GMAT MATH DECONSTRUCTED

GMAT MATH DECONSTRUCTED

The always necessary DISCLAIMER – I've tried to create as precise, correct and effective a guide as I possibly could. Nonetheless, there is no guarantee whatsoever that this guide adequately reflects the GMAT test, or that your performance will improve by using it. USE AT YOUR OWN RISK, and good luck!

CHAPTER 0 - Formalities

IntroductionThe GMAT can be challenging. We all know that. What follows is a teaching document that will describe a set of effective tools and approaches to the math part of the test. With any luck, reading this, and practicing these tools will make GMAT math a lot easier, and quicker for you.

What sets this document apart is that beyond teaching you the math you need to know, it also tries to provide some approaches and techniques to problem solving on the GMAT that should make everything quicker and less fraught.

Key conceptsI'm going to try and convince you of two things.

First, that nearly every GMAT question falls into one of several mathematical categories. There is a limited set of tools and abilities the GMAT requires you to know in each of those categories. This document will clearly delineate those categories, and the tools you need to know in order to solve problems in each category. I would argue that if you can consistently identify which category a questions belongs to, and choose the most suitable tool in that category to apply to the problem, you will do well on the test.

Second, that definitions on the test are key, and the ability to test a definition is fundamental to success on the test. If the test tells me that n is even, understanding the definition of an even integer, and being able to test whether something is, or is not, an even integer, are critical to my ability to solve the problem.

Obviously, we'll come back to these points often.

Also, I'm going to review a lot of basic math. If you lost the trail of mathematics in primary school, this might be an opportunity to pick it up again. We'll be doing some basic, basic stuff, but while we're at itwe'll learn some effective tools for everyone even the pretty math savvy. So it'll be worth everyone's time.

That being said there is one thing I won't review here. You need to know the multiplication table to 10.Sorry. There's not much to this. Memorize that crap. You don't get a calculator on the test, and even if you did, it wouldn't help you realize that 42 is 6x7. You need to have that in your head immediately when you see that number on a GMAT question. That kind of skill will make problem solving infinitely

GMAT Deconstructed 1 mathdeconstructed.com

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easier for you. Realizing that 72 has something to do with 8 and 9 means you're already thinking about possible ways to solve the problem, and that is a key skill on this test.

Why GMAT math is challengingNews flash - You learnt most of the math on the GMAT by the time you got to high school. Very few questions on it are profoundly complex mathematically. No really! So why is it so difficult, then?

Reason #1 - TimeIf I gave you all the time in the world, you would correctly solve virtually every GMAT problem that ever existed. Unfortunately, you only have 75 minutes to solve 37 questions, which brings us to our first math problem.

How many minutes do we have, on average, to solve each of the questions? This can also be phrased as – How many minutes do we have PER question?

When we have a sentence like that – How many A per B? - we are being asked to take all of our As anddivide them equally between all of the Bs, or in mathspeak – A/B. The division line is simply a graphical notation replacing the word “per”. Here we are specifically dealing with minutes per Question, and so we would write – Minutes / Question, and since there are 75 minutes, and 37 questions what we really need to calculate here is – 75/37.

Great, but the result isn't very obvious. It's not a nice round number, and at least some of us have no idea where to start hacking away at this problem. Here's one approach that is really effective on the GMAT. Let's get some idea of how these two numbers RELATE to each other. Is 75 many times bigger than 37? or is it just a little greater than 37?

Getting a broad handle on how two quantities relate on the GMAT is a powerful skill because it gives us a clue about the scale of the expected result. Is this going to be closer to 10 or closer to 1? Figuring that out is already a big step towards getting an answer, and in some cases, will be all that is required to solve the problem!

You may have figured out by now that 2x37 = 74. Which is really really close to 75. That means that when we divide 75 by 37 we'll get something very close to 2.

Well, here is another important question. Is it more than 2 or less than 2? I can see that easily as a result of our attempt to gauge the relative size of the numbers. If I had exactly two minutes for each ofthe 37 questions, I would have 74 minutes to take the test. Since I have 75, it means I have a little more than 2 minutes for each question on the test.

Well, 2 minutes is not a lot of time, which is the primary reason the GMAT is as challenging as it is.

We'll be focusing on a problem solving approach that will help you tackle GMAT problems more efficiently, as well as techniques to help you speed up your arithmetic. That being said, we're not going to spend much time on time management strategies. As you practice the approach and tools in



this document, you'll get more efficient, and time should slowly become less of an issue for you. I recommend focusing on effectively employing the tools you will learn here before worrying about your solving speed.

One strategy caveat. There is apparently a meaningful penalty for not completing a section on the GMAT, so you should absolutely figure out how to answer ALL the questions in a section. Ideally, you'll get to the point where you simply have the time to do that, but if you find yourself in a pinch, make sure to guess the last questions you are faced with to make sure you complete the section. We'll also talk about quick ways to improve your guessing odds, if you run into a situation where that is necessary.

Reason #2 - Layering of conceptsYou won't typically see a question as simple as the “how many minutes per question” one on the GMAT. It would be a little too easy. One of the classic approaches the GMAT takes to creating questions is layering concepts. You might start with a geometry question about triangles, only to find out that it is, at it's heart, an exponent problem, or and algebra one, or a counting question. By combining these two different concepts in one question the GMAT is creating a higher level of difficulty.

The approach we'll practice here focuses on identifying what topic each of the stages of the question relates to, and what the best relevant tool to solve that part is. Doing this will help you to methodicallywork through the different stages of the problem, and solve it within the time you are allotted.

Reason #3 - No ExcelMost of us haven't used pencil and paper, or for that matter a calculator to do arithmetic in forever. We're super super rusty. And then you take the GMAT and there's no spreadsheet to add up numbers for you, no calculator to divide or multiply.

Further, because you don't have much time, the ability to do calculations in your head, becomes radically important, because it saves you the bunch of time that is wasted by switching between the computer screen, and the piece of paper you're scratching out your notes on.

We'll spend some time early on learning some useful basic arithmetic techniques that help carry out all these calculations mentally, but there is no magical alternative to practice. Once we go through the arithmetic part – start making a habit of calculating things mentally. Try for a dozen times a day. Make up problems if you have to. The single most important skill for GMAT success is confidence in your mental calculating skills.

A Map of the UniverseGood. Enough talk. It's time to take a look at some actual math.

The basic elements of all math problems on the GMAT are, as expected, numbers. Numbers are grouped into several groups (also called Sets).



Counting (or Natural) numbers. Surprise surprise – these are the numbers we use to count things. 1,2,3, etc...

Add a Zero to that group of numbers, and you get the next set -

Whole numbers. Yup. These are just all the group of counting numbers, with zero added to them. 0,1,2,3, etc...In math, we represent Sets with curly brackets - {}. You can think of Sets as boxes with things in them. This box – {0} has a zero in it. This box {0,1} has a zero and a one. This box {1,1} has two ones. This box has a pencil in it {}. We'll talk about them more later on, but the reason we're touching on it here is that even though this is a basic concept, the GMAT pays attention to it (this from a really old GMAT question):

{Whole Numbers} – {Natural Numbers} = ?

What they are asking here is simple – If we take a box with all the Whole numbers in it, and remove allof the Natural numbers from that box, what are we left with? If you review the descriptions of these sets above you'll quickly realize this just leaves zero in the box. Therefore the answer is {0}. (Note, it is still a zero, inside a box. Not just the number – 0).

Ok. Now let's add all the negative versions of the Counting numbers to the group of whole numbers. So if we had 0, 1, 2, 3, etc... now we also have -1, -2, -3, etc. This gives us a very important group, at least as far as the GMAT is concerned -

Integers - … -3, -2, -1, 0, 1, 2, 3, …. A huge number of GMAT questions is devoted to integers. This is probably because while they are very very simple numbers, they work in very very specific ways. We'll spend the whole of Chapter 1 reviewing these and seeing how the GMAT likes to play with integers.

Now, some of you may have noticed that there are gaps between the integers. For example, my half eaten cookie cannot properly be described as 1 cookie, nor as 0 cookie. It is one cookie, now divided into 2 parts, or – ½ a cookie. ½ is part of the group of Rational numbers.

DEFINITION - A Rational number is a number that can be described as a ratio between two integers.

This is a definition. Definitions are really really important in math generally, and on the GMAT in particular. Most importantly, when we have a definition, we have to make sure we know how to test whether something falls under this definition.

To test whether a number is a rational number, I have to find a ratio between two integers that represents that number.

Is ½ a rational number? Well, it is the ratio between 1 (which is an integer) and 2 (which is also an integer). I can represent this number as a ratio between two integers, and therefore it IS a rational number.



So is ¾ (the ratio between 3 and 4, both integers) and 7/5. So is 2½. Because I can represent it as the ratio between 5 (an integer) and 2 (also an integer). If this last example confuses you, don't worry. We'll dive into fractions in great detail in Chapter 2.

So here's a question. Is 4 a rational number?

Well, I'm glad you asked. How can we find out??? We can test it according to the definition! Can we find two integers, the ratio of which represents 4? I'll give you a second to think about that one.Clearly, the answer is yes – here are a couple of such pairs of integers – 4 and 1 (4/1 = 4), 16 and 4 (16/4 = 4). I can definitely represent 4 as a ratio of two integers, and therefore it is, by definition, a rational number (Try the same test for 0).

