# In a range of numbers, count the numbers divisible by a specific integer

## Algorithm goal

In a range \(5\) to \(15\), there are \(4\) numbers divisible by \(n = 3\): \(6\), \(9\), \(12\) and \(15\). Compute this for a generic \(n\) (all non-negative numbers)

In Codility, there is a similar problem 'CountDiv'.

## Explanation

A brute-force method will have a complexity that depends strictlyon the size of the input number.

Notice however that the count of numbers divisible by\(3\), up to \(15\), is \(5\) (\(3, 6, 9, 12, 15\)). If we do that with respect to a range, we need to just take the difference between counts up to the end of the range, and the counts up to the start of the range (excluding the number itself). (this is © from www.scala-algorithms.com)

Also if the number \(0\) is in range, count is increased by 1 because \(0\) is divisible by all numbers (except \(0\), of course).

## Scala Concepts & Hints

- Range
The

`(1 to n)`

syntax produces a "Range" which is a representation of a sequence of numbers.`assert((1 to 5).toString == "Range 1 to 5") assert((1 to 5).reverse.toString() == "Range 5 to 1 by -1") assert((1 to 5).toList == List(1, 2, 3, 4, 5))`

## Algorithm in Scala

13 lines of Scala (version 2.13), showing how concise Scala can be!

## Test cases in Scala

```
assert(countDivisibles(5 to 15, divisor = 3) == 4)
assert(countDivisibles(0 to Int.MaxValue, divisor = Int.MaxValue) == 2)
assert(countDivisibles(0 to 0, divisor = 3) == 1)
assert(countDivisibles(1 to Int.MaxValue, divisor = Int.MaxValue / 2) == 2)
```

```
def countDivisibles(range: Range, divisor: Int): Int = ???
```