Assuming that x is a positive integer, how would I return the first n multiples of the number x?

Here is what I have so far:

function multiples(x, n){
  var arr=[];
  for (var i=1; i<=x; ++i)
    arr.push(n*i);
  return arr;
}


console.log(
  multiples(2, 5)
  );

Assuming that x is a positive integer, how would I return the first n multiples of the number x?

Here is what I have so far:

function multiples(x, n){
  var arr=[];
  for (var i=1; i<=x; ++i)
    arr.push(n*i);
  return arr;
}


console.log(
  multiples(2, 5)
  );

What I want it to return is this: multiples(2, 5) // [2, 4, 6, 8, 10]

But what it actually returns is this: [5, 10]

• javascript

Share Improve this question Follow asked Dec 30, 2019 at 16:48 Cody WirthCody Wirth 7111 silver badge66 bronze badges 2

• 3 You mistakenly swapped x and n. What you want is multiples(5, 2). – ASDFGerte Commented Dec 30, 2019 at 16:50
• 1 That's when you know it is time to take a break, thank you! – Cody Wirth Commented Dec 30, 2019 at 16:51

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4 Answers 4

Sorted by: Reset to default 3

You switched the x and the n in the for loop

//changed var to const & let

function multiples(x, n) {
  const arr = [];
  for (let i = 1; i <= n; i++)
    arr.push(x * i);
  return arr;
}

console.log(multiples(2, 5));

Using ES6 spread operator you can do like this:

function multiples(x, n) {
  return [...Array(n)].map((_, i) => x * ++i);
}

console.log(multiples(2, 5))

Using ES6 Array.from(...) you can do like this:

function multiples(x, n) {
  return Array.from(Array(n)).map((_, i) => x * ++i);
}

console.log(multiples(2, 5))

x swap n

function multiples(x, n){
  var arr=[];
  for (var i=1; i<=n; ++i)
    arr.push(x*i);
  return arr;
}


console.log(
  multiples(2, 5)
);

You had the right idea, but swapped x and n in your loop:

for (var i = 1; i <= n; ++i)
    arr.push(x * i);

const multiples = (x, n) => Array(n).fill().map((_, i) => x * ++i);

console.log(multiples(2, 5));