The divisibility by 11 rule says 


    An integer number is divisible by  11  if and only if the alternate sum of its digits is divisible by  11.

    See my lesson  Divisibility by 11 rule  in this site.



So, in order for the 3-digit number 41x, where x is a missed digit (or a "hidden" digit),  was divisible by 11, the alternate sum 

    4 - 1 + x

must be multiple of 11.


It is clear that we must (and can) to consider the only case when 

   4 - 1 + x = 11,

which gives  x = 11 - 4 + 1 = 8.


Indeed, the number 418 is a multiple of 11:  418 = 11*38.

Solve it by the same way.


The equation  3 - y + 5 = 0  gives  y = 3 + 5 = 8.


Indeed, the number 385 is a multiple of 11:  385 = 11*35.


Here I used  0  as a unique appropriate multiple of 11.

O-o-o, I finally got that z is the "tens" digit in this 4-digit number.


OK,  then we need to apply the divisibility rules for 3 and 2.

The "divisibility by 2 rule" is just satisfied, since the last digit of the number ("ones" digit) is even.


The "divisibility by 3 rule" requires the sum of the digits is multiple of 3:

    4 + 1 + z + 2  is divisible by 3

or, which is the same,  7+z must be divisible by 3.

So, the sum 7+z must be 9, or 12, or 15, which gives the possibilities for z to be equal 2, 5, 8.


Let us check three numbers  4122, 4152 and 4182 for divisibility by 6.


The answer is:  The numbers  4122,  4152 and  4182  all are multiples of 6.