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A number \(N\) is green if \(6N\) has none of the digits \(0,1,2,3,4\).

Looking at one-digit numbers, right away we know that \(1\) is a green number. That doesn't help though, since the problem already tells us that every green number has either a \(1\) or some other digit in it. Now, let's consider two-digit numbers.

Let's start off by considering the units digits of \(6N\).

If \(N\) has units digit 1, then \(6N\) has units digit 6.

If \(N\) has units digit 2, then \(6N\) has units digit 2.

Filling it out for all the digits, we have

1 -> 6

2 -> 2

3 -> 8

4 -> 4

5 -> 0

6 -> 6

7 -> 2

8 -> 8

9 -> 4

Using this list, we know that all green numbers must end in 1, 3, 6, or 8.

A quick check though shows that out of 1, 3, 6, and 8, only 1 is a green number. So this confirms that green numbers do indeed have to contain a 1 or some other number, but the problem already gave us that.

So, let's start looking through the two-digit numbers.

The possible two-digit green numbers are going to look like \(A1, A3, A6, A8\).

*note: whenever I use \(A\) here, I mean it as a digit. \(A1\) is a number with \(A\) in the tens place and \(1\) in the ones place, **not** \(A\cdot1\).

Now, let's see what \(6\) times these numbers gives us.

This is what \(A1 \cdot 6\) looks like. There is no digit \(A\) for which this works, because no value of \(6A\) has none of the digits \(0,1,2,3,4\). Let's move on to the case of \(A3 \cdot 6\).

So, we need a digit \(A\) such that \(6A+1\) has none of the digits \(0\to4\). Wait, \(A=9\) works! \(93\cdot6=558\), which has no digits \(0\to4\), so **\(93\) is a green number** (note that \(A=1\) also works in this case too, but that doesn't help us because the problem already tells us that \(1\) is one of the two numbers).

We already have two green numbers now: \(1\) and \(93\). Since the problem says that every green number has either a \(1\) or some other digit in it, we now know that the other digit has to be \(9\) or \(3\). Since the answer choices has no option for \(3\), the answer has to be \(\boxed{(E)\text{ } 9}\).

But, what if they didn't give us the answer choices? Well, I guess we just have to keep going through the cases.

Here's what \(A6 \cdot 6\) looks like. Again, the only possible \(A\) are \(A=1,9\). And we can confirm, \(16\) and \(96\) are green numbers, since \(16\cdot6=96\) and \(96\cdot6=576\). That means that the two numbers are indeed \(1\) and \(9\), and we can be safe with E as our answer.

Here's a solution that uses the choices: can you think of any multiples of \(6\) that doesn't use any of the digits 0-4? Try and list out a few. Once you have a multiple 6 that satisfies that, that's a possibility for \(6N\). Can you figure out which one is the answer now?

While you try that out, I'll work out a solution that is more precise

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