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My approach is you have the numbers: 102+108+114+...+990+996, that are all divisible by 6.
Thus there is a total of 996=102+(n-1)6 i.e there are in total n=105 numbers between 100-1000 that are divisible by 6.
My problem now is that these numbers include numbers with 0's: 990, and numbers with repetitions: 114.

Can anyone help me on how to solve this?

2 Answers 2

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A number is divisible by $6$ if it's divisible by $2$ and by $3.$ It's divisible by $3$ if the sum of the digits is divisible by $3.$ So we have to pick three different digits from $1,2,3,4,5,6,7,8,9$ and arrange them so that the last digit is even.

The nonzero digits fall into three residue classes:
$1,4,7$ leave a remainder of $1$ when divided by $3$;
$2,5,8$ leave a remainder of $2$ when divided by $3$;
$3,6,9$ leave a remainder of $0i$ when divided by $3.$

In order for the sum of our three digits to be divisible by $3,$ we must either pick them all from the same class, or else one from each class. And of course the final digit has to be $2,4,6,$ or $8.$

How many numbers ending with $2$? The other two digits can be in the same class ($1$ way, namely, $5$ and $8$) or one from each of the other two classes ($3\times3=9$ ways), so there are $10$ ways to choose the other two digits, and either one can come first, so there are $20$ numbers ending with $2$ that satisfy the conditions. Likewise there are $20$ ending with $4,6,$ and $8,$ for a total of $\boxed{80}.$

Or you could just count them by hand.

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A number is divisible by $6$ if it is divisible by both $2$ and $3$.

A number is divisible by $2$ if its unit digit is divisible by $2$.

A number is divisible by $3$ if its digit-sum is divisible by $3$.

For numbers that end with $2$:

  • Choose any pair from $[13,16,19,34,37,46,49,58,67,79]$
  • Arrange this pair in any order

So there is a total of $10\cdot2!=20$ numbers that end with $2$.

For numbers that end with $4$:

  • Choose any pair from $[17,23,26,29,35,38,56,59,68,89]$
  • Arrange this pair in any order

So there is a total of $10\cdot2!=20$ numbers that end with $4$.

For numbers that end with $6$:

  • Choose any pair from $[12,15,18,24,27,39,45,48,57,78]$
  • Arrange this pair in any order

So there is a total of $10\cdot2!=20$ numbers that end with $6$.

For numbers that end with $8$:

  • Choose any pair from $[13,16,19,25,34,37,46,49,67,79]$
  • Arrange this pair in any order

So there is a total of $10\cdot2!=20$ numbers that end with $4$.

Bottom line answer: $20+20+20+20=80$.