June 2007 LSAT Logic Game 1 Setup

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So the first game of the June 2007 LSAT starts off as follows:

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"A company employee generates a series of five-digit product codes in accordance with

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the following rules:"

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All right. So let's go through the rules one at a time. The first rule tells us: "The codes

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use the digits 0, 1, 2, 3, and 4, and no others."

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So rule #1 sets forth our digits 0, 1, 2, 3, and 4, and no others.

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Rule 2 tells us that: "Each digit occurs exactly once in any code."

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All right. So we have a five-digit product code. We have five digits, each digits showing

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up exactly once. So what we see here is a game that's going to be 1:1. So you notice

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again we have five product codes, five digits: 0, 1, 2, 3, and 4, and we're trying to place

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each of these digits exactly once in this five-digit code.

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The third rule tells us: "The second digit has a value exactly twice that of the first

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digit."

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So if the first digit was X, the second digit would be exactly twice that, which would be

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2X. And you should notice that there are only two possibilities from rule 3 because either

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our first digit is going to be 1 and our second digit would be exactly twice that, which would

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be 2, or our first digit would be 2, exactly twice that would be 4. Notice it's not possible

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to have 3 as the first digit because exactly twice 3 is 6 and 6 is not an option. And obviously

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we can't have 0 show up first because twice 0 is 0 and each digit only appears once.

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Which brings us to the four and final rule here that: "The value of the third digit is

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less than the value of the fifth digit."

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So basically what that tells me is that we cannot have 4 appear third and we cannot have

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0 appear fifth. But taking that and applying it now to our two scenarios that we've established,

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again we know from our first option, we've already placed 1 and 2, we have 0, 3, and

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4 remaining. Whereas in Option 2 we've already placed 2 and 4, we have 0, 1, and 3 remaining.

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Now in our first scenario, since we can't have 4 appear third, third would have to be

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either 0 or 3. And since we can't have 0 appear fifth, the fifth position would have to be

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either 3 or 4.

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And then, in Option 2, we've already placed 4, but now 3 is the highest digit remaining

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so we know 3 can't be third, which means it has to be either 0 or 1. And then the fifth

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position would be either 1 or 3.

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And that is the setup for this game.

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Now that we have this setup, all of our deductions, let's turn our attention to the questions.

June 2007 - Logic Game 1 Setup

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