We want to generate all the numbers of three digits where:

• The sum of their digits is equal to 10.

• Their digits are in increasing order (the numbers may have two or more equal contiguous digits)

The numbers that fulfill the two above constraints are: 118, 127, 136, 145, 226, 235, 244, 334

Make a function that receives two arguments:

• The sum of digits value

• The desired number of digits for the numbers

The function should output an array with three values: [1, 2, 3]

1. The total number of possible numbers

2. The minimum number

3. The maximum number

The example given above should be:

HowManyNumbers.FindAll(10, 3) == [8, 118, 334]   // as List<Integer>

If we have only one possible number as a solution, it should output a result like the one below:

HowManyNumbers.FindAll(27, 3) == [1, 999, 999]

If there are no possible numbers, the function should output the empty array.

HowManyNumbers.FindAll(84, 4) == []

The number of solutions climbs up when the number of digits increases.

HowManyNumbers.FindAll(35, 6) == [123, 116999, 566666]

• Number of tests: 112
• Sum of digits value between 20 and 65
• Amount of digits between 2 and 17

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