You can also now see, that every integer is also a rational number, but the opposite is not true. There are many rational numbers that are not integers (for example – ½).

Are there, you may ask, numbers that cannot be represented by the ratio of two integers? And that would be a brilliant question. One that many ancient mathematicians spent a great deal of time contemplating. It turns out that there are, and these group of numbers are called -

Irrational numbers – for example – π, √2 and many others.

By definition these are numbers that CANNOT be represented by a ratio of two integers, and thereforeno rational number can be irrational or vice versa. They are mutually exclusive groups.

Proving that a number CANNOT be represented by a ratio of two integers is far more complicated thanproving that it CAN. Fortunately, you don't have to worry about doing that on the GMAT.

The GMAT uses irrational numbers in a fairly simple way. For example:

Q - 5⋅√2−3⋅√2=?(A) 2(B) 15⋅√2(C ) 30(D) 2⋅√2(E ) 4

The irrational number √2 can essentially be replaced by whatever random symbol we choose. The question remains the same – 5$ - 3$ = ?

This might be a little simpler, right? If you have 5 dollars and someone takes away 3 dollars, what do you have left? 2 dollars. By the same coin (all puns utterly intended) if you had 5 square roots of 2, and someone took away 3 of your square roots of two, you'd be left with 2 square roots of two, or in this case – Answer (D). (Also, 5X-3X = 2X, etc.)



In many cases, irrational numbers will be used like this. We'll be working around them. Treating them like shapes, or icons, that don't have a real mathematical meaning to us.

We'll deal with the smaller number of cases where we actually need to do math with irrational numbers (mostly dealing with square roots and such) later, but the only other thing you need to know about irrational numbers for now is that you need to memorize the approximate values of the following three irrationals:

√2 ≈ 1.41, √3 ≈ 1.73 and π ≈ 3.14

The reason you should memorize these values is that they often pop up on the GMAT in approximation problems. The GMAT leverages the time crunch it creates in a group of problems wherecalculating the actual result would take an inordinately long amount of time, but where you could figure out the solution with a quick and rough approximation. For example, figuring out the exact value of 3⋅√3 is not easy. But if we remember that √3 ≈ 1.73, it's easy to see that 3⋅√3 is less than 6 (which is 3x2) and more than 4.5 (which is 3x1.5). We've greatly narrowed down the range of possible results, without resorting to long drawn out calculations that we might not even know how togo about.

Approximations in general are an extremely powerful tool on the GMAT. Whenever you are faced with a value or a variable in a GMAT question, getting a general idea of what this element is, and a ballpark of how big or small you might expect it to be is very helpful in increasing your solving speed, correctness, and if you're guessing answers, your probability of success. We saw its effectiveness in our very first problem (how many minutes per question?) and will continue to see its usefulness as wepractice using this important tool.

Key topics covered- Power of approximations- Importance of definitions- Types of numbers- Irrational number values to memorize



CHAPTER 1 – The Integers and other basics

The integers are one of the most important groups of numbers on the GMAT. Though they seem simple, they have a wide range of characteristics for problems to be built around. In this chapter we'll plumb the depths of the integer world as well as get some basics and vocabulary out of the way.

Arithmetic VocabularyThere are four basic arithmetic operations, and they carry some vocabulary baggage with them:

Sign Operation Components Result

+ Plus Addition Terms Sum

- Minus Subtraction Addend / Subtrahend Difference

X Times Multiplication Factors Product

/ Division Numerator / Denominator Quotient

The plus sign (+) represents addition. The things we are adding together are called terms, and the result of the addition is called a sum.

The minus sign (-) represents subtraction. Don't worry about the names of the things we are subtracting, just remember that the result of the subtraction is a difference.

The times sign (x) represents multiplication. The things we are multiplying are called factors, and the result of the multiplication is a very important word on the GMAT – a product, as in, for example - “The product of 4 and 6 is 24.”

This division sign (/), not surprisingly, represents division. Again, we're not going to worry about what the things we are dividing are called right now, but the result of the division is a quotient.

We don't frequently use these words in everyday conversation, so naturally, the GMAT uses them very often. You know, just to make things difficult. If you aren't scoffing at how obvious this all is, take a second to review and memorize these terms.

Great. So about those Integers. Integers can be divided in many different ways. The first and most straightforward is to divide them into positive and negative integers.

Positive / NegativeA positive integer is one that is greater than 0. A negative integer is one that is less than 0. Simple enough.



But what does that make 0 then? Positive or negative? Well, it makes it neither. 0 is neutral. Which allows the GMAT to define groups like the “non-negative numbers”, which simply means the positive numbers and 0.

One important piece of information to know about positive and negative numbers, is how they behave when they are multiplied or divided:

multiplication / divisionFirst factor Second factor Result

+ + +

+ - -

- + -

- - +

Note that when both factors have the same sign, whether positive or negative, the product, or quotient will end up positive. If the signs are opposite, the product or quotient will be negative.

Q – If a quotient st

is positive, which of the following must be true?

(A) s > 0(B) t > 0(C ) st > 0(D) s+t > 0(E ) s−t > 0

A – First off, let's take a second to ask a very important question – what KIND of question is this? WhatTOPIC does it pertain to? Well, we see the words quotient and positive right there in the question, which should set our spidey sense tingling that this is a multiplication / division of positive and negative values question.

Great! We have a table for that!!! If s/t is positive that tells us that one of two things is true. Either s and t are BOTH positive, or they are BOTH negative.

If we look at answer choice A – is it true that s is greater than 0? It COULD be true (is both s and t are positive), but it could also NOT BE true (if both s and t are negative). The question asks which of the following MUST be true (as in ALWAYS), A is not NECESSARILY always true, and therefore it CANNOT bethe answer.

Ditto for B for the exact same reasons. If we know the table we are using well, we'll immediately realize that C is ALWAYS true – if the quotient is ALWAYS positive, the product will also ALWAYS be positive.So the answer is (C).



Let's take a quick look at D and E just to see what is going on – we know that both S and T are the same sign (positive or negative), but we don't know which. If they're both positive then their sum will also be positive, but if they're both negative – the sum will be negative! Again, there is no certainty that D will be true. We also don't know the RELATIVE size of s and t. So we have no idea whether the difference between s and t will be positive (which is the same as saying that s is greater than t), or the opposite. Again – nothing certain here.

Cool! Moving right along, let's look at another important way to divide integers – into even integers and odd integers.

Even / OddDEFINITION – An even number is a number that is divisible by 2.

(Important note: Definitions are the basic tools of our problem solving efforts – you'll notice below I leverage them to derive information from statements in the GMAT questions. You'll be doing yourself aservice to embed them in your brain solidly!)

Wait. What on earth does “divisible by” mean?

DEFINITION – Integer n is divisible by k if n/k is an integer. (See – the definition has given us the test – If when we divide n by k we get an integer, then we can say – n is divisible by k. Conversely, if the GMAT tells us in a question that n is divisible by k – what they are really saying is that if you divide n byk you get an integer. We'll see why that's valuable information in a moment).

So, is 9 divisible by 3? Let's check. How do we check??? We look to see if 9/3 is an integer. Simple as that. That's why definitions are so effective. They give us a simple test to see if something fits the definition.

9/3 = 3. Since 3 is an integer, by definition, 9 is divisible by 3.How do we know if X is even? We check if it is divisible by 2. How can we tell if it is divisible by 2? if X/2 is an integer, then X is divisible by 2, and by definition is even.

This may all seem a little complicated right now (it'll become much clearer shortly), but the important point here is that whenever we have a definition of a term, it will define a test. A way to figure out if something falls under that definition or not. The test is the important part here. We need to understand what it is we need to check, and remember it.

Some examples of even numbers are – 2, 12, 148, 40,343,555,226. Since these numbers are all divisible by 2, it means that they have the form – 2k, where k is an integer. Why is this so? Because when we take a number with the form 2k, and divide it by 2 to check whether it is even, we get 2k/2 = k, which IS an integer. Therefore any even number can be presented in this fashion:

n=2k, where k is an integer.



DEFINITION – An odd number is a number that is NOT divisible by 2.

Odd numbers can be represented by the form m = 2k+1, where k is an integer. (If you are curious about why this might be the case do the test for divisibility by 2 on m, that is – check to see if m/2 is aninteger, assuming that m=2k+1, where k is an integer).

Question, then. Is 0 even or odd? That's a simple thing to check. We need to find out if 0 is divisible by2. We test that by checking whether 0/2 is an integer. 0/2 is equal to 0 (in fact 0 divided by anything, other than 0, equals 0). 0 is, in fact, an integer, which means that 0 is divisible by 2 and therefore even.

What do we get when we add an even number to an even number? An even number. Here's why. Let'scall one of the numbers 2r and one of the numbers 2s (where r and s are integers). Then the sum of these two numbers is 2r+2s = 2(r+s). Since r and s are both integers, r+s must also be an integer, and thus the sum is of the form 2k (k here equals r+s), where k is an integer, also known as – even.

