GMAT Number Theory - Articles, Videos & Problems

May 2021

GMAT Factors Problem – 700 Level GMAT Question

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.

Important Notes About Factors

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.

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.

Answer

So when we’re adding 17 to our number all we need to see is that, hey, 15 doesn’t go into 17, it’s not going to get all the way up there. 17 fits perfectly. 19? guess what? It’s too big and we’re going to have a remainder. So our answer here is B, only 17.

For other problems like this, other factorials, and what have you, please check out the links below and we will see you next time. If you enjoyed this GMAT problem, try your hand at this Science Fair Problem.

Apr 2021

GMAT Prime Factorization (Anatomy of a Problem)

By: Rich Zwelling (Apex GMAT Instructor)
Date: 1st April 2021

First, if you’d prefer to go straight to the explanation for the solution to the problem given in the last post, continue to the end of this post. But we’ll start with the following Official Guide GMAT problem as a way to talk about GMAT Prime Factorization. Give the problem a shot, if you can:

How many prime numbers between 1 and 100 are factors of 7,150?

A) One
B) Two
C) Three
D) Four
E) Five

One of the things you’ll notice is the linguistic setup of the problem, which is designed to confuse you immediately (a common theme on GMAT problems). They get you panicking right away with mention of a large range (between 1 and 100), and then they compound your frustration by giving you a rather large value (7,150).

Don’t let that convince you that you can’t do the problem. Because remember, the GMAT is not interested in large calculations, memorization involving large numbers, or weird arcana. Chances are, if you find yourself thinking about a complicated way to do a problem, you’re taking the wrong approach, and there’s a simpler way.

Try to pick out the most operative signal words, which let you know how to address the problem. We are dealing with prime numbers and also the topic of finding factors. The language of the problem may make you nervous, thinking that we must consider a slew of prime numbers up to 100. But the only primes we are really interested in are those that are actually factors of 7,150.

So let’s focus our attention there. And we can do so with a prime factor tree. Does this bring back memories?

Now, in the case of 7,150, we don’t have to break it down into prime numbers immediately. Split the number up into factors that are easy to recognize. In this case, the number ends in a zero, which means it is a multiple of 10, so we can start our tree like this:

prime factorization on the GMAT Notice that the advantage here is two-fold: It’s easier to divide by 10 and the two resulting numbers are both much more manageable.

Splitting up 10 into it’s prime factorization is straightforward enough (2 and 5). However, how do we approach 715? Well, it’s since it ends in a 5, we know it must be divisible by 5. At that point, you could divide 715 by 5 using long or short division…

…or you could get sneaky and use a NARRATIVE approach with nearby multiples:

750 is nearby, and since 75/5 = 15, that must mean that 750/5 = 150. Now, 750 is 35 greater than 715. And since 35/5 = 7, that means that 715 is seven multiples of 5 away from 750. So we can take the 150, subtract 7, and get 143.

Mathematically, you can also see this as:

715/5 = (750-35)/5 = 750/5 – 35/5 = 150 – 7 = 143

So as stands, here’s our GMAT prime factorization:

Now, there’s just the 143 to deal with, and this is where things get a bit interesting. There are divisibility rules that help make factoring easier, but an alternative you can always use is finding nearby multiples of the factor in question.

For example, is 143 divisible by 3? There is a rule for divisibility by 3, but you could also compare 143 against 150. 150 is a multiple of 3, and 143 is a distance of 7 away. 7 is not a multiple of 3, and therefore 143 is not a multiple of 3.

This rule applies for any factor, not just 3.

Now we can test the other prime numbers. (Don’t test 4 and 6, for example. We know 143 is not even, so it’s not divisible by 2. And if it’s not divisible by 2, it can’t be divisible by 4. Likewise, it’s not divisible by 3, so it can’t be divisible by 6, which is a multiple of 3.)

143 is not divisible by 5, since it doesn’t end in a 5 or 0. It’s not divisible by 7, since 140 is divisible by 7, and 143 is only 3 away.

