How to round up number to nearest 100/1000 depending on number, in JavaScript?

Question

I have a number that can be in the 2 digits, like 67, 24, 82, or in the 3 digits, like 556, 955, 865, or 4 digits and so on. How can I round up the number to the nearest n+1 digits depending on the number?

Example:

roundup(87) => 100,
roundup(776) => 1000,
roudnup(2333) => 10000

and so on.

Answer 1

You could take the logarithm of ten and round up the value for getting the value.

function roundup(v) {
    return Math.pow(10, Math.ceil(Math.log10(v)));
}

console.log(roundup(87));   //   100
console.log(roundup(776));  //  1000
console.log(roundup(2333)); // 10000

For negative numbers, you might save the sign by taking the result of the check as factor or take a negative one. Then an absolute value is necessary, because logarithm works only with positive numbers.

function roundup(v) {
    return (v >= 0 || -1) * Math.pow(10, 1 + Math.floor(Math.log10(Math.abs(v))));
}

console.log(roundup(87));    //    100
console.log(roundup(-87));   //   -100
console.log(roundup(776));   //   1000
console.log(roundup(-776));  //  -1000
console.log(roundup(2333));  //  10000
console.log(roundup(-2333)); // -10000

Answer 2

 const roundup = n => 10 ** ("" + n).length

Just use the number of characters.

Answer 3

You can check how many digits are in the number and use exponentiation:

const roundup = num => 10 ** String(num).length;
console.log(roundup(87));
console.log(roundup(776));
console.log(roundup(2333));

Answer 4

You can use String#repeat combined with Number#toString in order to achieve that :

const roundUp = number => +('1'+'0'.repeat(number.toString().length));

console.log(roundUp(30));
console.log(roundUp(300));
console.log(roundUp(3000));

Answer 5

//Math.pow(10,(value+"").length)   


console.log(Math.pow(10,(525+"").length))
console.log(Math.pow(10,(5255+"").length))

I came up with another solution that does'nt require a new function to be created