Here's what happens when you use basic arithmetic operations with odds and evens:

Addition / Subtraction Multiplication1st term 2nd term Result 1st factor 2nd factor Product

Even Even Even Even Even Even

Even Odd Odd Even Odd Even

Odd Even Odd Odd Even Even

Odd Odd Even Odd Odd Odd

let's start with the addition / subtraction table. First off, why are these two operations (addition and subtraction) included in the same table? Well, subtraction can be thought of as adding a negative termto the first term. Positivity, or negativity have no bearing on divisibility by 2. If a number is divisible by 2 (i.e. even) than the negative of that number is very much equally divisible by two. Thus, as far as evenness is concerned, subtraction and addition are identical operations.

The key thing to notice here is that to get an even sum or difference, the terms MUST BE the same. Either both must be even, or both must be odd. If one of the terms is odd and the other even – the sum or difference will always be odd.

Here's a quick side question. Why doesn't it matter whether the first term is odd and the second even,or vice versa? Well, for addition that's pretty clear. 2+3 is exactly the same as 3+2. In fancy pants words, addition is commutative. But subtraction is not. 3-2 is not the same as 2-3. 3-2=1, while 2-3= -1.

But if we look at a couple more examples, we might get what is going on here – 7-3=4, 3-7= -4.12-5=7, 5-12= -7. When it comes to subtraction – flipping the order of the terms results in the same value but of opposite sign. If we have a-b = c, then we know that b-a will be equal to -c !



In short hand we write: (a-b) = -(b-a). But for the test it is a great result to memorize. Need to flip the terms around a subtraction for some reason? No problem, you'll get the same value, but of the opposite sign.

Now since signs are irrelevant to oddness or evenness, as far as that is concerned, we can flip around the order of the subtraction all we want, and the resulting even or oddness will stay the same.

Here's an example of how the “flipping the subtraction order” concept can be powerful:

Q - If c-d = 5 then 5d – 5c = ?

(A) −25(B) 5(C ) −5(D) 25(E ) 0

Pretty straightforward here if you remember the flipping thing. 5d-5c = 5(d-c). d-c is c-d flipped, therefore d-c = -(c-d) = -5. So then, 5x(d-c) = 5x(-5) = -25, and the answer is (A).

Clearly we could rearrange algebraically, and substitute (c-d=5, so c=5+d, substitute into 5d-5c, and you get 5d-5(5+d) = 5d-25-5d = -25), but that takes longer, is more confusing, and more prone to errors. It's important to have this solving skill as a fallback, and we'll study it in depth in the Algebra chapter, but the concept of flipping the order of a subtraction is a useful and powerful one.

Now let's look at the multiplication table. This one is really interesting. Notice that the only way to get an odd product is if all the factors are odd. The moment we get even one single even factor into that product, it immediately means that the product is divisible by 2 and therefore even.

3x3x3x3x3x3x3x3x3x3x3x3 is odd, but add even one little 2 in there, and 3x3x3x3x3x3x3x3x3x3x3x3x2 is even.

All of the above results are true for any even or odd number. We won't take the time to methodically prove these, but it is important to get really comfortable with them because so many of the GMAT even and odd questions relate to these characteristics. For example:

Q - Integers c and d are not both odd. Which of the following MUST be odd?

(A) 5c(B) c+d(C ) c⋅d(D) c −d(E ) 2 (c+d ) −1



A - The approach to multiple choice questions on the GMAT that I'll espouse here, is to first off the bat, get an idea of what topic the question pertains to. As we move through the chapters, we'll obviously gather a bigger toolbox of topics and solving implements, but at this point this step is easy. The first word in the question is “Integers”, so it clearly has to do with integers, and the first sentence in the question ends with the word “odd”, which clearly points us in the direction of even/odd questions.

Well. So far, I've only given you one tool to deal with even and odd integer questions. The table above.So it's safe to assume we're going to use it in solving this problem, but there's one step we need to do first. We need to tease out all the information this question gives us about c and d.

It tells us that they are “not both odd”. What does that mean? Only one, very limited thing. It is not the case that both a and b are odd. They can be any other combination of odds and evens. To be very methodical about this just this first time, let's lay it out:

C D Possible?

Even Even Yes

Even Odd Yes

Odd Even Yes

Odd Odd No!!

So this piece of information is not very limiting. Nonetheless, it is critical to clearly understand exactly what it means, or it becomes impossible to figure out what is going on here.

Next, the question asks, “which of the following MUST be”. This is a very common phrasing on GMAT questions, and it indicates that some of the options we will find in the answer choices MAY sometimesbe even, but not ALWAYS. We're looking for the answer that is ALWAYS even.

Answer choice A only involves c. If c is even, the result will be even, if c is odd, the result will be odd. This expression (5c) can clearly turn out to be EVEN, and therefore it is not true that it MUST be ODD.

We can also rule out the additions and the subtractions. A quick glance at the +/- table shows us that if both c and d are even (which is possible, but NOT CERTAIN) their sum or difference will be even, while if one of them is odd and the other even (which is ALSO POSSIBLE), the sum or difference will beodd. From what we know about c and d (NOT BOTH ODD), we find that it is POSSIBLE that the sum or difference between c and d will be odd, but it is NOT CERTAIN.

Answer choice D involves the product of c and d. As we mentioned when we reviewed the multiplication table – the only way to get an odd product of two numbers, is if both numbers are odd. Since we are certain that c and d are not both odd (it says so in the question!) we can be certain that cd is definitely even. Clearly this product can NEVER BE ODD.



Which leaves us with E as the correct answer BY DEFAULT.

But let's look at E and see why we could have jumped straight to it and called it a day. The first part of the expression – 2(c+d), is clearly even. Because c and d are integers, their sum is also an integer. Any number that can be written in the form 2k where k is an integer is divisible by 2 which means it is even. So if the front part is even, and then we add 1 to it, clearly E will always be odd!

Of course, this is the generalized form of an odd number that we learned about – 2k+1, where k is an integer.

Since it will lead us to cover very interesting ground, let's dig a mite deeper into the reasons why this represents an odd number.

FACT - In a sum, if one of the terms is divisible by integer k, for the sum to be divisible by k, the other term MUST ALSO be divisible by k.

Um, what? let's look at an example to help us understand. If I have a sum, e.g. a+b. If a is divisible by 5, for the sum (a+b) to be divisible by 5, b MUST ALSO be divisible by 5.

5+2 = 7. 5 is divisible by 5, obviously, but 2 isn't divisible by 5, and therefore 7 isn't divisible by 5.

5+25 = 30. 5 and 25 are BOTH divisible by 5, and therefore 30 is as well.

But, you might say, 3+7 = 10 and 10 is divisible by 5 despite the fact that neither 3 nor 7 is divisible by 5. Good point. Note that the fact above only describes what must be the case IF WE ALREADY KNOW that one of the terms is divisible by the number we are checking for. If neither of the terms are divisible by that number it does not mean that the sum will not be divisible by it. This fact only pertains to cases where one of the terms IS divisible by an integer we are checking.

More examples: Neither 3 nor 7 are divisible by 5, and their sum (10) is. Neither 3 nor 9 are divisible by 5, and their sum (12) isn't. This fact is utterly useless if neither of the terms are divisible by the number we are testing.

But, 3 and 9 are both divisible by 3, and therefore their sum (12) must be. 3 IS divisible by 3, but 7 IS NOT, which necessarily means their sum (10) will NOT be divisible by 3.

First term divisible by k Second term divisible by k Sum divisible by k

Yes Yes Yes

Yes No No

No Yes No

No No ????



How does this relate to 2k+1 representing odd numbers? Well, 2k is clearly even (i.e. divisible by 2). 1 on the other hand is not divisible by 2. If one of the terms in a sum is divisible by a number, and the other is not, then the sum is not divisible by that number, and so in this case the sum – 2k+1 is not divisible by 2 and therefore necessarily odd.

Let's take a short detour to see how the table above comes into play in GMAT questions:

Q - If n=21! + 21, then n is divisible by which of the following?

I. 7II. 19III. 21

(A) None(B) I only(C) II only(D) I and III(E) II and III

A – Before anything else – let me point out that this is clearly a DIVISIBILITY question. No need to dig very deep to figure that one out. Next, let me introduce you to the mathematical operator starring in this question – some of you may be wondering, what on earth is 21! ??

This is the symbol for the FACTORIAL. It represents the product of all the positive integers up to the number indicated. Thus 5! would be – 1x2x3x4x5 = 120. We'll reunite for a longer introduction to the factorial much later in the book, but for now all we need to understand is the following:

While 21! is a rather large number, we know one thing for absolutely sure – it is divisible by every integer from 1 to 21. n is the sum of 21! and 21. If 21! AND 21 are both divisible by a certain integer, then n MUST ALSO be divisible by that integer. 21 is divisible by 3, 7 and 21, 21! is also. So n is certainly divisible by 3, 7, and 21 (both terms are divisible by all these numbers). 21! is in fact divisible by 19, but 21 IS NOT. If one term of a sum IS divisible by a certain integer, and the other IS NOT, then the sum IS NOT divisible by that number. Therefore, of the three options n is ONLY divisible by 7 and 21.

So the answer is D.

Ok. So you may wonder why I haven't shown you a division table regarding evens and odds (we've hadaddition, subtraction and multiplication)? Good question. let's just take a quick example -

4/2 = 2. So here even / even = even.

But 6/2 = 3. So here even / even = odd.



