*After a too-long hiatus, Alex the Archangel (a.k.a. Knewton’s Lead Verbal Developer), is back to reveal another one of the SAT testmakers’ evil tricks. Prepare to step up your SAT prep game!*

Oh, hello, how are you doing toda–EVIL! I sense evil in our presence! The wind is whistling, the wolves are on the prowl, a new season of Jersey Shore is in progress. It must be time for the:

**SAT EVIL TRICK OF THE WEEK #5: The Time Vampires**

Something happens to the writers of the math section of the SAT when they sit down to write questions:Maybe some of the questions they write are happy questions, nice ones about puppies and frozen yogurt and sharing pizza with friends. But these are *too easy for the SAT*, they are told. *We need some questions that will really fool people*, *make them waste their time on silly calculations*. Then the SAT writers sharpen their canine teeth, and they say, “Oh, you mean **time vampires**! I can do that.” And then they cackle and cackle, all the while writing math problems–which is actually rather difficult to do.

What are **time vampires**, you ask?

Simply put, time vampires are questions that invite you to spend far more time on them than you have to. They suck your precious seconds away, drop by drop.

Luckily, time vampires have a major weakness: There is almost always a big shortcut to get the question correct quickly and relatively painlessly–**IF **you can spot it.

Some of the time consuming tasks that time vampires exploit include:

- Long strings of numbers or variables for you to add, divide or multiply.
- Complex algebra that makes you rewrite long variable equations over and over.
- Equations that include radicals, which are a pain to keep track of if you don’t deal with them.
- Equations that use large exponents, which get massively large if you don’t wrangle them.
- Probability questions with lots of combinations or permutations of possible outcomes.
- Geometry questions that invite you to wander around, solving every little angle or line length.
- Problems that include large data sets like “all the numbers from 1-100″ or “between 700 and 800.”
- Problems that ask you to find a late (like the 101
^{st}) element in a pattern.

Today, we’ll talk about some time vampires on which you could use the pattern recognition strategy to solve them quickly.

1)Â Â Â Â Â 1,1,2,2,2,3,3,3,3,….

All positive integers appear in the sequence above, and each positive integer *b* appears in the sequence *b+1 *times. Each integer in the pattern is greater than or equal to the one before it. If the integer 11 appears in the sequence as the *x*th term, what is the value of *x*?

This question **wants** you to spend the next three to five minutes writing out all numbers in this pattern and then counting them. That would take forever.

But wait; take a look at this pattern: there are going to be **2** 1′s, **3** 2′s, **4** 3′s, etc. and you’re going to **proceed **until you hit the first 11. How many 10′s would there be right before this 11 comes? There would be 11! There’s the shortcut! At this point, the pattern disappears. Really, you are just being asked to add the numbers 2 through 11–the number of 1′s through the number of 10′s–and then (don’t forget) to add one more to represent that first number 11 in the sequence.

Adding 2 through 11 isn’t that bad (especially on a calculator), but there’s even a shortcut to that. We call it the Rainbow Method, because you connect the numbers at the opposite ends of the pattern. 2 + 11 = 13. 3 + 10 = 13. 4 + 9 = 13. Hmmm…Â Basically, we have 5 groups of 13, which all adds up to 65. Don’t forget to add that extra 1 (which represents the first 11 in the sequence).** 66. **Done and done.

Now, if you didn’t see that shortcut right away, don’t worry; that’s not really the point anyway.

What’s important is that we *spotted *this time vampire coming, and immediately starting looking for ways to conquer the problem by cutting extra work. Even if you just wrote out the **number **of numbers, rather than the numbers themselves, you would have saved a lot of time on this problem. When you see a pattern, use it to your advantage!

Let’s try one where the pattern is a little better hidden:

For all positive integers, let &x be defined **by &x =**^{1}/2. Which of the following is equal to 36?

a)Â Â Â Â Â &2 + &4

b)Â Â Â Â Â &3 + &5

c)Â Â Â Â Â Â &4 + &6

d)Â Â Â Â Â &5 + &7

e)Â Â Â Â Â &7 + &9

What makes this a time vampire is that you are being asked to plug many different numbers into a formula that is a bit ugly. So…Â make it prettier! If you spot the fact that (x+1)(x-1) is the difference of two squares, and really equals x^{2 }-1, things get a lot quicker. Basically, you only have to take each number next to a &, square it, subtract 1, and then split **it** into 2.

Let’s try A: 2^{2 }-1 = 3, Â so we get 3/2. 4^{2 }-1 = 15, and 15/2 = 15/2. 3/2 + 15/2 = 18/2. That’s 9. Too small.

Let’s try B: 3^{2 }-1 = 8, and 8/2 = 4. 5^{2 }**- 1** = 24, and 24/2 = 12. 4 + 12 = 16.

At this point, you should start to see a pattern. Hmmm… 9, 16, these are perfect squares. It looks like every time you increase by 1, you get to the next perfect square. That would make D the right answer; just to be sure, let’s check.

Let’s try D:Â 5^{2 }-1 = 24, and 24/2 = 12. 7^{2 }-1 = 48, and 48/2 = 24. 12 + 24 = 36. Perfection.

With choice-driven questions like this, you can be even **more** of a vampire-slayer by starting with the answer choice C and then adjusting based on whether your answer is too big or small. But then you might not have spotted this sweet perfect square pattern.

See, isn’t math fun? I mean, EVIL?!!!

x+1)(x-1 ↩

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