# June 2007 - Sec 1 - Game 1 - Q4

## Video Transcript:

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Moving on to Question 4 now, Any of the following pairs could be the third and fourth
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digits respectively of an acceptable product code except?
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Again, let's use our previous work here. First in Question 1, three and four are zero and
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three. So based on that, I can eliminate (B), because again we're looking for what could
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be true except; we're looking for what cannot be true. The answer choice (B) is clearly
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shown by Question 1.
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Looking at Question 3, our previous work, we see three, zero and one, zero so we can
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eliminate (C) and (D). Again, those two coming directly from Question 3, which means we only
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have to check zero and one and three and four.
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You should notice that (A), zero and one, is clearly a possibility in option two here.
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Again, three would be zero, four would be one, five would be three. I will draw that
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out very quickly just so you can see it.
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And you notice that does not violate any of our conditions. We only had each digit once.
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The third digit is less than the fifth digit and the second digit is exactly twice that
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of the first digit, which proves that (A) could be true as well, so (E) here would be
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Just to prove to you that (E) would be invalid if three and four were three and four respectively,
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we would know that one and two would have to come first and second to satisfy our third
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condition that the second digit has a value exactly twice that of its first digit but
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what that does is it leaves only zero for the fifth spot which violates our fourth condition
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that the third digit must be less than the fifth digit, so (E) is invalid and (E) would
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