Not to mention that often when dividing integers they are not evenly divisible (that is, the result is notand integer). Something more complex is going on here, and we'll get to the bottom of it a little later in the chapter. Meanwhile, all we'll say is that there is no hard and fast division rule like with addition, subtraction and multiplication.

Decimal notation and DigitsMany of the numbers on the GMAT are noted in decimal notation. This is a fancy word for the usual way we write numbers, for example:

1 3 2 7 . 9 8 2thousands hundreds tens units decimal tenths hundredths thousandths

point

This number, 1327.982 can also be written as – 1x1000 + 3x100 + 2x10 + 7x1 + 9x0.1 + 8x0.01 + 2x0.001, which is to say – one in the thousands place, three in the hundreds place, etc.

The numbers in each of the spaces around the decimal point are called digits. The digits are the group of whole numbers less than 10, or in other words – the whole numbers from 0 to 9. Once we know that a number on the GMAT is a digit, it really limits the options of what that number can be. For example:

Q – In the two digit number MN, N > M. If N is odd, how many options exist for the value of N?

(A) 1(B) 2(C ) 3(D) 4(E ) 5

A – Notice – MN the two digit number, and N is odd. All indicators that we are dealing with a number properties question, that has to do with DIGITS.

Now, since M and N are digits they are whole numbers smaller than 10. In addition, M cannot be 0, because if that was the case, MN would not, in fact, be a two digit number. So M is at least 1, which means N is at least 2. But N is odd, so the lowest possible value for N is 3. That leaves 4 total options for the value of N – 3, 5, 7 and 9, so:The answer is D.

As we mentioned a decimal number like 1234 can be written algebraically as 1x1000 + 2x100 + 3x10 + 4x1. This fact can be useful in digit questions where we have to add together numbers. For example:



Q – If positive integers E and F are both two digit numbers, and have the same digits but in reverse order, which of the following CANNOT be the sum of E and F.

(A) 181(B) 154(C ) 121(D) 77(E ) 33

A - Clearly this question is an integer question (it's the third word in the question). It is also a digits question. Unlike the previous digits question, this one deals with the two digit numbers, not with the digits themselves. It helps to try and visualize the two digit numbers. let's say the tens digit of E is a, and the units digit is b. We could then write that E = 10 x a + b. F in that case would be F = 10 x b + a. Then the sum of E and F, E+F = 10xa + b + 10xb + a = 11xa + 11xb = 11x (a+b). This reveals something interesting – the sum MUST be a multiple of 11 (it is equal to 11 times an integer). The sum of any twodigit integers that have the same digits but in reverse would always be a multiple of 11.

We can also see that answers D and E are multiples of 11, and with the additional knowledge we'll garner in a later chapter, we'd also immediately notice that 121 is 11 x 11. So three of the 5 answer choices are clearly multiples of 11. Are any of them NOT multiples of 11? We'll learn some fast tools tofigure that out later in this chapter, but in the meantime – 154 is 121 + 33. Since we know 121 is divisible by 11, and 33 is divisible by 11, 154 must be divisible by 11 (remember that from a little earlier?). 181 on the other hand is 154 + 27. 154 is divisible by 11, but 27 is NOT, which means the sum of the two numbers is NOT divisible by 11, and therefore CANNOT be the sum of E and F.

The answer is A.

RoundingOne last note about decimals. We can speak about rounding a decimal to a certain place. For example if we round a decimal to the nearest tenth we would ignore all places to the right of the tenths place, and change the digit in the tenths place to represent the closest value to our original number. For example if we wanted to round 11.48 to the nearest tenth, we would have two options – write it as 11.4 or 11.5. But clearly 11.5 is closer to 11.48 than 11.4 is. So 11.48 rounded to the nearest tenth is 11.5.

The RULE for rounding is simple, if the digit in the place to the right of our rounding place is 5 or greater we add 1 to the digit in our rounding place (in the example above we added 1 to the 4 in the tenths place). If the digit in the place to the right of our rounding place is 4 or less, we leave the digit in our rounding place unchanged (note – we do NOT decrease the digit!).

let's take our decimal from the beginning of this section – 1327.982 and round it to both the nearest hundredth, tenth and ten.



Hundredth – Since we're rounding to the nearest hundredth we look at the thousandths place, the digit there is a 2, so we would round down. That means the digit in the hundredths place remains unchanged. 1327.982 rounded to the nearest hundredth is 1327.98.

Tenth – We look to the hundredths place – the digit there is an 8, so we round up. The digit in the tenths place is a 9, but 9+1 is 10... and 10 is not a digit. What this means is that we change the digit in the tens place to 0, and carry the 1 to the units place. 1327.982 rounded to the nearest tenth is 1328.0 (we don't actually have to write the .0, but I've left it in there to remind us we rounded to the nearest tenth).

Ten – We look to the units place, the digit there is a 7, so we round up. We add 1 to our tens place digit, and ignore all the rest. So 1327.982 rounded to the nearest ten is 1330.

Arithmetic tricksBefore we dive into the most important section of this chapter, let's discuss some mental math tricks that can help dramatically increase your solving speed without resorting to writing on paper. Learning and practicing these tricks is valuable, and important. Whenever you are practicing GMAT problems, be on the lookout for opportunities to practice these methods.

Subtraction1) Step by step:Clearly, subtraction problems like: 256 – 21 are not the most troubling. We can make things a little easier by breaking down the process into steps. The step by step method is one of the keys to simplifying mental math. We can start by subtracting 20 from 256, which leaves us with 236, then subtract the 1 = 235.

Here is a slightly more tricky example: 1256 – 832. We can first subtract 800, leaving us with 456 – 32, a more simple problem which equals 424.

2) Adjustments:In some very specific cases like 4926 – 399 a simple adjustment can make life more easy. Subtracting 400 from 4926 is easy – it equals 4526. But we're supposed to only subtract 399... so we need to add 1back, equaling – 4527. Obviously we can do this with many other types of situations like: 4926 – 729. Here an easier subtraction would be 4926 – 726. That equals 4200. But we need to subtract 3 more, so the result would be 4197. Still, a simpler process, that can be done mentally.

3) Adding up:The most powerful subtraction method, is in fact to flip the script. Take for example 11201 – 3985. That looks pretty tough. But let's add to the second number instead of trying to subtract. We'll try to make that second number a nice round one. If we add 15 to 3985 we get 4000. Nice and round. How much do we need to add to 4000 to get to 11201? That's a fairly easy problem, it's 7201. So the difference between 11201 and 3985 equals 15 + 7201. Together 7216. That's a quick and powerful method to carry out any subtraction in your head.



Multiplication1) Step by stepThis is where the step by step method gets even more useful.Let's start with a simple multiplication to see how this works – 8x11. 11 equals 10 + 1. so 8x11 = 8(10+1), or 8x10 + 8x1, or 80+8 = 88.

Now let's scale up – 23x15. In cases where both factors are more than one digit, we need to choose which one we will break apart. We CANNOT break apart both. In this case 23 is the trickier factor, the one easier to break apart. So we get 20x15 + 3x15. 3x15 is 45. We don't actually have to calculate 20x15. We can lose the trailing zeros and just calculate 2x15 = 30. Then we add back the zero we threw out to get 300. So 23x15 is 300+45 = 345. Pretty nifty.

Important note: As I mentioned before, I'm skipping a rudimentary prerequisite to this multiplication work, which is knowing the multiplication table to 10 by heart. You need to have full, comfortable, control of the multiplication table to 10 to the level that every number on that table is HARDWIRED to it's factors in your brain, and every time you see 56, 7x8 is hovering right there near it. If you're not there, you need to work to get there, because there is NOTHING that will affect improvement in your solving speed like solid command of the multiplication table. MEMORIZATION

2) Dropping zerosDropping the zeros makes it easy to calculate things like 3000x19. 3x19 is (break apart 19) 57. We dropped 3 zeros, so when we put them back we get 57000.

3) AdjustmentsHere too adjustments are useful. Multiplying 29x75 is more difficult than multiplying 30x75. That's 2250 (drop the zero, break apart 75). But 2250 is 30 times 75, and we only need 29 times. So we have to subtract one 75 (step by step) and we get 2175. All calculated mentally.

Note again that we CANNOT adjust BOTH numbers. If the problem is 29x39 we can't calculate 30x40 and try and figure it out from there... that's TOO complicated. Just pick one of the two to adjust.

With that under our belt, let's dive into the heart of integers – DIVISIBILITY.

Factors and divisibilityHere's some math you don't need me to explain to you:

4 x 6 = 24

This expression can be written in English four different ways (at least):The product of 4 and 6 is 24.4 and 6 are factors of 24.24 is divisible by 4 and 6.24 is a multiple of 4 and 6.and they all mean the same thing.



Every integer has a set of factors. These are the numbers it is a multiple of. The numbers that it is divisible by. In the case of 24 we can write them like this:

241, 2, 3, 4, 6, 8, 12, 24

As you can see – these factors come in pairs - 1x24 = 24, 2x12 = 24, 3x8 = 24, etc. Now, any integer can be presented in this way:

361, 2, 3, 4, 6, 9, 12, 18, 36

But, wait! Hang on a minute! Where's the pair for 6??! In this particular case where - 4x9 =36, but 6x6 = 36, 6 is its own pair. This is because 36 is the square of an integer. That is – there is an integer (6)that when squared (multiplied by itself) equals our number (36).

