javascript how to tell if one number is a multiple of another

You can use the modulus to find the remainder after a division and then if the remainder is equal to zero then it's a multiple.

//x and y are both integers
var remainder = x % y;
if (remainder == 0){
//x is a multiple of y
} else {
//x is not a multiple of y
}

If the numbers your using could be to 2dp, the modulus should still work, if not, multiply both by 100 first then carry out the above check.

This avoids JavaScript precision issues (if your factor y has less than 5 decimal places).

function isMultiple(x, y) {
    return Math.round(x / y) / (1 / y) === x;
}

The above algorithm will fail for very small factors <= 1e-5. The following is a solution that will work for all values that would be <= Number.MAX_SAFE_INTEGER with the decimal point removed, i.e. most values that would have up to 16 digits and all values that would have up to 15 digits, i.e. most non-scientific values.

If you are working with scientific values, use big.js.

function decimalPlaces(x) {
    const str = ''+ +x;
    const dindex = str.indexOf('.');
    const eindex = str.indexOf('e');
    return (
        (dindex < 0 ? 0 : (eindex < 0 ? str.length : eindex) - dindex - 1) +
        (eindex < 0 ? 0 : Math.max(0, -parseInt(str.slice(eindex + 1)))));
}

function isMultiple(x, y) {
    const xplaces = decimalPlaces(x);
    const yplaces = decimalPlaces(y);
    if (xplaces > yplaces) {
        return false;
    }
    const pfactor = Math.pow(10, yplaces);
    return Math.round(pfactor * x) % Math.round(pfactor * y) === 0;
}

[
    [2.03, 0.01],
    [2.03, 0.0123],
    [2.029999999999, 0.01],
    [2.030000000001, 0.01],
    [0.03, 0.01],
    [240, 20],
    [240, 21],
    [1, 1],
    [4, 2],
    [6, 3],
    [6, 4],
    [1.123456, 0.000001]
].forEach(([number, multiple]) => {
    const result = isMultiple(number, multiple);
    console.log(`isMultiple (${number}, ${multiple}) =`, result);
});

In javascript there is the remainder operator (similar to most languages with a c-like syntax).

Let x = length and y = price and z = product of x*y

var remainder = (z % x) / 100;

if (remainder === 0) {
   // z is a multiple of x
}

To get the closest x multiple to your result z you could round the result up (or down) using ceil or floor functions in the Math library.

if (r >= x / 2) {
    a = Math.ceil(z/x)*x;
}
else {
    a = Math.floor(z/x)*x;
}

Then round to two decimal places

Math.round(a / 100) * 100;

Use the % (modulus) operator in Javascript and PHP, which returns the remainder when a is divided by b in a % b. The remainder will be zero when a is a multiple of b.

Ex.

//Javascript
var result = userLength * basePrice;     //Get result
if(result % patternLength){              //Check if there is a remainder
  var remainder = result % patternLength; //Get remainder
  if(remainder >= patternLength / 2)      //If the remainder is larger than half of patternLength, then go up to the next mulitple
    result += patternLength - remainder;
  else                                    //Else - subtract the remainder to go down
    result -= remainder;
}
result = Math.round(result * 100) / 100;  //Round to 2 decimal places