Posted on
04
Aug 2021

## GMAT Abstract Data Sufficiency Problem

#### Abstract Data Sufficiency Problems & Scenarios

Hi guys! Abstract data sufficiency problems tend to really lend themselves to running scenarios – It doesn’t matter if it’s an abstract inequality or a number theory problem, really anytime you’ve got variables thrown into the question stimulus on a DS problem, scenarios is a good way to go. Now your scenarios can be discrete actual numbers that you throw in there, but you can also leverage rules and have more conceptual-level scenarios. We’re going to take a look at both in this problem.

#### Problem Introduction

We’re being asked here for the evenness or oddness of n which is an integer. At first blush, we’re going to say, “Well, if we have the evenness or oddness of any expression involving n and n alone, we should be able to backtrack it to n.” If you don’t see that then you might fall into the trap of having to go much more deeply into it and figure out “Well, what if n is this, what if n is that?” But notice here that because we’re dealing with evens and odds there are a set of identities that govern every possible addition, or multiplication, subtraction, or division of evens and odds. So, as long as there’s nothing complicating it the expression itself will be enough.

#### Statement 1

Taking a look at the introduced information, number one gives us n2 + 1 is odd that means that n2 is even. How do we know without numbers? If n2 + 1 is odd then adjusting it down by one, removing that one, means we’re definitely going to get to an even, because the number line is just even, odd, even, odd, even, odd all the way up. So, we have n2 is even, and only even times even gives us an even. Odd times odd doesn’t, odd times odd gives us an odd.

So, n must be even if the square of it leads us to an even. Notice again, that we don’t need to do any of that, it’s enough just to say we’ve got n in an expression, and we have its evenness and oddness.

#### Statement 2

Number two works the same way. 3n + 4 is even that’s enough, no more to do, but if we want to we can adjust that 3n – 4 as even down by 4 notches (odd, even, odd, even). So 3n is even and then we know that n divided by 3, that is what is an even divided by 3, will give us n. An even divided by an odd is going to always be an even, for the same reason even times an odd is always going to be an even.

Run some scenarios here, start out with an even number; let’s do 6, 50, and 120. Divide each by 3; 2 (6/3=2), 50 divided by 3 doesn’t work, 40 (120/3=40). So on the two that do work, we get to even numbers. 50 is not allowed to be used as a scenario because we’re told that n – 3 has to be an integer which means, that 3n must also be an integer; that is 3n is a multiple of 3. Since 50 is not a multiple of 3 it’s not a potential 3n. Take a minute with that one, because it’s kind of looking at everything in reverse.

So here we have two different expressions that both give us evenness and oddness, they both work independently. The answer choice is D – each alone is sufficient.

If you enjoyed this problem, try your hand at these Data Sufficiency Problems GMAT Trade Show Problem & Area of a Triangle Problem.

Posted on
28
Jul 2021

## GMAT Trade Show Problem – Data Sufficiency

#### GMAT Trade Show Problem Introduction

Today we’re going to take a look at the Trade Show Problem and this is a GMAT Data Sufficiency problem with averages as the focal point. But really the concept of average is distracting from this problem. So, if we take a look at the question stimulus, we want to figure out what we need, but we need to synthesize some of the information there to understand what we know.

We’re being asked whether or not it gets above a certain threshold an average of 90, and over six days that’s going to be over a total of 540 points. Notice how I did it mathematically, you can represent it graphically as a rectangle, but 90 times 6 is that 540 points. We know though that all of our days at a minimum are 80 which means we can build up from that piece of knowledge. We have 80 x 6 = 480 points and we want to know if we have more or less than 60 points above that minimum that we’re already working with that’s what we need.

#### Solving the GMAT Problem

Ways we might get it include any number of slices and dices for the performance of the rest of the days and the difficulty of this problem in large part will be dependent on how convoluted the GMAT gives us the introduced information on number one and two.

When we look at number one, we’re told that the final four days average out to a hundred. Once again, like with other average problems, each of the individual four days the performance doesn’t matter. We can just say each is exactly 100 and make that assumption, which means each is 20 over – we’re 80 points over the mean. Because we want to know whether we are more or less than 60 points, this knowledge that we’re 100 points tells us “Yes, definitively. We are over that average of 90, we’re over that surplus of 60 points.” So, number one is sufficient.