What about 11? Here you have two options:

Think of 143 as 110+33, which is 11*10 + 11*3 → 11*(10+3) → 11*13
If you know your perfect squares well, you could think of 143 as 121+22

→ 11*11 + 11*2 → 11*(11+2) → 11*13

Either way, you should arrive at the same prime factorization:

Notice that I’ve marked all prime numbers in blue. This result shouldn’t be a surprise, because notice that everything comes relatively clean: there are only a few prime numbers, they are relatively small, and there is just one slight complication in solving the problem (the factorization of 143).

So what is the answer? Be very careful that you don’t do all the hard work and falter at the last second. There are five ends to branches in the above diagram, which could lead you prematurely to pick answer choice E. But two of these branches have the same number (5). There are actually only four distinct primes (2, 5, 11, 13). The correct answer is D.

And again, notice that the range given in the question stem (1 to 100) is really a linguistic distraction to throw you off track. We don’t even go beyond 13.

Next time, we’ll talk about the fascinating topic of twin primes and how they connect to divisibility.

Find other GMAT Number Theory topics here:
Odds and Ends (…or Evens)
Consecutive Integers (plus more on Odds and Evens)
Consecutive Integers and Data Sufficiency (Avoiding Algebra)
GMAT Prime Factorization (Anatomy of a Problem)
A Primer on Primes

GMAT Prime numbers article with questions

Mar 2021

A Primer on Primes

By: Rich Zwelling (Apex GMAT Instructor)
Date: 30th March 2021

As I said in my previous post, GMAT Prime Numbers are my favorite topic. This is because not only are they inherently interesting mathematically but they show up in unexpected circumstances on GMAT problems, even when the term “prime” is not explicitly mentioned.

But before we get to that, I thought it would help to review a basic definition:

If you’ve gone through school, you’ve likely heard the definition of a prime as “any number that can be divided only by 1 and itself.” Or put differently, “any number that has only 1 and itself as factors.” For example, 3 is a prime number, because 1 and 3 are the only numbers that are factors of 3.

However, there is something slightly problematic here. I always then ask my students: “Okay, well then, is 1 prime? 1 is divisible by only 1 and itself.” Many people are under the misconception that 1 is a prime number, but in truth 1 is not prime.

There is a better way to think about prime number definitionally:

*A prime number is any number that has EXACTLY TWO FACTORS*

By that definition, 1 is not prime, as it has only one factor.

But then, what is the smallest prime number? Prime numbers are also by definition always positive, so we need not worry about negative numbers. It’s tempting to then consider 3, but don’t overlook 2.

Even though 2 is even, it has exactly two factors, namely 1 and 2, and it is therefore prime. It is also the only even prime number. Take a moment to think critically about why that is before reading the next paragraph…

Any other even number must have more than two factors, because apart from 1 and the number itself, 2 must also be a factor. For example, the number 4 will have 1 and 4 as factors, of course, but it will also have 2, since it is even. No even number besides 2, therefore, will have exactly two factors.

Another way to read this, then, is that every prime number other than 2 is odd.

You can see already how prime numbers feed into other number properties so readily, and we’ll talk much more about that going forward. But another question people often ask is about memorization: do I have to memorize a certain number of prime values?

It’s good to know up to a certain value. but unnecessary to go beyond that into conspicuously larger numbers, because the GMAT as a test is less interested in your ability to memorize large and weird primes and more interested in your reasoning skills and your ability to draw conclusions about novel problems on the fly. If you know the following, you should be set (with some optional values thrown in at the end):

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, (41, 43)

Thankfully, you’ll notice the list is actually pretty manageable.

(And an interesting note that many people forget that 27 is actually not prime. But don’t beat yourself up if this happens to you: Terence Tao, one of the world’s leading mathematicians and an expert on prime numbers, actually slipped briefly on national television once and said 27 was prime before catching himself. And he’s one of the best in the world. So even the best of the best make these mistakes.)

Now, here’s an Official Guide problem that takes the basics of Prime Numbers and forces you to do a little reasoning. As usual, give it shot before reading the explanation:

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

A) 10⁹
B) 10⁸C) 10⁷
D) 10⁶
E) 10⁵

Explanation

For this one, you have a little hint going in, as we’ve provided you with the necessary list of primes you’ll use to find the product.

