function(a){return a==ar[0];});
}

Yes you can, but you need to use the "Logical Operators" like the && or || to check more than 1 statement like (x<1 && y>0).
You can use this as a quick easy reference:
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Logical_Operators

If you have more than three values, it might be more convenient to create a function for use on an array:
function allEqual(arr) {
return arr.every(function (x, i) {
return i === 0 || x === arr[i - 1];
});
}
allEqual([1, 1, 1])
ES6:
function allEqual(...arr) {
return arr.every((x, i) => i === 0 || x === arr[i - 1]);
}
allEqual(1, 1, 1)

As an addition to #vp_arth's answer you could even add a method to the Array prototype
Array.prototype.isHomogeneous = function(){
return Array.prototype.every.call(this, function(c,i,a){ return c === a[0];})
}
So you could do
[1,2,3].isHomogeneous() = false
[1,1,1].isHomogeneous() = true