The second way that we can know this is a made-up word is to be familiar with what the SAT is allowed to ask us about when it comes to math. The SAT can only ask about the concepts in the toolbox in this book, and the toolbox doesn’t cover tri-factorability.
So it must just be that we’re supposed to figure out what the word means from reading the question. If we know what the words "product" and "consecutive" mean, then we know what a tri-factorable number is. Remember, as always, that the most important skill on the SAT is reading carefully.
Now, let’s proceed.
We know that all of these questions can be answered in less than 30 seconds, and we know that it would clearly take a lot longer than 30 seconds to attempt to tri-factor each of the first 1,000 integers.
So there must be another way to go.
It’s important to remember that we can always just list out the answers to these kinds of questions that ask how many numbers in a certain set satisfy a certain condition. Either we’ll start listing them and realize there’s only a small number of them, or we’ll start listing them and realize there’s a certain pattern they all follow, and then we can predict the final number from the pattern.
So how can we figure out a tri-factorable number? Since they’re made by multiplying consecutive integers, I’d start with the smallest positive integer and see what happens:
1 * 2 * 3 = 6
So if we start with 1 as the first positive integer, we arrive at the product 6, which must be the first tri-factorable number.
The next tri-factorable number will start with 2:
2 * 3 * 4 = 24
From there we can basically get on a roll:
3 * 4 * 5 = 60
4 * 5 * 6 = 120
5 * 6 * 7 = 210
6 * 7 * 8 = 336
7 * 8 * 9 = 504
8 * 9 * 10 = 720
9 * 10 * 11 = 990
10 * 11 * 12 = 1320
Oops—notice that that last number is bigger than 1000! So there are 9 numbers that work: the one I get when I multiply three consecutive numbers starting with 1, the one I get when starting with 2, starting with 3, with 4, with 5, 6, 7, 8, and finally 9, and that's it.
Notice that this question, like so many other ‘challenging’ SAT Math questions, involves nothing more than careful reading, basic arithmetic, and a willingness to play around with familiar concepts in strange ways. This is typical for the SAT, as we’ve seen many times by now and will see again.
Notice also that there are many ways to mess this question up. We might miscount, or accidentally overlook one or more of the tri-factorable numbers. Any one of those tiny mistakes will cause us to miss the question completely.
Page 468, Question 17
This is another one of those questions that we could try to answer in a few different ways. Many people will attempt an algebraic solution, or you could also try guessing and checking. There’s also another approach we’ll talk about after I go over the algebraic one.
For call A, the cost will be $1 + $0.07(t-20).
For call B, the cost will be $0.06(t).
So just set them equal and solve:
$1 + $0.07(t-20) = $0.06(t) (initial setup)
1 + .07t - 1.4 = .06t (distribute $0.07 on the left)
1 - 1.4 = -.01t (combine like terms)
.4 = .01t (simplify the expression on the left)
40 = t (isolate t)
The faster approach is a little more holistic, and would probably not be tried by most test-takers, even trained ones. Still, I thought we should talk about it. To work this problem out without really writing out any algebra, we could realize that the $1 for the first 20 minutes works out to 5 cents per minute, which is one cent per minute less than the other rate. After those 20 minutes are up, the rate goes to 7 cents per minute, which is one cent per minute more than the other rate. So to get everything to equal out, you'll need 20 more minutes of talking at the higher rate after the initial 20 minutes of talking at the lower rate. 20 + 20 is 40, so 40 is the answer.
Just to be clear, both approaches are equally valid, of course. I just wanted to introduce the second one as an exercise of sorts, to keep calling your attention to the fact that the College Board usually sets up SAT Math questions so that they can be attacked quickly and easily if we know how to look at them.
Page 468, Question 18
This question is rated 5 out of 5 for difficulty, but, of course, we know that that doesn’t mean much. All it means is that a lot of untrained people missed it. As we’ll see in this explanation, this is yet another SAT Math question that requires nothing more than careful reading and careful thinking.
We’re told the perimeter is p, and we know that each square has a side of k. The perimeter consists of 16 sides, so the perimeter’s length is 16k.
The area of each square must be k2, and there are 10 squares. So the area is 10k2.
Since the question says that p and a are equal, we know that