For now all you need to remember is that squares of integers have an odd number of factors, whereas all other integers have an even number of factors. We'll delve into this a little more later.

PrimesDEFINITION: A prime number is a positive integer that has only two factors – 1 and itself.

What this means is that primes are divisible by no other integer, which makes them really useful basic elements when we investigate integers and divisibility. They are the bedrock. You cannot break them down further.

It also means that 1 is NOT a prime – since it is NOT divisible by two factors, but rather, only by 1 - Itself.

Here is a list of the first few prime numbers. They're worth MEMORIZING.

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37

Some notes: Since every even number is divisible by 2, and therefore not a prime, the ONLY EVEN PRIME is 2 (it is divisible ONLY by 1 and itself).

Since every number ending in a 5 is divisible by 5, 5 is the only prime that has a units digit of 5. So all primes greater than 5 can only have a units digit of 1,3,7 or 9.



Ok. So why are these buggers so useful to us? Because the GMAT fricking LOVES THEM!! Also, they allow us to do a prime number factorization of any integer as follows:

242 12

2 6 2 3

Basically, we split the original number into any two factors that multiply together to give us the number – in this case – 2 and 12. Since 2 is a prime number we don't have to keep going down that branch. 12 is not a prime – so we break it down into 2 and 6 (we could have gone with 3 and 4 as well and got to the same result). 2 is a prime so we stop that branch right there. 6 is not – so we break it into 2 and 3, which are both primes, and therefore we are done. We have factored 24 into its prime factors which are – 2x2x2x3. We can also represent them this way – 23x3. If this freaks you out right now, ignore it. By the time we're done here, you'll be wondering why you ever freaked out in the first place. I promise.

Let's do that factorization one more time – just to prove that whatever path we take, we'll get to the same place -

244 6

2 2 2 3

So – 24 = 4x6. Neither is prime, so we continue both branches. 4 = 2x2 both of which are prime, so we can stop there. 6 = 2x3 which are also both prime, so we're done. Oh and look – 2x2x2x3.

In some other version we might have gotten 3x2x2x2, or 2x3x2x2 but these are literally identical representations to 2x2x2x3. Basically the components of 24 are three 2s and one 3.

FACT – Any number that divides 24 will only have some subset of those prime factors, and never moreof any prime factor than 24 had.

Q – Integer n =693011⋅k

where k is an integer. Which of the following is a possible value of k.

(A) 225 (B) 165 (C) 135 (D) 126 (E) 36

A - For n to be an integer, everything in the denominator (the bottom of the fraction) must cancel out with something in the numerator (the top part). Since we're dealing with integers, breaking everythingdown to prime factors will help here. 6930 = 11x7x5x3x3x2. Both 11s cancel out, so in order for everything in the denominator to cancel out, k can only contain the following prime factors – 7, 5, two3s and a 2. 225 (ans A) has two 5s in the factorization. If k had two 5s in its factorization, the second would not cancel out with anything on top and therefore n would not be an integer, so this answer is NOT a possible value. 165 has 11 as a factor, but since the only 11 in the numerator has already



canceled out, we'd be left with something in the denominator, and that can't happen. 135 has three 3s in the factorization (we're “allowed” only two), and 36 has two 2s (we're “allowed” only one). So the answer must be D.

But let's double check – 126 = 2x3x3x7. That's legit, all of those prime factors will cancel out with factors in the numerator.

Important note – Calling your attention to the fact that I DID NOT CALCULATE THE VALUE OF n. No oneasked for it, and nobody cares. Don't waste time doing calculations you weren't asked to do. This might seem trivial to some of you, but I've seen plenty of cases where students spent their time figuring out some pretty fancy results, that no one asked for.

Alright. Now, let's do this prime number factorization to 36, which you may recall is a square of an integer (6x6 = 36).

366 6

2 3 2 3

Notice that the prime factors here come in pairs. Since a square of an integer can be presented as thatinteger multiplied by itself, each one of its prime factors will appear twice. This means that:

FACT – If a number is the square of an integer, every prime factor in the number will appear an even number of times. (note – this could be 4 or 6 times, not necessarily just twice. For example 576 is 24 squared, but since 24 is 2x2x2x3, 2 will appear in the prime factorization of 576 six times).

Greatest Common Denominator / Least Common MultiplePrime factorization allows us to be a little faster in figuring out a couple of other characteristics of integers.

For any group of integers we can define:DEFINITION – The Greatest Common Denominator of integers n and k, is the LARGEST number that is a factor of both n and k. (You can expand the definition to apply to any group of integers q,r,s,t,... )

DEFINITION – The Least Common Multiple of integers n and k, is the SMALLEST integer that is divisibleby both n and k.

We can see how it would be easy to find small common denominators for a pair (or group of integers) – the integer 1 is a common denominator of ALL numbers. 2 is a common denominator of ANY group of even integers. It gets a little more difficult to figure out what the LARGEST common denominator is.

For example – what's the largest common denominator of 24 and 36? Is it 6? No. 8 is larger, and 12 is even larger. Is it the largest???



A good way to find the GCD quickly is to take ALL OF THE SHARED PRIME FACTORS of the two integers.THAT is the GCD. In the case of 24 (2x2x2x3) and 36 (2x2x3x3) they share two 2s and one 3 (36 only has two 2s, while 24 has three 2s – so the number of 2s they SHARE is two). 2x2x3 = 12, which is the GCD.

Let's try for larger numbers – I'll just throw some random ones out there -

72 = 2x2x2x3x3 and 120 = 2x2x2x3x5

Shared prime factors are – three 2s, and a 3. That's it. So the GCD is 2x2x2x3 = 24. That was pretty quick, no?

You won't usually have to calculate an GCD on the GMAT directly, but you may end up needing it to solve a different part of the question – and so this method is useful.

How about LCM? Finding multiples of two integers is super easy – for one, you can just multiply the two numbers and you'll get a common multiple (by definition!). But this multiple would be pretty big. (for example – in the case of 120 and 72... practice doing this in your head, I'll wait... 100x72 is 7200, plus 20x72 is 2x72=144 and add a 0 → 1440, so together 8640 – pretty large number). Since the LCM must be divisible by both integers, it has to contain ALL of their respective prime factors (so they can cancel out in the divisibility check and we get an integer), but we also want it to be the SMALLEST possible such integer, so if there are any SHARED prime factors, we'll just make sure not to DOUBLE THEM, that is – to only repeat them once in the LCM.

Let's go back to 72 and 120 – they both have three 2s and a 3 in their prime factorizations. So the LCMwill also have to have three 2s and at least one 3. BUT IT DOESN'T HAVE TO HAVE SIX 2s! Three suffice to make it divisible by both 72 and 120. Now 72 has a second 3 in it. The LCM will have to have two 3s in order to be divisible by 72. 120 also has a 5 in it. The LCM will also need this 5 to be divisible by 120.So three 2s, two 3s, and a 5 → 2x2x2x3x3x5 = 360. Hey that's WAAAAY smaller than 8640, and in fact –360 is divisible by 120 (120x3) and by 72 (72x5). Hooray!

Divisibility rulesWe've discussed the technique for finding a prime number factorization - you find a factor (any kind offactor) that the number is divisible by, and keep dividing till you get to primes. Ok, but how do you know where to start? What's 567 even divisible by, if anything?

Well, there's a pretty simple set of tests / rules that will allow you to quickly find out if a number is divisible by the integers 2 through 11. Here they are:



A number is divisible by If

2 The number is EVEN (which is the same as saying, the UNITS DIGIT is EVEN). Examples: 12, 250435698, 32892348238

3 * The sum of the number's DIGITS is divisible by 3. For example – the sum ofthe digits of 567 is 5+6+7=18. 18 is divisible by 3, and therefore so is 567.

4 ** When you divide the number by 2 it's still EVEN. Tip – You can just check that the number formed by THE LAST TWO DIGITS is still even after dividing it by 2. For example the number formed by the last two digits of 42084, 84, equals 42 when divided by 2, which is still even. Therefore 84 and 42084 are both divisible by 4.

5 The number ends in a 5 or a 0. For example 75680 ends with a zero and is therefore divisible by 5.

6 The sum of the number's DIGITS is divisible by 3 AND the number is EVEN.Because this means it's divisible by 3 AND by 2, and therefore by 6. For example, the sum of the digits of 42084 is 18, which is divisible by 3. 42084 is also even, and therefore divisible by 6.

7 There isn't a great, quick method for checking divisibility by 7, but see below for a generalized hack for divisibility testing.

8 ** When you divide the number by 4 it's still EVEN. Tip – You can just check that the number formed by THE LAST THREE DIGITS is still even after dividing it by 4. For example the number formed from the last three digits of 75680 is 680. 680 divided by 4 is 170, which is still even. Therefore 680 as well as 75680 are divisible by 8.

9 * The sum of the number's DIGITS is divisible by 9. For example, the sum of the digits of 567 is 18 which is divisible by 9, thus 567 is divisible by 9.