Number two gives us the opposite information, it talks about the minimum, and, in aggregate, that doesn’t let us know directly whether or not we make those 60 points. That is it’s possible but it’s also possible that we don’t, because we’re dealing with a minimum rather than a maximum or rather we’re dealing with information that can lie on either side of what we need. Therefore 2 is insufficient. Our answer here is A.

I hope that was useful. GMAT nation stay strong, keep averaging. You guys got this! I believe in you. If you want to test your GMAT Data Sufficiency skills, check out the Science Fair problem.

Posted on
22
Jul 2021

## GMAT Quant Syllabus 2021-2022

Author: Apex GMAT
Contributor: Altea Sollulari
Date: 22 July, 2021

We know what you’re thinking: math is a scary subject and not everyone can excel at it. And now with the GMAT the stakes are much higher, especially because there is a whole section dedicated to math that you need to prepare for in order to guarantee a good score. There is good news though, the GMAT is not actually testing your math skills, but rather your creative problem solving skills through math questions. Furthermore, the GMAT only requires that you have sound knowledge of high school level mathematics. So, you just need to practice your fundamentals and learn how to use them to solve specific GMAT problems and find solution paths that work to your advantage.

The Quantitative Reasoning section on the GMAT contains a total of 31 questions, and you are given 62 minutes to complete all of them. This gives you just 2 minutes to solve each question, so in most cases, the regular way of solving math equations that you were taught in high school will not cut it. So finding the optimal problem solving process for each question type is going to be pivotal to your success in this section. This can seem a daunting start, so our expert Apex GMAT instructors recommend that you start your quant section prep with a review of the types of GMAT questions asked in the test and math fundamentals if you have not been using high school math in your day to day life.

## What types of questions will you find in the GMAT quant?

There are 2 main types of questions you should look out for when preparing to take the GMAT exam:

### Data Sufficiency Questions

For this type of GMAT question, you don’t generally need to do calculations. However, you will have to determine whether the information that is provided to you is sufficient to answer the question. These questions aim to evaluate your critical thinking skills.

They generally contain a question, 2 statements, and 5 answer choices that are the same in all GMAT data sufficiency questions.

Here’s an example of a number theory data sufficiency problem video, where Mike explains the best way to go about solving such a question.

### Problem Solving Questions

This question type is pretty self-explanatory: you’ll have to solve the question and come up with a solution. However, you’ll be given 5 answer choices to choose from. Generally, the majority of questions in the quant section of the GMAT will be problem-solving questions as they clearly show your abilities to use mathematical concepts to solve problems.

Make sure to check out this video where Mike shows you how to solve a Probability question.

## The main concepts you should focus on

The one thing that you need to keep in mind when starting your GMAT prep is the level of math you need to know before going in for the Quant section. All you’ll need to master is high-school level math. That being said, once you have revised and mastered these math fundamentals, your final step is learning how to apply this knowledge to actual GMAT problems and you should be good to go. This is the more challenging side of things but doing this helps you tackle all the other problem areas you may be facing such as time management, confidence levels, and test anxiety

Here are the 4 main groups of questions on the quant section of the GMAT and the concepts that you should focus on for each:

### Algebra

• Algebraic expressions
• Equations
• Functions
• Polynomials
• Permutations and combinations
• Inequalities
• Exponents

### Geometry

• Lines
• Angles
• Triangles
• Circles
• Polygons
• Surface area
• Volume
• Coordinate geometry

### Word problems

• Profit
• Sets
• Rate
• Interest
• Percentage
• Ratio
• Mixtures

Check out this Profit and Loss question.

### Arithmetic

• Number theory
• Percentages
• Basic statistics
• Power and root
• Integer properties
• Decimals
• Fractions
• Probability
• Real numbers

Make sure to try your hand at this GMAT probability problem.

## 5 tips to improve your GMAT quant skills?

1. Master the fundamentals! This is your first step towards acing this section of the GMAT. As this section only contains math that you have already studied thoroughly in high-school, you’ll only need to revise what you have already learned and you’ll be ready to start practicing some real GMAT problems.
2. Practice time management! This is a crucial step as every single question is timed and you won’t get more than 2 minutes to spend on each question. That is why you should start timing yourself early on in your GMAT prep, so you get used to the time pressure.
3. Know the question types! This is something that you will learn once you get enough practice with some actual GMAT questions. That way, you’ll be able to easily recognize different question types and you’ll be able to use your preferred solution path without losing time.
4. Memorize the answer choices for the data sufficiency questions! These answers are always the same and their order never changes. Memorizing them will help you save precious time that you can spend elsewhere. To help you better memorize them, we are sharing an easier and less wordy way to think of them:
5. Make use of your scrap paper! There is a reason why you’re provided with scrap paper, so make sure to take advantage of it. You will definitely need it to take notes and make calculations, especially for the problem-solving questions that you will come across in this GMAT question.
• Only statement 1
• Only statement 2
• Both statements together
• Either statement
• Neither statement
Posted on
21
Jul 2021

