The "Even" Dilemma

If one of three consecutive numbers ends with 0 then the result will not be 2 at the last digit.

So, we have $2\times4\times6$ which result will end with 8 and $4\times6\times8$ which result will end with 2.

Three numbers must end with 4,6,8 consecutively.

Now look at this, $50^3=125000$ $60^3=216000$ $70^3=343000$ We can assume that the three numbers must be $54\times56\times58$ or $64\times66\times68$

As the solution, only answer is $64\times66\times68 = \boxed{287232}$

We have ${58}^{3} = 195 112$ . Which means that the consecutive even numbers should be close to 58.

Now we need the condition that the last digit is 2.

And I got $64 \times 66 \times 68 = 287 232$ .

I know the method understated is a bit vague but it is the way I did it.

Since the required number is a 6-digit number, it is bound to be a product of three 2-digit even numbers.

Let's say the first number is n. Then the corresponding numbers are (n+2) and (n+4).

On careful scrutiny, we find that if n>60, we'll get closer to our answer. In the end, we get 64.66.68 = 287232.

I am a novice to such problems and would ecstatically welcome refined solutions.