### Practice the Questions, Learn the Principles

Check out the above problem. On our free GRE Forums we have set up a thread for you to attempt this problem and discuss your approaches.

To use practice questions effectively for the GRE, you have to get a little out of your comfort zone. Try problems cold. Imagine you have no idea what's coming up next because on the GRE, you won't. While it is possible to learn all the math principles tested on the GRE, adapting on the fly to deal with the unknown is a hallmark of success on Quant.

For instance, on the above problem, use your existing math knowledge to attempt to solve the problem. Work through it step-by-step, but if you encounter a roadblock, do not give up! Punch through difficult moments by trying an alternative approach, working backwards, gaming out the answer choices.

You might be familiar with the concepts of static friction and kinetic friction. Static friction is what keeps a stationary object at rest. Kinetic friction is what slows down a moving object. Imagine you're trying to push a box down your driveway. What's harder? Getting the box to start to move or keeping the box moving once you've given it a good push? In real-life circumstances, it is always harder to move a stationary object than it is to apply enough pressure to keep a moving object going.

#### The same principle applies on the GRE!

If you don't know the "right way" to do a problem, try doing anything you can to get yourself one step closer to the solution. Truly stumped? Skip it and come back to it later. No second-thoughts. Keep moving. Read on below for a discussion for how to apply this approach to the problem above.

### Algebra? Not Again!

What do we remember from 10th grade Algebra 2? Perhaps just the basics, like simplifying or evaluating equations. Even if fundamentals like "what you do to one side you have to do to the other" are all you know, you might be surprised how much you can accomplish on the GRE. At its core, the GRE is not a math test per se. Instead, it is a problem solving and reasoning test. This question is a perfect example.

First, let's figure out what the question is asking. We're looking for the least possible sum of and c. What could we do with the equation above to find this sum? Could we add or subtract from both sides?

7(– 2) = 17(c – 2)
7(– 2) – 17(c – 2) = 0

This doesn't seem to have gotten us any closer to c. Let's try something else. What if we divide both sides by 7 or 17.

7(– 2) = 17(c – 2)
7(– 2) ÷ 17 = c – 2

The problem is 7 and 17 aren't factors or multiples of each other. They don't reduce. In fact, they're both prime! That's a clue.

Okay, maybe we're feeling stuck. Let's go back to the question to see what else we know. We know and are integers. That means they're whole numbers. What else do we know? We know b ≥ > 2. This means b and are at least 3.

1. 4
2. 12
3. 14
4. 24
5. 28

If they're both at least 3, that knocks out (A). can't be 4 because it would have to be at least 6. Why did they throw 4 in there? There's always a reason. It's because for the sum to be 4, and would have to each be 2. That would satisfy the equation because both sides would end up as 0.

7(2 – 2) = 17(2 – 2)
7(0) = 17(0)
0 = 0

But this is forbidden because b ≥ > 2. However, we might still learn something from this dead end. We want these two sides to be equal to each other. How could we get these two sides to be equal to each other? We can't use 2 for and c, so what would the next number up from 2 be.

7 times what equals 17 times what?
7 times 17 equals 17 times 7.
7 • 17 = 17 • 7

Can we compare this to our initial problem?

7(b – 2) = 17(c – 2)
7 • 17 = 17 • 7

b – 2 corresponds to 17. c – 2 corresponds to 7.

b – 2 = 17
c – 2 = 7
b = 19
= 9

So what's c? Using 19 and 9, = 28. This is answer choice (E). Now you might ask yourself, "But I know b + c could be 28, but I want the least possible sum of b and c. How do I know 28 is the least possible sum?" We know this because, remember, 17 and 7 are both prime! That observation came in handy!

17 and 7 do not have any common factors, so there will not be any other, smaller values of and that satisfy the equation. Thus, answer choice (E) is the credited response.

What else could we have noticed just from evaluating the answer choices?

1. 4
2. 12
3. 14
4. 24
5. 28

4 was out because b ≥ > 2. That takes care of (A). 12 is the mean of 7 and 17. That doesn't have anything to do with this situation. It's a trap. Get rid of (B). 14 is double 7, another trap. Get rid of (C). 24 is 17+7, another trap. Get rid of (D). Only (E) is left, and it is the correct answer!

Does that look too easy? Sure, but problem-solving shortcut approaches like that are the bread and butter of the GRE. The bottom line is, you should always come at GRE problems from different angles, and whatever happens, don't give up!

Especially when you're practicing, try to figure out how to get to the solution by any means necessary. The correct answer is correct no matter your approach.

### There's More!

When you're done with this problem, try out some of our other Questions of the Week on our free GRE Forums. We have Quant and Verbal problem sets. Don't forget to register to ask your own questions and receive expert answers and advice for GRE (and GMAT) preparation. We look forward to seeing you there!