10 The number ends with a 0. Examples – 40, 400, 75680.

11 ^ When you take the leftmost digit of the number, and then alternately subtract and add the value of the following digits from this number until you reach the Units digit the resulting sum is divisible by 11. Example, because that is the least clear explanation possible – To test the number 2374438 for divisibility by 11, I take the leftmost digit (2) and alternately subtract and add the other digits – so I get 2-3+7-4+4-3+8 = 11 (see how the minuses and pluses alternate?). 11 is divisible by 11, clearly, and therefore so is 2374438. In fact 2374438 = 11 x 215858. Of course I'm just exaggerating for effect here, you're never going to see this kind of complexity in the GMAT, but this is a really quick test for 3 and 4 digit numbers, for example 1001 → 1-0+0-1 = 0 Zero is divisible by 11, and therefore so is 1001. Same for 198 → 1-9+8 = 0, so 198 is divisible by 11. Real quick test.



General notes – How come I claim that 0 divisible by 11? Well, I claim more than that – 0 is divisible byANY number. Here's why. For 0 to be divisible by k, 0/k must be an integer. But 0 divided by ANY (non-zero) NUMBER is ALWAYS 0, which is an integer. Therefore 0 is divisible by all non-zero numbers.

Specific notes – (skip this part if you're not interested in more mathematical proofs of why these tests work. You don't need this for the GMAT, but I know plenty of people get curious).

* - Why do these tests for 3 and 9 work? Well let's take three digit number ABC, it can be written as – 100xA + 10xB + C, which is the same as – 99xA + 9xB + (A + B + C). 99xA is obviously divisible by 3 and 9, as is 9xB. If (A+B+C) is also divisible by 3 or by 9, then each of the terms of the sum would be divisible by 3 or by 9, and therefore the entire sum (the original number) would be divisible by 3/9 (per the rule that says that if each term of a sum is divisible by a number, so is the sum itself).

** - Why do these tests work for 4 and 8? Any number in the world can be written as [all the digits except the last two] x 100 + [the number formed by the last two digits]. Since 100 is divisible by 4, if the number formed by the last two digits is divisible by 4 – so is the sum, which is the original number.Unfortunately, 100 is NOT divisible by 8, but 1000 IS. So we can write [all the digits except the last three] x 1000 + [the number formed by the last THREE digits]. 1000 is divisible by 8, so if the number formed by the last three digits is divisible by 8 so is the sum, which is the original number.

^ - Why does this 11 rule work? Well.... I'll give an example for a 5 digit number (ABCDE), but this can be generalized to any length of a number. ABCDE can be written as 10000xA + 1000xB + 100xC + 10xD + E (stop me if you know where I'm going already ;) ). This is equal to 9999xA + A + 1001xB – B + 99xC + C + 11xD – D + E. 9999, 1001, 99 and 11 are all divisible by 11, and therefore so are all the respectiveterms. That leaves us needing to check whether A – B + C – D + E is divisible by 11, and if so, the wholesum / original number will be. NOTE that -11 and 0 are both divisible by 11, and therefore acceptable results.

General division hackWell, what if you need to check if a number is divisible by 13 then?? What if you need to actually carryout the division, not just check for divisibility? Here's a quick way to do both, that is essentially long division, but done in a way that will allow you to move faster through the calculation.Is 5,839,197 divisible by 13? (Not a real GMAT question)

We'll proceed by throwing out large chunks of the number that we KNOW are divisible by 13, if we end up left with something that is ALSO divisible by 13, then the whole thing is divisible by 13. If not – the whole number is NOT divisible by 13. Sound simple?

So 4x13 is 52. The leftmost two digits are 58. Close enough. Throw away 52 from 58, and you're left with 639,197 (that is subtract 5,200,000 from 5,839,197, but we don't have to keep looking at the whole number to do our calculation). 5x13 is 65 which is greater than 63 (the start of 639,197), so we'll just have to chuck 52 out again, and we'll be left with 119,197, well, that's nice because 10x13 is 130, and therefore 9x13 is 117 (subtract 13 from 130). Chuck 117 out from 119 and we're left with 2,197, throw out 13 from 21 and we have 897, throw out 78 (6x13) we get 117, which we already



established is divisible by 13 (9x13), and therefore the whole number is divisible by 13.

Now note, we didn't actually get a result of the division here. To do that we'd have to “hold on” to the results of each step that we “threw out”. That is – in the first step we threw away 400,000 times 13, then 40,000, then 9,000, 100, 60 and 9. For a total of 449,169 x 13. There's your result. Obviously, on the test we won't be engaging in such large divisions, but this hack allows you to do some longer division problems in your head, instead of on the paper, and thus speeds up your arithmetic.

Obviously, this technique can also be used to test for divisibility by 7.

Let's take a quick look at a typical divisibility problem:

Q – If A is a positive integer, less than 20, and B = 5,422 + A, for how many values of A would B be divisible by 3 and by 4?

(A) 0 (B) 1 (C) 2 (D) 3 (E) 5

A – For B to be divisible by 3 the sum of the digits would have to be divisible by 3. 5+4+2+2 = 13. So A can be 2, 5, 8, 11, 14 and 17. But B also needs to be divisible by 4. For that to happen, the number created by the last two digits must be divisible by 4. The number created by the last two digits WITHOUT A is 22. So A can be 2, 6, 10, 14, and 18 to create a two digit number divisible by 4, thus making B divisible by 4. If A is 2 or 14 both of these divisibility rules apply, so there are 2 values of A that would match.

The answer is C.

Of course, there are 19 possible values for A, so there are only 19 possible values for B. In any span of 19 values there are either 1 or 2 values that are divisible by 12 (and thus divisible by 3 AND by 4). In this case the span is between 5,423 and 5,441, and there are two values – 5,425 and 5,436. If the range was between 5,426 and 5,444 there would only be ONE value (5,436). Either way, you can see that the only number of possible values for a suitable A are 1 or 2, so the only possible answer choicesto begin with are B or C (in case you're at the end of the test and have to guess).

ConsecutivesA set of consecutive integers, is a group of integers that follow each other in immediate order, such as 3,4,5,6 and 7, or -2, -1, 0, 1, 2 and 3.

These groups of consecutive integers are a simple form of series or progression, which we'll delve into more deeply much later, but they have several characteristics that are important to note right here.

Let's look at one set of consecutive numbers -

3 4 5 6 7 8



This group can be looked at as a set of centered pairs (I'll call them couplets to denote they are the same distance from the center of the set). You might notice that each couplet adds up to the same number. 3+8 = 11, 4+7 = 11, 5+6 = 11. Because this is the case, we can quite easily find the THE SUM of a set of consecutive integers by multiplying THE NUMBER OF COUPLETS x THE SUM OF EACH COUPLET. In this case there are three couplets, the sum of each is 11, so the sum of all the integers in the set is 3 x 11 = 33 (you can check to see that this is correct).

Turns out this holds for sets with an ODD number of integers as well (because the middle integer is counted as HALF a couplet).

Now, since the sum of each of the couplets is the same, we can just calculate it by taking the first and last integer in the set of consecutives. The number of couplets is simply the total number of elements in the set, divided by 2.

So for EVERY SET OF CONSECUTIVE INTEGERS we can say that the SUM off all the integers in the set is:

(first integer+last integer) ⋅(number of integers in the set)

2

We can try this on the other set I listed at the top of the section: -2, -1, 0, 1, 2, 3. There are 6 integers. The first is -2, the last is 3, so the sum is: (-2+3)x6/2 = 3.

The AVERAGE of the numbers in this consecutive set is simply THE MIDDLE NUMBER IN THE SET. If the number of integers in the set is odd, great. If not – there is NO middle number, and we have to average the two middle numbers to get the overall average. But wait a minute – since each couplet in the sequence has the same sum, averaging any of them will provide us with the overall average!!A series of consecutive integers can also be presented algebraically. For example if the first number in the series is x, then the entire series might look like – x, x+1, x+2, x+3, x+4. Approaching consecutive number problems like this is often time consuming, and so you should only revert to it if no other method we discuss here works first.

Q – What is the sum of the integers from 1 to 100?

(A) 6,000 (B) 5,050 (C) 5,000 (D) 2,525 (E) 2,500

A – (First + Last)x(number of integers)/2. But HOW MANY INTEGERS ARE THERE HERE? 100? 99? 101?We can run a quick test by checking 1 through 3 and 1 through 5, which might convince us fairly certainly that there are 100 numbers here, but what about the numbers from 100 to 115? Are there 15? 14? 16?

Fortunately there's a good formula for that. The number of numbers in a set of CONSECUTIVE INTEGERS = Last # – First # +1. So in the case of 1 to 100 → 100-1+1 = 100. In the case of 100 to 115 →115-100+1 = 16 (!)



And this works for negative numbers too. In the set -2, -1, 0, 1, 2, 3 we'd get – 3-(-2)+1 = 6 which is exactly the number of numbers in that set.

Returning to our problem we substitute accordingly and get – (1+100)x(100-1+1)/2 = 101x50 = 5050 !

The answer is B.

More importantly, we just calculated the sum of 100 consecutive integers in less that a minute!

Cool. Now, we can also discuss other sets of consecutives, such as a set of CONSECUTIVE EVEN INTEGERS (16, 18, 20, 22, 24, 26) or CONSECUTIVE ODD INTEGERS (-17, -15, -13, -11, -9, -7). If you check you'll notice the same rules apply to these sets as well (the couplets all add up to the same number), and therefore the rules for calculating their AVERAGE and SUM are the same.

