As a GMAT instructor, I frequently find myself perusing the GMAT Official Guide, dare I say, “for fun”. The OG is a terrific indication of the types of questions you can expect to see on the GMAT, and the solution is usually a great method to get to the right answer. However, sometimes I find myself surprised at the official answer because I would solve the question in a completely different way and get to the answer in significantly less time than the OG method.

This is not really unusual, as there are many ways to solve most questions, in particular quant problems. Some people are better at algebra, and can find an algebraic solution to almost any question. Others excel at mental math, and can quickly try two or three different answer choices to see which is correct without trying to prove the general case. Obviously, the best option is to be proficient at all of these options, but identifying which one will yield the fastest result is also important to avoid going down the proverbial rabbit hole and spending excess time to get the correct answer (you want me to solve this cubic equation? How about I just plug in C and see if it works).

Whenever I see an inelegant solution to a stated problem, I always wonder if there’s a faster way to get to the correct answer. I was struck by such a question this week, so I figured I’d share it with you all.

The product of all prime numbers less than 20 is closest to which of the following powers of 10?

(A) 10^9

(B) 10^8

(C) 10^7

(D) 10^6

(E) 10^5

Now this seems like a standard GMAT question, the numbers are going to get unwieldy and we’re going to have to use some kind of number property to avoid getting bogged down in 5 minutes of rote mathematics, right? (Right??) Well it turns out the OG solution is: The prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17 and 19. Multiply them to get 9,669,690. This is closest to 10^7.

Uhm… they remember that they don’t allow calculators on this test, right?

I’m pretty good at math, and I wouldn’t subject myself to the tedious calculation of multiplying 8 different numbers in my head, so I’d need paper and time (or at least a really big abacus). Surely there must be a better way!? Luckily, there is (and don’t call me Shirley)!

Instead of serially multiplying all these numbers, we can selectively pair them together to find numbers that are very close to 10, 100 or even 1,000. If we can approximate these relatively well, we can get the right answer in less than two minutes. Considering the question is asking which answer is closest, we already know we’ll have to estimate at some point, so we may as well start early. I’d rewrite all the numbers and scratch them off as we go through each of them (like GMAT Bingo).

2 3 5 7 11 13 17 19

These are the numbers. If we set ourselves a rough goal of not being off by more than ~10%, we should end up at the correct answer, especially if some of our answers overshoot the actual value and others undershoot it. The very first one I’d eliminate is 11. 11 is basically 10. That leaves us 7 numbers to deal with, and 10^1 as a running tally.

2 3 5 7 11 13 17 19

Any strategy will work here, but my goal will be to eliminate the ugly numbers first. 19 is an ugly number, it’s a lot harder to work with than, say, 2, so I’m more or less going from biggest to smallest, combining as necessary. 19 is almost 20, so if I multiply it by 5, I get ~100. Even better, we’re slightly undershooting 100, whereas 11 slightly overshot 10. This will get us exactly where we want to go pretty quickly. Scratch off 5 and 19. Running tally: 10^3 (previous 10^1 * current 10^2).

2 3 5 7 11 13 17 19

The next number to get rid of would be 17. 17 is very close to 1/6 of 100, so if we had 6, we could just multiply them together. Luckily, we have 2 and 3, so if we multiply these 3 numbers together, we’ll get very close to 100. Scratch these off and update the running tally: 10^5 (previous 10^3 * current 10^2).

2 3 5 7 11 13 17 19

Looking at what’s left, we only have 13 and 7, which multiplied together gives just a little less than 100. Clearly, we can eliminate these two choices and update the running tally: 10^7 (previous 10^5 * current 10^2).

2 3 5 7 11 13 17 19

And that’s how approximating can save you profuse amounts of time on the quant section. As the Veritas Prep arithmetic strategy warns: It’s not about doing the math. This type of question is both tedious and error-prone if the traditional method of multiplying numbers over and over again is employed. It’s much more efficient to use your knowledge of logic and pattern-recognition to unlock the correct answer. This is true regardless of the official answer provided. Obviously, the OG strategy will always work, but it may not always be the fastest way (or even seem sane).

Remember my old mantra of all roads leading to Rome. If the proposed solution in the book doesn’t make sense, even if it’s the Official Guide, try and find a more elegant solution. There are no problems that can’t be solved in multiple ways; it all depends on which method is easiest for you. If you constantly search for a valid alternative solution to the one put forth in front of you, you’ll develop a deeper understanding of the question and the issue being tested. This in turn will make you that much better prepared for test day.

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*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.*