# June 2007 LSAT, Game 1, #3

Here's our setup for Game 1 of the June 2007 LSAT, and if you don't already have the test, you'll find it here. Question 3 says "If the third digit of an acceptable product code is not zero, which one of the following must be true?" Just like Question 1, this question requires us to make a new diagram that incorporates the new condition (third digit can't be 0) with all of the original conditions. The question couldn't be easier, if we simply apply the two worlds we developed in the initial setup. Remember, the initial rules conspired together to leave us with only two templates for completing the game. Those two templates looked like this:

World 1:   1      2     0/3   __   3/4  (where the third spot and the fifth spot can EACH be 3, but they can't SIMULTANEOUSLY be 3).

World 2:   2      4     0/1    __   1/3  (where the third spot and the fifth spot can EACH be 1, but they can't SIMULTANEOUSLY be 1).

The new condition says the third spot CAN'T be 0. Let's apply that first to World 1.  If the third spot can't be 0, then the only alternative for the third spot in World 1 is 3.  And we can go further. If the third spot in World 1 has to be 3, then the last spot in World 1 can only be 4 (because the fifth spot has to be higher than the third spot). And we can go further still. If the third spot is 3 and the last spot is 4, then only 0 is left for the fourth spot. So it looks like this:

Revised World 1 (where the third spot can't be zero):   1      2      3      0      4

We can do the same thing for World 2. If the third spot can't be 0, then the only alternative for the third spot in World 2 is 1. And we can go further. If the third spot in World 2 is 1, then the last spot in World 2 can only be 3 (because the fifth spot has to be higher than the third spot). And we can go further still. If the third spot is 1 and the last spot is 3, then only 0 is left for the fourth spot. So it looks like this:

Revised World 2 (where the third spot can't be zero):   2      4      1      0      3

Now we just have to pick the answer choice that must be true in BOTH of these worlds.

A)  The second digit has to be 2 in our revised World 1, but not in our revised World 2. So it's not our answer.

B)  The third digit has to be 3 in our revised World 1, but not in our revised World 2. So it's not in our answer.

C)  The fourth digit has to be 0 in our revised World 1, and also has to be 0 in our revised World 2. So this is going to be our answer.

D)  The fifth digit has to be 3 in our revised World 2, but not in our revised World 1. So it's not our answer.

E)  The fifth digit can't be 1 in either revised World 1 or revised World 2. So it's not our answer. Our answer is C.