How to divide number into integer pieces that are each a multiple of n?

Question

Had a hard time coming up with a concise title for this. I'm sure there are terms for what I want to accomplish and there is no doubt a common algorithm to accomplish what I'm after - I just don't know about them yet.

I need to break up a number into n pieces that are each a multiple of 50. The number is itself a multiple of 50. Here is an example: Divide 5,000 by 3 and end up with three numbers that are each multiples of 50:

• 1,650
• 1,700
• 1,650

I also would like to have the numbers distributed so that they flip back and forth, here is an example with more numbers to illustrate this: Divide 5,000 by 7 and end up with 7 numbers that are each multiples of 50:

• 700
• 750
• 700
• 750
• 700
• 700
• 700

Note that in the above example I'm not worried that the extra 50 is not centered in the series, that is I don't need to have something like this:

• 700
• 700
• 750 <--- note the '50s' are centered
• 700
• 750 <--- note the '50s' are centered
• 700
• 700

Hopefully I've asked this clearly enough that you understand what I want to accomplish.

Update: Here is the function I'll be using.

var number = 5000;
var n = 7;
var multiple = 50;

var values = getIntDividedIntoMultiple(number, n, multiple)

function getIntDividedIntoMultiple(dividend, divisor, multiple)
{
    var values = [];
    while (dividend> 0 && divisor > 0)
    {
        var a = Math.round(dividend/ divisor / multiple) * multiple;
        dividend -= a;
        divisor--;
        values.push(a);
    }

    return values;
}

Answer 1

var number = 5000;
var n = 7;

var values = [];
while (number > 0 && n > 0) {
    var a = Math.floor(number / n / 50) * 50;
    number -= a;
    n--;
    values.push(a);
}  // 700 700 700 700 700 750 750

Edit

You can alternate Math.floor and Math.ceil to obtain the desired result:

while (number > 0 && n > 0) {
    if (a%2 == 0)
        a = Math.floor(number / n / 50) * 50;
    else
        a = Math.ceil(number / n / 50) * 50;
    number -= a;
    n--;
    values.push(a);
}  // 700 750 700 750 700 700 700

Answer 2

// i - an integer multiple of k
// k - an integer
// n - a valid array length
// returns an array of length n containing integer multiples of k
// such that the elements sum to i and the array is sorted,
// contains the minimum number of unique elements necessary to
// satisfy the first condition, the elements chosen are the
// closest together that satisfy the first condition.
function f(i, k, n) {
  var minNumber = (((i / k) / n) | 0) * k;
  var maxNumber = minNumber + k;
  var numMax = (i - (minNumber * n)) / k;
  var nums = [];
  for (var i = 0; i < n - numMax; ++i) {
    nums[i] = minNumber;
  }
  for (var i = n - numMax; i < n; ++i) {
    nums[i] = maxNumber;
  }
  return nums;
}

So your second example would be

f(5000, 50, 7)

which yields

[700,700,700,700,700,750,750]