## GMAT 3D Geometry Problem

In this problem we’re going to take a look at 3D objects and in particular a special problem type on the GMAT that measures the longest distance within a three-dimensional object. Typically, they give you rectangular solids, but they can also give you cylinders and other such objects. The key thing to remember about problems like this one is that effectively we’re stacking Pythagorean theorems to solve it – we’re finding triangles and then triangles within triangles that define the longest distance.

This type of problem is testing your spatial skills and a graphic or visual aid is often helpful though strictly not necessary. Let’s take a look at how to solve this problem and because it’s testing these skills the approach is generally mathematical that is there is some processing because it’s secondary to what they’re actually testing.

#### GMAT 3D Geometry Problem Introduction

So, we have this rectangular solid and it doesn’t matter which way we turn it – the longest distance is going to be between any two opposite corners and you can take that to the bank as a rule: On a rectangular solid the opposite corners will always be the longest distance. Here we don’t have any way to process this central distance so, what we need to do is make a triangle out of it.

Notice that the distance that we’re looking for along with the height of 5 and the hypotenuse of the 10 by 10 base will give us a right triangle. We can apply Pythagoras here if we have the hypotenuse of the base. We’re working backwards from what we need to what we can make rather than building up. Once you’re comfortable with this you can do it in either direction.

#### Solving the Problem

In this case we’ve got a 10 by 10 base. It’s a 45-45-90 because any square cut in half is a 45-45-90 which means we can immediately engage the identity of times root two. So, 10, 10, 10 root 2. 10 root 2 and 5 makes the two sides. We apply Pythagoras again. Here it’s a little more complicated mathematically and because you’re going in and out of taking square roots and adding and multiplying, you want to be very careful not to make a processing error here.

Careless errors abound particularly when we’re distracted from the math and yet we need to do some processing. So, this is a point where you just want to say “Okay, I’ve got all the pieces, let me make sure I do this right.” 10 root 2 squared is 200 (10 times 10 is 100, root 2 times root 2 is 2, 2 times 100 is 200). 5 squared is 25. Add them together 225. And then take the square root and that’s going to give us our answer. The square root of 225 is one of those numbers we should know. It’s 15, answer choice A.

Okay guys for another 3D and Geometry problem check out GMAT 680 Level Geometry Problem – No Math Needed! We will see you next time.

Posted on
14
Jul 2021

## GMAT Geometry Problem

Hey guys, top level geometry problems are characterized typically by stringing a whole bunch of different rules together and understanding how one thing relates to the next thing, to the next thing. Until you get from the piece of information you started with to the conclusion. We’re going to start out by taking a look at this problem using the z equals 50° and seeing how that information goes down the line.

But afterwards we’re going to see a super simple logical pathway utilizing a graphic scenario that makes the z equals 50° irrelevant. To begin with we’re being asked for the sum of x and y and this will come into play on the logical side. We need the sum not the individual amounts but let’s begin with the y. We have a quadrilateral and it has parallel sides which means the two angles z and y must equal 180°. That’s one of our geometric rules. If z is 50° that means y is 130° and we’re halfway there.

Next we need to figure out how x relates and there are several pathways to this. One way we can do it is drop. By visualizing or dropping a third parallel line down, intersecting x, so on the one hand we’ll have 90 degrees. We’ll have that right angle and on the other we’ll have that piece. Notice that the parallel line we dropped and the parallel line next to z are both being intersected by the diagonal line going through which means that that part of x equals z. So we have 50° plus 90° is 140°. 130° from the y, 140° from the x, gives us 270°.

Another way we can do this is by taking a look at the right triangle that’s already built in z is 50° so y is 1 30°. now the top angle in the triangle must then be 180° minus the 130° that is 50°. it must match the z again we have the parallel lines with the diagonal coming through then the other angle the one opposite x is the 180° degrees that are in the triangle minus the 90° from the right triangle brings us to 90° minus the 50° from the angle we just figured out means that it’s 40° which means angle x is 180° flat line supplementary angles minus the 40° gives us 140° plus the 130° we have from y again we get to 270°.

