A-Rule-of-Divisibility-by-7.js

/* A number m of the form 10x + y is divisible by 7 if and only if x − 2y is divisible by 7. In other words, subtract twice the last digit from the number formed by the remaining digits. Continue to do this until a number known to be divisible by 7 is obtained; you can stop when this number has at most 2 digits because you are supposed to know if a number of at most 2 digits is divisible by 7 or not.

The original number is divisible by 7 if and only if the last number obtained using this procedure is divisible by 7.

Examples:
1 - m = 371 -> 37 − (2×1) -> 37 − 2 = 35 ; thus, since 35 is divisible by 7, 371 is divisible by 7.

The number of steps to get the result is 1.

2 - m = 1603 -> 160 - (2 x 3) -> 154 -> 15 - 8 = 7 and 7 is divisible by 7.

3 - m = 372 -> 37 − (2×2) -> 37 − 4 = 33 ; thus, since 33 is not divisible by 7, 372 is not divisible by 7.

4 - m = 477557101->47755708->4775554->477547->47740->4774->469->28 and 28 is divisible by 7, so is 477557101. The number of steps is 7.

Task:
Your task is to return to the function seven(m) (m integer >= 0) an array (or a pair, depending on the language) of numbers, the first being the last number m with at most 2 digits obtained by your function (this last m will be divisible or not by 7), the second one being the number of steps to get the result.

Forth Note:
Return on the stack number-of-steps, last-number-m-with-at-most-2-digits 

Examples:
seven(371) should return [35, 1]
seven(1603) should return [7, 2]
seven(477557101) should return [28, 7]

*/

// solution

function seven(m) {
    var steps = 0;
    while (m > 99){
      m = Math.floor(m / 10) - 2 * (m % 10);
      ++steps;
    };
    return [m, steps];
}