And the language given (“closest to”) is a huge hint that you can estimate:

2*3*5*7*11*13*17*19 ~= ??

Since powers of 10 are involved, let’s try to group the numbers to get 10s as much as possible. The following is just one of many ways you could do this, but the universal easiest place to start is the 2 and the 5, so let’s multiply those. We’ll mark numbers we’ve accounted for in red:

(2*5)*3*7*11*13*17*19 ~= ??

10*3*7*11*13*17*19 ~= ??

Next, we can look at the 19 and label it as roughly 20, or 2*10:

10*3*7*11*13*17*19 ~= ??

10*3*7*11*13*17*20 ~= ??

10*3*7*11*13*17*2*10 ~= ??

We could also take the 11 and estimate it as another 10:

10*3*7*11*13*17*2*10 ~= ??

10*3*7*10*13*17*2*10 ~= ??

At this point, we should be able to eyeball this. Remember, it’s estimation. We may not know 17*3 and 13*7 offhand. But we know that they’re both around or less than 100 or 10². And a look at the answer choices lets us know that each answer is a factor of 10 apart, so the range is huge. (In other words, estimation error is not likely to play a factor.)

So it’s not unreasonable in the context of this problem to label those remaining products as two values of 10²:

10*3*7*10*13*17*2*10 ~= ??

10*(10²)*10*(10²)*2*10 ~= ??

And at this point, the 2 is negligible, since that won’t be enough to raise the entire number to a higher power of 10. What do we have left?

10¹*(10²)*10¹*(10²)*10¹ ~= 10⁷

The correct answer is C.

Next time, we’ll get into Prime Factorizations, which you can do with any positive integer.

Mar 2021

Consecutive Integers and Data Sufficiency (Avoiding Algebra)

By: Rich Zwelling (Apex GMAT Instructor)
Date: 25 March 2021

Last time, we left off with the following GMAT Official Guide problem, which tackles the Number Theory property of consecutive integers. Try the problem out, if you haven’t already, then we’ll get into the explanation:

The sum of 4 different odd integers is 64. What is the value of the greatest of these integers?
(1) The integers are consecutive odd numbers
(2) Of these integers, the greatest is 6 more than the least.

Explanation (NARRATIVE or GRAPHIC APPROACHES):

Remember that we talked about avoiding algebra if possible, and instead taking a narrative approach or graphic approach if possible. By that we meant to look at the relationships between the numbers and think critically about them, rather than simply defaulting to mechanically setting up equations.

(This is especially helpful on GMAT Data Sufficiency questions, on which you are more interested in the ability to solve than in actually solving. In this case, once you’ve determined that it’s possible to determine the greatest of the four integers, you don’t have to actually figure out what that integer is. You know you have sufficiency.)

Statement (1) tells us that the integers are consecutive odd numbers. Again, it may be tempting to assign variables or something similarly algebraic (e.g. x, x+2, x+4, etc). But instead, how about we take a NARRATIVE and/or GRAPHIC approach? Paint a visual, not unlike the slot method we were using for GMAT combinatorics problems:

___ + ___ + ___ + ___ = 64

Because these four integers are consecutive odd numbers, we know they are equally spaced. They also add up to a definite sum.

This is where the NARRATIVE approach pays off: if we think about it, there’s only one set of numbers that could fit that description. We don’t even need to calculate them to know this is the case.

You can use a scenario-driven approach with simple numbers to see this. Suppose we use the first four positive odd integers and find the sum:

_1_ + _3_ + _5_ + _7_ = 16

This will be the only set of four consecutive odd integers that adds up to 16.

Likewise, let’s consider the next example:

_3_ + _5_ + _7_ + _9_ = 24

This will be the only set of four consecutive odd integers that adds up to 24.

It’s straightforward from here to see that for any set of four consecutive odd integers, there will be a unique sum. (In truth, this principle holds for any set of equally spaced integers of any number.) This essentially tells us [for Statement (1)] that once we know that the sum is set at 64 and that the integers are equally spaced, we can figure out exactly what each integer is. Statement (1) is sufficient.