#### Graphic Solution Path

Now here’s where it gets really fun and really interesting. We can run a graphic scenario here by noticing that as long as we keep all the lines oriented in the same way we can actually shift the angle x up. We can take the line that extends from this big triangle and just shift it right up the line until it matches with the y. What’s going to happen there, is we’re going to see that we have 270° degrees in that combination of x and y and that it leaves a right triangle of 90°, that we can take away from 360° again to reach the 270°.

Here the 50° is irrelevant and watch these two graphic scenarios to understand why no matter how steep or how flat this picture becomes we can always move that x right up and get to the 270°. That is the x and y change in conjunction with one another as z changes. You can’t change one without the other while maintaining all these parallel lines and right angles. Seeing this is challenging to say the least, it requires a very deep understanding of the rules and this is one of those circumstances that really points to weaknesses in understanding most of what we learn in math class in middle school, in high school. Even when we’re prepping only scratches the surface of some of the more subtle things that we’re either allowed to do or the subtle characteristics of rules and how they work with one another and so a true understanding yields this very rapid graphic solution path.

#### Logical Solution Path

The logical solution path where immediately we say x and y has to be 270° no matter what z is and as you progress into the 80th, 90th percentile into the 700 level on the quant side this is what you want to look for during your self prep. You want to notice when there’s a clever solution path that you’ll overlook because of the rules. Understand why it works and then backtrack to understand how that new mechanism that you discovered fits into the framework of the rules that we all know and love. Maybe? I don’t know if we love them! But they’re there, we know them, we’re familiar with them, we want to become intimate. So get intimate with your geometry guys put on some al green light some candles and I’ll see you next time.

If you enjoyed this problem, try other geometry problems here: GMAT Geometry.

Posted on
24
Jun 2021

## The Basics of GMAT Combinatorics

By: Apex GMAT
Contributor: Svetozara Saykova
Date: 24th June 2021

Combinatorics can seem like one of the most difficult types of questions to come across on the GMAT. Luckily there are not many of them within the exam. Still these questions make up the top level of scoring on the test and therefore it is best if you are well equipped to solve them successfully, especially if you are aiming for a 700+ score. The most important rule to follow when considering this question type is the “Fundamental Counting Principle” also known as the “Counting Rule.” This rule is used to calculate the total number of outcomes given by a probability problem.

The most basic rule in Combinatorics is “The Fundamental Counting Principle”. It states that for any given situation the number of overall outcomes is equal to the product of the number of each discrete outcome.

Let’s say you have 4 dresses and 3 pairs of shoes, this would mean that you have 3 x 4 = 12 outfits. The Fundamental Counting Principle also applies for more than 2 options. For example, you are at the ice cream shop and you have a variety of 5 flavors, 3 types of cones and 4 choices for toppings. That means you have 5 x 3 x 4 = 60 different combinations of single-scoop ice creams.

The Fundamental Counting Principle applies only for choices that are independent of one another. Meaning that any option can be paired with any other option and there are no exceptions. Going back to the example, there is no policy against putting sprinkles on strawberry vanilla ice cream because it is superb on its own. If there were, that would mean that this basic principle of Combinatorics would not apply because the combinations (outcomes) are dependent. You could still resort to a reasoning solution path or even a graphical solution path since the numbers are not so high.

Let’s Level Up a Notch

The next topic in Combinatorics is essential to a proper GMAT prep is  permutations. A permutation is a possible order in which you put a set of objects.

Permutations

There are two subtypes of permutations and they are determined by whether repetition is allowed or not.

• Permutations with repetition allowed

When there are n options and r number of slots to fill, we have n x n x …. (r times) = nr permutations. In other words, there are n possibilities for the first slot, n possibilities for the second and so on and so forth up until n possibilities for position number r.

The essential mathematical knowledge for these types of questions is that of exponents

To exemplify this let’s take your high school locker. You probably had to memorize a 3 digit combination in order to unlock it. So you have 10 options (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) for 3 available slots. The total number of locker passwords you can have is 103 = 1,000.

• Permutation without repetition allowed

When repetition is restricted in the given GMAT problem, we would have to reduce the number of available choices for each position.

Let’s take the previous example and add a restriction to the password options – you cannot have repeating numbers in your locker password. Following the “we reduce the options available each time we move to the next slot” rule, we get 10x9x8 = 720 options for a locker combination (or mathematically speaking permutation).

