# Test Prep Strategies: Picking Numbers

### By James Rodkey

(Test Prep (ACT/SAT/SSAT), ESOL, English Builder, Latin tutor at The Edge Learning Center)

The current version of the SAT now has a greater focus on math – 800 of the total 1600 points compared to only 800 of 2400 previously – so scoring high on the math sections of the test is necessary for a good overall score. While there is no replacement for a solid fundamental knowledge of mathematical concepts, it’s also important to realize that the SAT is (largely) a multiple choice test, and you can often use this to your advantage.

The SAT is different from your high school math tests: you don’t get points for showing your work, there’s no partial credit, and usually you simply have to pick the correct answer from a list of 4 choices. These differences can sometimes mean that a straightforward, algebra-based approach to solving difficult questions is not always the most efficient way forward. Let’s look at a difficult algebra question from one of the practice exams provided by Khan Academy.

The polynomials f(x) and g(x) are defined above. Which of the following polynomials is divisible by 2x + 3 ?

A) h(x) = f(x) + g(x)

B) p(x) = f(x) + 3g(x)

C) r(x) = 2f(x) + 3g(x)

D) s(x) = 3f(x) + 2g(x)

This question seems rather intimidating, and many students won’t really have a good idea of how to tackle it and find the correct answer quickly. The “proper” way to solve this question using fundamentals of algebra is to do the following: First, notice that f(x) can be rewritten with a common factor of 2x removed:  Then, using substitution, rewrite f(x) as 2x g(x). Finally, evaluate the answer choices with this substitution in mind, and notice that answer choice B, f(x) + 3g(x), becomes 2x g(x) + 3g(x). Removing the common factor of g(x) leaves 2x + 3, which is the factor we’re looking for.

Luckily, this is not the only way to answer the question. For algebra questions like this one with a variable in the question stem and the answer choices, it’s often simpler to pick a real number for x and solve using actual numbers rather than variables.

So let’s pick a number for x and see if we can make this problem simpler. The domains of these functions are unrestricted, so we can choose any value we like for x. Let’s make the math as easy as possible and just say x = 1. Plugging 1 for x into the functions defined by the question gives us the following:

Now we have real number values for f(x) and g(x).  The question wants to know which answer choice is divisible by 2x + 3, which is now 5 since we picked 1 for x: 2(1) + 3 = 5. If we substitute f(x) = 12 and g(x) = 6 into the answer choices, we’re looking for a value that is divisible by 5.

A) h(x) = f(x) + g(x) = 12 + 6 = 18

B) p(x) = f(x) + 3g(x) = 12 + 3(6) = 30

C) r(x) = 2f(x) + 3g(x) = 2(12) + 3(6) = 42

D) s(x) = 3f(x) + 2g(x) = 3(12) + 2(6) = 48

Only answer choice B, 30, is divisible by 5, so it must be the correct answer.

Picking a small number for x in questions like this can turn a seemingly unsolvable puzzle into nothing more complicated than simple multiplication and addition of small integers. While this kind of approach won’t work on every SAT math question, it’s always worth considering whether you can pick a number to simplify a difficult algebra problem.