Though NOT the rule about the number of numbers! Here something a little more complicated is happening. If we just take the difference between 26 and 16 (in the first set), and add 1, we'll get 11. But there are only 6 numbers. This is because we're skipping a number for every one that is in the set. (we want 16, but we're skipping 17, want 18, but skip 19). We could try to use the equation we used before but we have to reflect this skipping. How would we do that? Well, here's the general approach – count ALL of the numbers between the first and last number of our set, then divide by the appropriate amount. Sort of. As we've seen – the number of integers from 16 to 26 is 11, and that is NOT divisible by 2. Here's why that happened.

16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26

The series of consecutive integers is divided into pairs of integers we ARE interested in, followed by integers we ARE SKIPPING, except at the end. There, we just ended on an integer we ARE using, without it's pair. That's why the calculation was off. Let's correct it.

RULE – In consecutive series where we skip some integers, IF THE FIRST AND LAST integers are integers we ARE USING, calculate the correct number of numbers as follows:

(last number−first number)deflator

+1

Where deflator is the number we are dividing by to get rid of the integers we skipped. In this example we're skipping every other integer, so we need to count HALF of the integers, and we'd divide by 2. IF we were skipping 2 integers for every one in the set, we'd divide by 3, etc.

So in this example – (26-16)/2 + 1 = 5+1 = 6. Which is correct. Again, this will always work AS LONG AS the first and last number in the consecutive integer sequence we're counting the size of ARE PART of the skipping sequence we are working with.



Now, why stop expanding there? We can talk about sets of CONSECUTIVE MULTIPLES OF THREE for example – 24, 27, 30, 33, 36, 39, 42 (same deal – all the couplets here add up to 66), and same calculations hold – the average is 33. The sum is 66x7/2 = 231.

For the number of numbers we'd have to divide by 3 (since we're skipping 2 numbers for each 1 we are using) – (42-24)/3 + 1 = 18/3 + 1 = 6+1 = 7. And in fact, there are 7 numbers in our set.

This works for any set of consecutive multiples of any integer.

We can also talk about sets of other consecutive things, such as this set of CONSECUTIVE PRIME NUMBERS – 7,11,13,17,19,23,29,31. Here however our system breaks down – the couplets don't add up to the same sum every time...

The reason is that all the previous consecutive sequences or sets were actually ARITHMETIC PROGRESSIONS, where the difference between each two consecutive elements of the set is THE SAME. In the set of consecutive primes this does not hold – the distance between 11 and 7 is not the same as the distance between 13 and 11. It is not simply THE WORD consecutive that imbues these series with the characteristics we've discussed but that the DIFFERENCE BETWEEN EACH ADJACENT ELEMENTS IS THE SAME IN THE WHOLE SET.

Let's look at a couple of example questions -

Q – How many integers that are divisible by 3 are there between 200 and 300 inclusive?

(A) 100 (B) 50 (C) 34 (D) 33 (E) 32

A – If we try to calculate the number of numbers between 200 and 300 and divide by 3 we'll hit a snag– 300-200 is 100 and that is NOT divisible by 3. But remember – we need our first and last numbers to be IN OUR SET. 200 is NOT divisible by 3. 201 though is (the sum of the digits is 3) as is 300. Now we can do our calculation (since our first and last numbers are part of the set) – (300-201)/3 + 1 = 99/3 + 1 = 33+1 = 34 numbers.

The answer is C.

(If you'd like, we can now quickly calculate the sum of multiples of 3 in this range – 34x(300+201)/2 = 17x501 = (5x17 add two zeros + 1x17) = 8500+17 = 8517).

Q – Integer n =99011⋅k

where k is a positive integer greater than 1. What is the value of k?

(1) k is NOT divisible by 3.(2) k is odd.



A – Oh. Hey. Wait a minute. What is this here question?? That's not a straight multiple choice problemsolving Q!

Nope. No it isn't. This is an exemplar of the much beloved DATA SUFFICIENCY questions on the GMAT math section, and we'll take a short detour to discuss these before solving this problem.

DATA SUFFICIENCY questions DO NOT ask you to solve a problem for a result. INSTEAD they present you with a question, and two pieces of additional information. Then they ask – which of the pieces of information allows you to ANSWER THE QUESTION THAT HAS BEEN POSED. All you have to do is deduce whether, YES that is SUFFICIENT DATA to answer the question, or NO, the DATA is not enough to decidedly answer the question. You don't actually need to solve anything if you can PROVE that it WOULD BE POSSIBLE (or NOT possible) to solve given the information available.The answer choices ALWAYS map as follows:

If the FIRST statement is sufficient to answer the question ON ITS OWN, but the second IS NOT, the answer is (A).

If the FIRST statement is NOT sufficient to answer the question ON ITS OWN, but the second one IS sufficient ON ITS OWN, the answer is (B).

If NEITHER the first or the second statements, ALONE, were sufficient to answer the question, BUT TOGETHER they ARE sufficient, the answer is (C).

If BOTH the first AND the second statements are sufficient to answer the question, EACH ON THEIR OWN, the answer is (D).

and, if NEITHER the first nor the second statement is sufficient ALONE, AND together the statements are INSUFFICIENT AS WELL, then the answer is (E).

One more time, hopefully for clarity:Only 1st statement is sufficient, ALONE → AOnly 2nd statement is sufficient, ALONE → BNeither 1 nor 2 alone are sufficient, but TOGETHER, yes → CBOTH 1 and 2 are sufficient ALONE → DNEITHER 1 nor 2 alone are sufficient, NOR together → E

Since, as you can see, these are more of a logical exercise than a straight math problem, I recommend sticking to the solving process I will suggest VERY STRICTLY. It's the most effective and efficient way to work through these problems.

You will have seen this workflow before. But note the points of emphasis I focus on – the process here can often make solving the problems a LOT easier.



1) The first thing to do in a data sufficiency question is to completely understand the QUESTION BEINGPOSED, and simplify it as much as possible. In this case, it is clear we are dealing with an integer question, and the specific question we need to answer is – what is the value of (integer) k ? What we need, in order to answer that question is a SINGLE value for k. If there is more than one possible value for k given the information that we have, we CANNOT answer the question (because it could be any ofthe possible values)!

We can do quite a bit of work to better understand what limits the possible values of k, BEFORE even looking at the pieces of information provided later in the question. First of all, we can reduce the fraction (we'll discuss reducing and expanding fractions in the next chapter!) and instead of 990/11k – we'd get – 90x11/11k = 90/k (the 11 from the denominator and numerator would cancel out). That makes things simpler already.

Also, since n is an integer, everything in the denominator of the fraction representing n will have to cancel out with something in the numerator. That means ALL of k's prime factors must be included in 90, which is 2x3x3x5. Those are the only prime factors k can have! In fact we can now list ALL the possible values of k! They are – 2,3,5,6,9,10,15,18,30,45 and 90. I wouldn't spend the time listing them out on the test, unless the information makes for a MUCH SHORTER list than this (like 4 or 5 options), but it is worth keeping in mind what the field of options is.

So, we've both simplified the question, AND learned some additional information, just by making sure we investigate the question ITSELF before we jump in.

2) NOW, we can look at the first statement ALONE, and check to see if it is sufficient to answer the question. This statement says that k is NOT divisible by 3. Put another way, that means that 3 is NOT a prime factor of k (or else it would be divisible by 3). This limits the number of options that k can be – the only prime factors it can have in this case are 2 and 5. But that still leaves 2, 5, and 10 as options! We CANNOT tell what the value of k is (it could be any of the three), so statement 1 is NOT SUFFICIENTon its own.

But wait! This tells us something about our possible answer choices! Since statement 1 ALONE is NOT SUFFICIENT, both answers A and D are ruled out! The only possibly correct answers to the question that remain are B, C and E.

3) Alright, now it's time to look at the second statement ALONE. This is critical. For just this moment, we need to completely SET ASIDE everything we analyzed and figured out from STATEMENT 1, because we're dealing with statement 2 ALONE. A very very very very common fallacy is dragging over information we gleaned from the first statement, into this step of the process. Just don't.

Statement 2 says – k is odd. That means 2 is NOT a prime factor of k, but that still leaves LOTS of options for k, just a few examples – 3,9,15 and 90. So clearly this statement alone is NOT SUFFICIENT to answer the question.



That RULES OUT one more answer choice – B CANNOT be the correct answer. The only two possibly correct answers that remain are C and E.

4) FINALLY, it's time to take BOTH statements together, and see if combined they provide sufficient information to answer the question. If k is NOT divisible by 3 (statement 1) and is odd (statement 2), it CANNOT have 2 or 3 as a prime factor. But from the question itself we know k can only have 2x3x3x5 as prime factors! Rule out 2 and 3, and your left with k=5 (the question states that k is greater than 1 which is why that option is ruled out). Taken TOGETHER the information is SUFFICIENT to determine a single value for k.

So the answer is C !

Here's a flow chart to summarize ALL the possible results of a data sufficiency question.

What a trip!!!! Yes, Data Sufficiency questions are a little different in that they test not only our mathematical abilities, but also logical analysis. That's why they're on the test!