To be more mathematically precise and derive a formula we use the factorial function (n!). In our case we will take all the possible options 10! for if we had 10 positions available  and divide them by 7!, which are the slots we do not have.

10! =  10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1

7! =  7 x 6 x 5 x 4 x 3 x 2 x 1

And when we divide them (7 x 6 x 5 x 4 x 3 x 2 x 1) cancels and we are left with 10 x 9 x 8 = 720.

Pro tip: Taking problems and deeply examining them by running different scenarios, and changing some of the conditions or numbers is a great way to train for the GMAT. This technique will allow you to not only deeply understand the problem but also the idea behind it, and make you alert for what language and piece of information stands for which particular concept.

So those are the fundamentals, folks. Learning to recognize whether order matters and whether repetition is allowed is essential when it comes to Combinatorics on the GMAT. Another vital point is that if you end up with an endless equation which confuses you more than helps, remember doing math on the GMAT Quant section is not the most efficient tactic. In fact, most of the time visualizing the data by putting it into a graph or running a scenario following your reasoning are far more efficient solution paths.

Feeling confident and want to test you GMAT Combinatorics skills? Check out this GMAT problem and try solving it. Let us know how it goes!

Posted on
13
May 2021

## GMAT Factors Problem

Hey guys! Today we’re going to take a look at one of my favorite problems. It’s abstract, it’s oddly phrased and in fact the hardest part for many folks on this problem is simply understanding what’s being asked for. The difficulty is that it’s written in math speak. It’s written in that very abstract, clinical language that if you haven’t studied advanced math might be new to you.

How this breaks down is they’re giving us this product from 1 to 30, which is the same as 30!. 30*29*28 all the way down the line. Or you can build it up 1*2*3*……*29*30.

#### The Most Difficult Part of The GMAT Problem

And then they’re asking this crazy thing about how many k such that three to the k. What they’re asking here is how many factors of three are embedded in this massive product. That’s the hard part! Figuring out how many there are once you have an algorithm or system for it is fairly straightforward. If we lay out all our numbers from 1 to 30. And we don’t want to sit there and write them all, but just imagine that number line in your head. 1 is not divisible by 3. 2 is not divisible by 3, 3 is. 4 isn’t. 5 isn’t. 6 is. In fact, the only numbers in this product that concern us are those divisible by 3. 3, 6, 9, 12, 15, 18, 21, 24, 27, 30.

Here it’s important to note that each of these components except the three alone has multiple prime factors. The three is just a three. The six is three and a two. The nine notice has a second factor of three. Three times three is nine and because we’re looking at the prime factors it has two. It’s difficult to get your head around but there are not three factors of three in nine when you’re counting prime factors.

Three factors of three would be 3 by 3 by 3 = 27. So notice that 3 and 6 have a single factor. 9 has a double factor. Every number divisible by 3 has one factor. Those divisible by 9 like 9, 18 and 27 are going to have a second factor and those divisible by 27, that is 3 cubed, are going to have a third factor. If we lay it out like this we see ten numbers have a single factor. Another of those three provide a second bringing us to thirteen. Finally, one has a third bringing us to fourteen. Answer choice: C.

#### GMAT Problem Form

So let’s take a look at this problem by writing a new one just to reinforce the algorithm. For the number 100 factorial. How many factors of seven are there? So first we ask ourselves out of the 100 numbers which ones even play? 7, 14… 21 so on and so forth. 100 divided by 7 equals 13. So there are 13 numbers divisible by 7 from 1 to 100. Of those how many have more than one factor of 7? Well we know that 7 squared is 49. So only those numbers divisible by 49 have a second factor. 49 and 98. There are none that have three factors of 7 because 7 cubed is 343. If you don’t know it that’s an identity you should know. So here our answer is 13 plus 2 = 15.

Try a few more on your own. This one’s great to do as a problem form and take a look at the links below for other abstract number theory, counting prime type problems as well as a selection of other really fun ones. Thanks for watching guys and we’ll see you soon.

If you enjoyed this GMAT factors problem, here is an additional number theory type problem to try next: Wedding Guest Problem.

Posted on
23
Apr 2021

## Standard Deviation – Clustering (Birds) Problem

Hey guys! Today we’re going to take a look at a DS problem that is a skills problem, focused on GMAT standard deviation.

