Let us take a breather by tackling a problem simple enough that its solution needs only a couple of conditional statements and some arithmetic, but not even one loop or anything more fancy. The difficulty is coming up with the conditions that cover all possible cases of this problem just right, including all of the potentially tricksy edge and corner cases, without being off-by-one anywhere.

Given three integers start, end, n, so that start <= end, count how many integers between start and end, inclusive, are evenly divisible by n. Note that either start or end can well be negative or zero, but n is guaranteed to be greater than zero. Here is the scheme for 1, 9, 3 input.

Sure, this problem could be solved as a one-liner with the lazy comprehension. Try to have code with no loops at all, but use only integer arithmetic and conditional statements to root out the truth.

Input: Three integers (number).

Output: Integer (number).

Examples:

assert.strictEqual(countDivisibles(7, 28, 4), 6);
assert.strictEqual(countDivisibles(-77, 19, 10), 9);
assert.strictEqual(countDivisibles(1, 999999999999, 5), 199999999999);

Preconditions:

• start <= end;
• n > 0.

The mission was taken from Python CCPS 109. It is taught for Ryerson Chang School of Continuing Education by Ilkka Kokkarinen