So the correct answer here is (E): only roman numerals I and III can work.
Notice that this question has really no resemblance to anything that ever happens in the average math class in school. Notice that with this solution we never really needed a calculator, or even a pencil. Notice, also, that there’s no other question in the entire book that calls for the same specific approach to that this one calls for. In other words, the lesson to learn from this question is NOT the idea of specifically attacking roman numeral questions by looking to see what kinds of subtraction can result in zero, because you’ll probably never see a question that tests that concept in this exact way again. Instead, the lesson to take away from this is, as always, that we need to practice thinking about SAT Math questions in terms of basic principles, definitions, and properties. We need to read everything carefully. And we have to learn to be flexible in our thinking in the way that the SAT Math section rewards over and over again.
Page 714, Question 8
This question is a marvelous example of how important it is to be aware of the limitations of what the SAT Math section can ask you. Most people try to approach this question by taking the 6th root of both sides of the equation, but, as trained test-takers, we need to know that the SAT can never ask us to take any root besides the square root. So taking the 6th root isn’t going to be the easiest way to go here (and anyway, the 6th root of 432 is an irrational number).
So now what?
Well, as is often the case, there are two basic ways to go here. We can try a kind of backsolving, brute-force sort of approach, or we can try using algebraic principles. The first kind of approach requires less familiarity with math but will be a bit more time-consuming for most people, since it’s basically just trying a whole bunch of hit-or-miss combinations of a and b values. The second approach will go a little faster but will be harder for most test-takers to think of or pull off.
If we were going to do the first approach (the hit-or-miss approach), we’d start out by listing all the factors for the quantities in the answer choices. Notice, by the way, that the choices have a lot of factors in common, since they’re all multiples of 6. That would suggest that concentrating on numbers like 2, 3, and 6 might not be a bad way to start.
Anyhow, to start working through our options, we’d take an answer choice like (A), figure out its factors (they’re 1, 2, 3, and 6), and then get out our calculators and start plugging in combinations of those factors for a and b in the original expression to see if we could come up with a value of 432.
This will be made more challenging and time-consuming by the fact that, to be really thorough, you have to try each factor for a and again for b, since they aren’t treated the same in the expression.
But it’s made a little easier by the fact that some combinations of factors will prove to be too small or too large, which might help you adjust your guessing and hit on the right answer more quickly. A lot of people who try these kinds of backwards solutions like to start with (C), since it’s the value in the middle, see how that goes, and then adjust up or down. In general I think that’s a fine approach, but in this case it might result in more work because the larger numbers probably also have more pairs of factors to work with. So it’s a toss-up, in my opinion—start where you like.
The second approach, as I mentioned, is a bit more mathematical, and is the one I would probably employ (not that that matters).
I would start out by realizing that I can simplify the expression on the left-hand side of the equation, so that the whole equation reads like this:
a3b2 = 432
Now that’s starting to look like something we might have seen at some point in algebra class—it’s starting to look a bit like a prime factorization, with a and b being the possible factors of 432.
So at this point, I’d break down 432 into its prime factors. I’d do this by using what my algebra teacher, Mrs. Turner, used to call a “factor tree.” They look something like this:
From this, we can see that the prime factorization of 432 is 24 * 33.
That’s almost what we’re looking for, except for one small problem: 432 is the same thing as 24 * 33, but we’re looking to express it as a3b2. It looks like the a can match up with the 3 in our factorization, because they’re both cubed, but how do we get b2 to match up with 24? We’ll have to realize that 24 is 16, and 16 is 42.
That means that a3b2 is 432 when a is 3 and b is 4. And that, in turn, means ab is 12, so (B) is correct.
As far as our answer choices are concerned, we see that we have a series of multiples of 6. In this situation we’d expect the right answer not to be a value on either end of the series—maybe that’s not a ton of help here, but at least it doesn’t suggest we’re wrong.