# June 2007 - Sec 1 - Game 1 - Setup

## Video Transcript:

0:04
So the first game of the June 2007 LSAT starts off as follows:
0:08
"A company employee generates a series of five-digit product codes in accordance with
0:13
the following rules:"
0:14
All right. So let's go through the rules one at a time. The first rule tells us: "The codes
0:17
use the digits 0, 1, 2, 3, and 4, and no others."
0:24
So rule #1 sets forth our digits 0, 1, 2, 3, and 4, and no others.
0:32
Rule 2 tells us that: "Each digit occurs exactly once in any code."
0:38
All right. So we have a five-digit product code. We have five digits, each digits showing
0:45
up exactly once. So what we see here is a game that's going to be 1:1. So you notice
0:50
again we have five product codes, five digits: 0, 1, 2, 3, and 4, and we're trying to place
1:00
each of these digits exactly once in this five-digit code.
1:07
The third rule tells us: "The second digit has a value exactly twice that of the first
1:13
digit."
1:14
So if the first digit was X, the second digit would be exactly twice that, which would be
1:21
2X. And you should notice that there are only two possibilities from rule 3 because either
1:36
our first digit is going to be 1 and our second digit would be exactly twice that, which would
1:43
be 2, or our first digit would be 2, exactly twice that would be 4. Notice it's not possible
1:54
to have 3 as the first digit because exactly twice 3 is 6 and 6 is not an option. And obviously
2:00
we can't have 0 show up first because twice 0 is 0 and each digit only appears once.
2:07
Which brings us to the four and final rule here that: "The value of the third digit is
2:11
less than the value of the fifth digit."
2:14
So basically what that tells me is that we cannot have 4 appear third and we cannot have
2:23
0 appear fifth. But taking that and applying it now to our two scenarios that we've established,
2:32
again we know from our first option, we've already placed 1 and 2, we have 0, 3, and
2:39
4 remaining. Whereas in Option 2 we've already placed 2 and 4, we have 0, 1, and 3 remaining.
2:47
Now in our first scenario, since we can't have 4 appear third, third would have to be
2:52
either 0 or 3. And since we can't have 0 appear fifth, the fifth position would have to be
2:58
either 3 or 4.
3:02
And then, in Option 2, we've already placed 4, but now 3 is the highest digit remaining
3:08
so we know 3 can't be third, which means it has to be either 0 or 1. And then the fifth
3:13
position would be either 1 or 3.
3:17
And that is the setup for this game.
3:19
Now that we have this setup, all of our deductions, let's turn our attention to the questions.