(And notice that I’m not even going to bother finding the integers. All I care about is that I can find them.)

Similarly, let’s take a graphic/narrative approach with Statement (2) by lining the integers up in ascending order:

_ + __ + ___ + ____ = 64

But very important to note that we must not take Statement (1) into account when considering Statement (2) by itself initially, so we can’t say that the integers are consecutive.

Here, we clearly represent the smallest integer by the smallest slot, and so forth. We’re also told the largest integer is six greater than the smallest. Now, again, try to resist the urge to go algebraic and instead think narratively. Create a number line with the smallest (S) and largest (L) integers six apart:

S—————|—————|—————|—————|—————|—————L

Narratively, where does that leave us? Well, we know that the other two numbers must be between these two numbers. We also know that each of the four numbers is odd. Every other integer is odd, so there are only two other integers on this line that are odd, and those must be our missing two integers (marked with X’s here):

S—————|—————X—————|—————X—————|—————L

Notice anything interesting? Visually, it’s straightforward to see now that we definitely have consecutive odd integers. Statement (2) actually gives us the same information as Statement (1). Therefore, Statement (2) is also sufficient. The correct answer is D.

And again, notice how little actual math we did. Instead, we focused on graphic and narrative approaches to help us focus more on sufficiency, rather than actually solving anything, which isn’t necessary.

Next time, we’ll make a shift to my personal favorite GMAT Number Theory topic: Prime Numbers…

Mar 2021

Consecutive Integers (plus more on Odds and Evens)

By: Rich Zwelling, Apex GMAT Instructor
Date: 23rd March, 2021

In our last post, we left you with a GMAT Official Guide Data Sufficiency problem to tackle regarding Odd/Even Number Theory. Here it is, if you didn’t get a chance to do it before:

If x and y are integers, is xy even?
(1) x = y + 1.
(2) x/y is an even integer.

Solution:

The title of today’s post gives a little hint. We discussed last time that for a product of integers to be even, all you need is a single even integer in the set. (Conversely, for the entire product to be odd, every integer must be odd.)

How does that affect how we interpret Statement (1)? Well, this is where taking a purely Algebraic approach can get you into trouble. Why not take a NARRATIVE APPROACH? What is the equation really telling us narratively about the relationship between x and y? Develop the habit of thinking about numbers narratively instead of purely algebraically, as this can make numerical relationships easier to understand.

Statement (1), in essence, is really telling us that y and x are consecutive integers. If I take y and add one to it to get x, they must be consecutive. So what are the implications for the number property of the question (even/odd)? Well, between any pair of consecutive integers, one must be odd and one must be even. I don’t know which is which, but it doesn’t matter in this case, because I care only about the product. Whether it’s Odd*Even or Even*Odd, the final product of xy will always be Even.

Even if you don’t see this narratively, you could use a scenario-driven approach and test simple numbers using the equation. If x = y + 1, try y = 2 and x = 3 to get xy = 6. Now, that’s just one instance of xy being even, so that doesn’t prove it’s always even. But then if I use y = 3 and x = 4 to get xy = 12, I again get an even result for xy. Using y = 4 and x = 5 would again yield an even xy result. Hopefully what I will realize, at this point, is that I am switching between x=odd, y=even and x=even, y=odd, and yet I still end up with xy=even.

Statement (1) is SUFFICIENT.

And remember: the GMAT is very fond of testing interesting properties of consecutive integers.

For Statement (2), we discussed that division is less amenable to hard-fast rules of odd/even properties. For that reason, you could definitely use a scenario-driven approach. But hopefully, this approach would lead you towards a consideration of our previously discussed undesired possibility. Here’s what I mean:

If x/y is even, does that guarantee that xy is even? If you’re picking numbers, you must pick ones that fit the statement. It’s tempting to pick ones that contradict the statement (i.e. proving the statement wrong), but remember that the question is up for debate, not the statement. The statement is given to you as iron-clad fact.

