Start from the Beginning of GMAT Questions to Understand the Pattern

Ron Point_GMAT TipsIf you’ve ever walked into a conversation that was in progress, you know how hard it can be to figure out what’s going on without starting at the beginning. People often timidly ask “What are we talking about?” or “Could you please start over?” in such situations. This is because being parachuted into an ongoing conversation can be quite disorienting.

Most of the time, you can eventually figure out what’s happening, but sometimes you missed an important point near the beginning and just can’t understand the situation. As frustrating as this situation may seem, imagine if, at the end of the conversation, everyone turned to you and asked you to give your detailed opinion on the debate!

On the GMAT, you will frequently be parachuted into a situation that is already in progress. This type of scenario discombobulates most people, because we’re used to a gradual progression starting from the beginning. Since you won’t be at the beginning, you will need to figure out the beginning and the end given what you know from your position in the middle. (In essence, you’re Malcolm). You may not immediately know how to solve the issue, but you can deduce the beginning by seeing where you are in the middle and attempting to reverse engineer the process.

In many ways, this is similar to the dichotomy between multiplication and division. They are, in effect, the exact same operation (multiplying by 2 is dividing by ½ and vice versa). However, people tend to find multiplication easier because you’re going forward. Going backwards is typically harder, in no small part because your brain is not used to going in that (one) direction. When you do something a hundred times a day, it becomes second nature. If you start something for the first time on the GMAT, it may seem almost impossible to solve.

Let’s look at an example of a problem that starts you off in the middle of the action:

A term an is called a cusp of a sequence if an is an integer but an+1 is not an integer. If a5 is a cusp of the sequence a1, a2,…,an,… in which a1 = k and an = -2(an-1 / 3) for all n >1, then k could be equal to:

  • 3
  • 16
  • 108
  • 162
  • 243

Sequences are excellent examples of this parachuting phenomenon because you typically need to have the previous entry in order to find the next element (like a scavenger hunt!). If you find a3, you should be able to find a4. But if you have a4, it’s a lot harder to identify a3. Since you tend to have the pattern, you have to start at the beginning to uncover the progression.

This particular sequence is made easier if you manipulate the algebra a little to get a more manageable form. Instead of the way the sequence is defined, change the pattern to an = -2/3 an-1. This small change highlights the fact that the new element is just the old element multiplied by -2/3. And since the question hinges on when the sequence changes from integers to non-integers, it’s really the denominator that will be of interest to us.

Since this is fairly abstract, let’s go through plugging in answer choice A to see what happens to the series. If k = 3, then the second element of the series would be -2/3 (3). This gives us just -2, and is still an integer. However, the next iteration, a3, would call for -2/3 (-2), which is 4/3, and not an integer. Indeed, this sequence is just calling for us to continually divide by 3, and then determine when the result will no longer be an integer. Clearly, answer choice A won’t be the right choice, as we just found that a3 was not an integer, and thus a2 would be the “cusp” as defined in the question.

Now, using the brute force approach of plugging in each answer choice will eventually yield the correct answer, but it can be tedious and time-consuming. A more logical approach would involve determining that we need a number that has many 3’s in its prime factors. Every time we divide by 3, we will get another integer, provided that we still have 3’s in the numerator. Once we’re left with a number that is not a multiple of 3, the sequence will spit out a non-integer, and the previous number will be the cusp.  Using the prime factorization of the four remaining answer choices, we get:

16 = 2^4

108 = 2 * 54 –) 2 * 2 * 27 –) 2^2 * 3^3

162 = 2 * 81 –) 2 * 3 * 27 –) 2 * 3^4

243 = 3 * 81 –) 3^5

So as we can see, one answer choice has three 3’s, the other has four and the final one has five (the seventh would be Furious). How many 3s do we actually need? Well if the fifth one must be the cusp, then we need to divide by 3 four separate times to get rid of all the 3s. After that, the fifth element will be an integer (also, an action movie), and the sixth element will be a non-integer. Since answer choice D is our educated guess, let’s double check our answer by executing the sequence on 162.

A1 = 162

A2 = -2/3 (162) = -108

A3 = -2/3 (-108) = 72

A4 = -2/3 (72) = -48

A5 = -2/3 (-48) = 32

A6 = -2/3 (32) = -64/3.

This is exactly what we wanted. We can see that each time we are multiplying the previous item by 2/3 and changing the sign. Once we get to 32, that is just 2^5 and dividing it by 3 will no longer yield an integer.

If you’d gone through the complete trial and error process, you’d quickly see that answer choices A and B are incorrect. Answer choice C, 108, comes pretty close, but cusps at A4, not A5. If you then pick answer choice D, 162, you find that you get to 108 on the second iteration, and you can skip the next four steps because you just did them. Finally, answer choice E is a tempting number to start testing with, because it is a perfect exponential of 3. However, you will get to an integer at A6, and thus you need a number with fewer 3s in the numerator.

On test day, you might be able to recognize patterns or you might have to bite the bullet and try each answer choice one by one. However, if you recognize that you need to determine what happens at the beginning before moving on to the middle and the end, you’ll have more success. You always need to understand the pattern, and that starts at the beginning. If you keep this strategy in mind, you won’t find yourself stuck in the middle (with you).

Plan on taking the GMAT soon? We have GMAT prep courses starting all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!

Ron Awad is a GMAT instructor for Veritas Prep based in Montreal, bringing you weekly advice for success on your exam.  After graduating from McGill and receiving his MBA from Concordia, Ron started teaching GMAT prep and his Veritas Prep students have given him rave reviews ever since.