#### Standard Deviation & Variance

What they’re asking here is do we have enough information to compute a standard deviation? It’s useful to think of standard deviation as clustering. If we have a whole series of points we can define how clustered or un-clustered the group of points is. That’s all that’s standard deviation, that’s all that variance is. So if we have all the points that works. What we should be on the lookout here for are parametric measurements. Especially things like the average number is, because while the average can be used to compute standard deviation, we need to know how each of the points differs from the average. But if we have each of the points we always get the average. That is, we can compute the average. So the average is a nice looking piece of information that actually has little to no value here. So let’s jump into the introduced information.

#### Statement 1

Number 1 BOOM – tells us that the average number of eggs is 4 and that’s great except that it doesn’t tell us about the clustering. If we run some scenarios here we could have every nest have 4 eggs or we could have 5 nests have 0, 5 nests have 8, or 9 nests have 0, 1 nest has 40. These are all different clusterings and we could end up with anything in between those extremes as well. So number 1 is insufficient.

#### Statement 2

Number 2: tells us that each of the 10 bird’s nests has exactly 4 eggs. What does this mean? We have all 10 points. They happen to all be on the average, which means the standard deviation is 0. that is there’s no clustering whatsoever. But 2 gives us all the information we need so B – 2 alone is sufficient is the answer here.

Hope this was useful guys, check out the links below for a video about how to compute standard deviation as a refresher, as well as other problems related to this one. Thanks for watching we’ll see you again real soon

If you enjoyed this GMAT problem, try another one next: Normative Distribution

Posted on
14
Apr 2021

## GMAT Percentage Problems

Hey guys, GMAT Percentage problem/s are commonplace on the GMAT and today we’re going to take a look at one that is straightforward but could very easily get you caught up with the math. In this problem, notice that there’s the word “approximately.” That always means there’s an Estimation Solution Path. We’ll take a look at that first but then we’re going to look at a Scenario Solution Path, which for many people is a lot more natural. In addition to seeing that word approximately you can see that there’s this massive spread within the answer choices. Once again pushing us towards an Estimation Solution Path.

#### Estimation Solution Path

So let’s dive in: The unemployment rate is dropping from 16% to 9% and your quick synthesis there should be: okay it’s being cut about in half or a little less than half. And monitoring that directionality is important. Additionally, the number of workers is increasing. So we have lower unemployment but a greater number of workers. So we have two things, two forces working against one another. If the number of workers were remaining equal then our answer would be about a 50% decrease or just under a 50% decrease, so like 45% or something like that. But because we’re increasing the number of workers, our decrease in unemployment is lower. That is we have more workers, so we have a larger number of unemployed so we’re not losing as many actual unemployed people and therefore our answer is B: 30% decrease.

#### Scenario Solution Path

If we want to take a look at this via Scenario, we can always throw up an easy number like 100. We begin with 100 workers and 16% are unemployed so 16 are unemployed. Our workers go from 100 to 120. 9% of 120 is 9 plus 0.9 plus 0.9 = 10.8% or 11%. What’s the percentage decrease from 16 to 11? Well it’s not 50, that’s too big. It’s not 15, that’s too small. It’s about 30 and the math will bear us out there.

So thanks for watching guys! Check out the links below for other GMAT percentage problem/s and we look forward to seeing you again real real soon.

Another GMAT percentage problem

Posted on
08
Apr 2021

## GMAT Factorial Problem: Estimation & Scenario Solution

#### GMAT Factorial Introduction

Factorials and divisibility, together. Two mathematical kids from opposite sides of the tracks, they come together and fall in love and they create this problem. Here we’re asked what numbers might divide some new number 20 factorial plus 17. As a refresher, a factorial is simply the number times each integer below it. So in this case, 20! is equal to 20 x 19 x 18 …. x 3 x 2 x 1. It’s a huge number. And it’s not at all possible to process in GMAT time. What we want to notice about any factorial is that it has as factors every number that it contains. So 20! is divisible by 17, it’s divisible by 15, it’s divisible by 13, 9, 2, what have you and any combination of them as well.

#### What The GMAT is Counting On You Not Knowing

When we’re adding the 17 though, the GMAT is counting on the idea that we don’t know what to do with it and in fact that’s the entire difficulty of this problem. So I want you to imagine 20! as a level and we’re going to take a look at this graphically. So 20! can be comprised by stacking a whole bunch of 15’s up. Blocks of 15. How many will there be? Well 20 x 19 x 18 x 17 x 16 x 14 times all the way down the line. There will be that many 15’s. But 20! will be divisible by 15. Similarly, by 17, by 19, by any number. They will all stack and they all stack up precisely to 20! because 20! is divisible by any of them.