So the first game of the June 2007 LSAT starts off as follows:
"A company employee generates a series of five-digit product codes in accordance with
the following rules:"
All right. So let's go through the rules one at a time. The first rule tells us: "The codes
use the digits 0, 1, 2, 3, and 4, and no others."
So rule #1 sets forth our digits 0, 1, 2, 3, and 4, and no others.
Rule 2 tells us that: "Each digit occurs exactly once in any code."
All right. So we have a five-digit product code. We have five digits, each digits showing
up exactly once. So what we see here is a game that's going to be 1:1. So you notice
again we have five product codes, five digits: 0, 1, 2, 3, and 4, and we're trying to place
each of these digits exactly once in this five-digit code.
The third rule tells us: "The second digit has a value exactly twice that of the first
So if the first digit was X, the second digit would be exactly twice that, which would be
2X. And you should notice that there are only two possibilities from rule 3 because either
our first digit is going to be 1 and our second digit would be exactly twice that, which would
be 2, or our first digit would be 2, exactly twice that would be 4. Notice it's not possible
to have 3 as the first digit because exactly twice 3 is 6 and 6 is not an option. And obviously
we can't have 0 show up first because twice 0 is 0 and each digit only appears once.
Which brings us to the four and final rule here that: "The value of the third digit is
less than the value of the fifth digit."
So basically what that tells me is that we cannot have 4 appear third and we cannot have
0 appear fifth. But taking that and applying it now to our two scenarios that we've established,
again we know from our first option, we've already placed 1 and 2, we have 0, 3, and
4 remaining. Whereas in Option 2 we've already placed 2 and 4, we have 0, 1, and 3 remaining.
Now in our first scenario, since we can't have 4 appear third, third would have to be
either 0 or 3. And since we can't have 0 appear fifth, the fifth position would have to be
And then, in Option 2, we've already placed 4, but now 3 is the highest digit remaining
so we know 3 can't be third, which means it has to be either 0 or 1. And then the fifth
position would be either 1 or 3.
And that is the setup for this game.
Now that we have this setup, all of our deductions, let's turn our attention to the questions.