June 2007 - Sec 1 - Game 1 - Q4

Moving on to Question 4 now, Any of the following pairs could be the third and fourth

digits respectively of an acceptable product code except?

Again, let's use our previous work here. First in Question 1, three and four are zero and

three. So based on that, I can eliminate (B), because again we're looking for what could

be true except; we're looking for what cannot be true. The answer choice (B) is clearly

shown by Question 1.

Looking at Question 3, our previous work, we see three, zero and one, zero so we can

eliminate (C) and (D). Again, those two coming directly from Question 3, which means we only

have to check zero and one and three and four.

You should notice that (A), zero and one, is clearly a possibility in option two here.

Again, three would be zero, four would be one, five would be three. I will draw that

out very quickly just so you can see it.

And you notice that does not violate any of our conditions. We only had each digit once.

The third digit is less than the fifth digit and the second digit is exactly twice that

of the first digit, which proves that (A) could be true as well, so (E) here would be

the correct answer.

Just to prove to you that (E) would be invalid if three and four were three and four respectively,

we would know that one and two would have to come first and second to satisfy our third

condition that the second digit has a value exactly twice that of its first digit but

what that does is it leaves only zero for the fifth spot which violates our fourth condition

that the third digit must be less than the fifth digit, so (E) is invalid and (E) would

be the correct answer.

Again, this was (A). Again, this was (E).