Here are a couple more tips about how to go about solving DS questions that we'll see in practice as we continue through the book.

1) Once you've read a statement – CHOOSE AN ANGLE OF ATTACK. Do you think the statement seems sufficient? Then you'll attempt a POSITIVE PROOF. You will try and show, by using that information andthe information from the question that you can positively provide an answer. That is what we did


Simplify and Analyze the QUESTION

Check statement 1 for sufficiency

Check statement 2 for sufficiency Check statement 2 for sufficiency

SUFFICIENT!Possible remaining answers: A, D

INSUFFICIENT!Possible remaining answers: B, C, E

SUFFICIENTAnswer is D !

INSUFFICIENTAnswer is A !

SUFFICIENTAnswer is B !

INSUFFICIENT!Possible remaining answers: C, E

Check BOTH statements TOGETHER

SUFFICIENTAnswer is C !

INSUFFICIENTAnswer is E !


when checking BOTH statements in the question above. We attempted (and succeeded) to prove that the information leads to only a SINGLE value of k being possible.

It doesn't matter if you are RIGHT or WRONG in your assumption. If you attempt a POSITIVE PROOF, and fail to prove your choice, you might then proceed to attempting to REFUTE that assumption, that is, you'll attempt to prove the data is NOT SUFFICIENT.

2) TO SHOW A STATEMENT IS INSUFFICIENT - we simply need to show COUNTER EXAMPLES. That is, we need to show that USING THE INFORMATION we can reach TWO DIFFERENT ANSWERS TO THE QUESTION. For example, in our case the question was – what is the value of k? When testing each statement on it's own, we showed that using the information we could answer the question IN MORE THAN ONE WAY! (In the first statement k could be 2, 5 or 10, in the second statement, many different values). By showing that the information can lead to MULTIPLE DIFFERENT ANSWERS to the question, we are PROVING that the information is NOT SUFFICIENT.

Well, that's certainly mind boggling. Don't worry yet. We'll review all of these methods as we meet more DS questions down the road!

RemaindersTo return to our regular programming, we'll end the integer section with the fascinating but utterly frustrating concept of remainder.

The idea is extremely simple. Say I divide 17 items into groups of 3. I'd get 5 complete groups of three items each, but I'd have 2 items LEFT OVER. We call those the REMAINDER.

The long division would look like this -

53 | 17 15 So we can write – 17 = (5x3)+2 OR 17 divided by 3 equals 5R2. ___ 2

Remainder questions are the rare case on the GMAT where they are often most simply solved by plugging in numbers.

A typical GMAT remainder question might look like this:

Q – The remainder when n is divided by 5 is one. What is the remainder when 3n is divided by 5?

(A) 0 (B) 1 (C) 2 (D) 3 (E) It is not possible to tell from the information provided.



A – Let's start with solving, and then look at some additional characteristics of remainders. So, let's pick a number that when divided by 5 leaves a remainder of 1. The easiest way to do this is to take what we are dividing by (5), and just add the remainder to it (1). 5+1 = 6 and 6 when divided by 5 leaves a remainder of 1. 3x6 is 18, and when we divide 18 by 5 we get three groups of 5 and 3 left over.

But wait a minute, is this true for ALL cases of numbers that leave a remainder of 1 when divided by 5? let's check 3 other cases, then move to the proof of the general rule.

16, 21, and 36 all have remainder 1 when divided by 5 (you can check this yourself, but basically, I tookMULTIPLES of 5 and added 1 to them). We'll multiply each by 3 and get 48, 63, and 108. Surely enougheach of these when divided by 5 give us a remainder of 3.

So the answer is D !

Here's the general proof – a number that gives remainder 1 when divided by 5 can be written GENERALLY as – 5k + 1 where k is an integer. Divide this number by 5, and we would have k groups of 5, and a 1 left over. If we wanted to represent a number that gives a remainder of 2 when divided by 7, it would simply by 7k+2. There you go. let's multiply 5k+1 by 3 and we'll get 15k+3. Now when we divide 15k into groups of 5, we'd get 3k groups of 5. That divides EVENLY into 5. No remainder. But we still have the 3 part left over, so the remainder would be 3! It's true for ANY case where the original number gave a remainder of 1 when divided by 5.

So there are two excellent methods of approaching these questions – represent the case generally – as in the 5k+1 approach, or just plug in suitable numbers.

Let's look at the case of EVENLY DIVISIBLE numbers. If a number is EVENLY DIVISIBLE, that means thereIS NO REMAINDER. Which is really saying that the REMAINDER IS ZERO.

Finally, what are all the possible values for the remainder when I'm dividing by a certain number? We'll use 5 again, because we're already so comfortable with it.

Well we know it can be 0 (if the number is divisible by 5), it can be 1 (we just saw), and we can easily find examples where the remainder is 2, 3, and 4 (simple examples 7, 8, and 9 respectively). Can the remainder when dividing by 5 actually be 5? NO. Because if I have five things left over, I'd just make another pile of 5, and that actually would mean there would be NOTHING left over (a remainder of 0). The same is the case for higher remainders. I can't have a remainder of 6 when dividing by 5 because I'd just take another group of 5 out of the remaining items, and the ACTUAL remainder would be 1.

FACT – The reminder when dividing integer n by integer k, is always LESS THAN k.



Q – What is the remainder when positive integer n is divided by 2?1) n is odd2) n is a multiple of 3

A – This is an even and odd question hidden in a remainder box, but clearly we're dealing with divisibility, so we know exactly where we are tool wise. And this is a DS question, so let's look at the question before diving into the statements.

If integer n is divided by 2 the remainder can ONLY BE one of two things – 0 if n is EVEN, and 1 if n is ODD. The question can literally be rephrased here to be – IS N EVEN?

Realizing this is a big deal, because look at statement 1 – If the question is “Is n even?” Then this is CLEARLY SUFFICIENT information to answer the question! To be fully formal, if n is odd, then when dividing by 2, there will always be remainder 1, which answers the original form of the question.

SO THE ANSWER CAN ONLY BE A or D.

We look at statement 2 – if n is a multiple of 3, is it even or odd? Well, it could be either! So this statement is NOT SUFFICIENT,

and the answer is A.

We can be more thorough if we like – and find two examples that give different results for statement 2– 6 is a multiple of 3, and gives remainder 0 when divided by 2. 9 is also a multiple of 3, but gives remainder 1 when divided by 2. Since we found two cases that FIT THE INFORMATION IN THE STATEMENT, but give OPPOSING ANSWERS to the question. Clearly this statement is NOT SUFFICIENT to answer the question. Bam!

Q – When the integer n is divided by 8, the remainder is 2. Which of the following is NOT a multiple of 4?

(A) n – 2 (B) n+2 (C) 4n (D) 3n (E) 2n

A – This is a classic remainder question where picking a number is the fastest – let's take 18 (when 18 is divided by 8 we get a remainder of 2). Now let's look at the answer choices – 18-2 is divisible by 4, 18+2 is also. 4x18 is clearly divisible by 4, as is 2x18=36.

So the answer must be D.

And in fact – 3x18 = 54, which is NOT divisible by 4!

To present this formally.... n can be presented as 8k+2 where k is an integer. Subtract 2 and you get 8k which is clearly divisible by 4. Add 2 and you get 8k+4 – both terms are div by 4, so the sum must be div by 4. 2n = 16k+4 – both terms are divisible by 4, and we don't even have to check 4n because that number is a multiple of 4, and thus clearly divisible by 4 (n is an integer). That leaves us with 3n =



24k+6. Here 24k is divisible by 4, but 6 IS NOT. If one of the terms of a sum IS divisible by a number, and the other term IS NOT, then the sum IS NOT divisible by that number.Alright. That's all the fun we have time for in this chapter. In our next chapter, we'll start talking about PARTS OF INTEGERS – fractions, decimals and percents!

Also. I attempt to dive into as much detail as I possibly can in these info chapters, but there are, of course, hundreds and hundreds of questions out there to practice, and it is REALLY IMPORTANT that you practice as much as possible using the ideas and techniques described here. At first I strongly recommend NOT timing yourself. Do a bunch of integer / number property problems and take your time practicing the approaches and techniques discussed. Once you feel comfortable with the tools, then you can step it up, and time yourself using them.

Take the time to analyze and categorize the question type (obviously, this is easier at this stage, because we only learned one broad topic – all the questions are going to be number property ones – but are they a divisibility question? An evens and odds? A remainder question? We've looked a different tools for solving each of those questions.) Whenever you can – try to subtract or multiply that calculation you need to do IN YOUR HEAD, and of course spend time getting comfortable with theData Sufficiency “algorithm” we looked at. You want the possible answer choices to be old hat and come to you automatically, and you want to make sure you've learned all you can from the question itself, before diving into the statements.

But for now... Do good work, and good luck!

Key topics covered- Math vocabulary- Integers:

Positive / NegativeOdd / Even

- Divisibility:Prime number factorizationDivisibility rule for sumsDivisibility rules for integers 2-11

- LCM / GCD- Series of consecutives:

Number of numbers in the seriesCalculating sum and average

- Arithmetic tricks- Decimal notation- Digits- Rounding- Remainders- Data Sufficiency questions / workflow

© 2019 Aner Moss



GMAT MATH DECONSTRUCTED

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