So what if, for example, x = 4 and y = 2? That works for Statement (2), because x/y would certainly be even (4/2 = 2). And that would lead you to xy = 8, which is of course even.

You could pick many such examples. But here’s where the undesired part comes into play. Is it possible for us to pick numbers here such that xy becomes odd? Well, for xy to be odd, both x and y would have to be odd. But if x and y were both odd, could x/y be even as Statement (2) says? Even if you can’t see the answer right away, try some numbers here, knowing that x and y must be integers:

x=5, y=3 → 5/3 (not even)
x=3, y=5 → 3/5 (not even)
x=3, y=7 → 3/7 (not even)
x=9, y=3 → 3 (not even)

You’ll see that no matter what numbers you try, you’ll never get an even result for x/y. From a number theory perspective, this is because to get an even result, you must retain a factor of 2 in the numerator of the fraction. But we don’t even start with a factor of 2 if we have only odd numbers to begin with.

In conclusion, there’s no way that x and y can both be odd in Statement (2), meaning xy is not odd, and that guarantees xy is even. Statement (2) is also SUFFICIENT.

The correct answer is D. Each statement is SUFFICIENT on its own.

For next time, give the following Official Guide problem a shot, and use it as a chance to practice a NARRATIVE APPROACH:

Mar 2021

Odds and Ends (…or Evens)

By: Rich Zwelling, Apex GMAT Instructor
Date: 18th March, 2021

Last time, we signed off with an Official Guide GMAT problem that provided a nice segue into Number Theory, specifically today’s topic of GMAT Odds and Evens. Now we’ll discuss the solution. Here’s the problem, in case you missed it and want to try it now:

If x is to be chosen at random from the set {1, 2, 3, 4} and y is to be chosen at random from the set {5, 6, 7}, what is the probability that xy will be even?

A) 1/6
B) 1/3
C) 1/2
D) 2/3
E) 5/6

Method #1 (Certainly passable, but not preferable)

Since there are so few numbers involved here, you could certainly take a brute-force approach if pressed for time and unsure of a faster strategy. It doesn’t take long to map out each individual product of x*y systematically and then tally up which ones are even. Here I’ll use red to indicate not even and green to indicate even:

1*5 = 5
1*6 = 6
1*7 = 7

2*5 = 10
2*6 = 12
2*7 = 14

3*5 = 15
3*6 = 18
3*7 = 21

4*5 = 20
4*6 = 24
4*7 = 28

Since we know that all probability is Desired Outcomes / Total Possible Outcomes, and since we have 8 even results out of 12 total possible outcomes, our final answer would be 8/12 or 2/3.

However, this is an opportune time to introduce something about odd and even number properties and combine it with the method from our “undesired” probability post…

Method #2 (Far preferable)

First, some number theory to help explain:

You might have seen that there are rules governing how even and odd numbers behave when added, subtracted, or multiplied (they get a little weirder with division). They are as follows for addition and subtraction:

Even ± Odd = Odd

Odd ± Odd = Even

And with multiplication, the operative thing is that, when multiplying integers, just a single even number will make the entire product even. So for example, the following is true:

Odd * Odd * Odd * Odd * Odd * Odd * Odd * Odd * Odd * … * Odd = Odd

But introduce just a single even number into the above product, and the entire product becomes even:

Even * Odd * Odd * Odd * Odd * Odd * Odd * Odd * Odd * Odd * … * Odd = Even

This makes sense when you think about it, because you are introducing a factor of 2 to the product, so the entire product must be even.

When considering the above rules, you could memorize them, or you could turn to SCENARIO examples with simple numbers to illustrate the general pattern. For example, if you forget what Odd * Odd is, just multiply 3 * 5 to get 15, which is odd, and that will help you remember.

This can also help you see that introducing just a single even number makes an entire product even:

3 * 3 * 3 = 27

2 * 3 * 3 * 3 = 54

So how does all of that help us get the answer to this problem faster?

Well, the question asks for the desired outcome of xy being even. That presents us with three possibilities:

1. x could be even AND y could be odd
2. x could be odd AND y could be even
3. x could be even AND y could be even

This is why I, personally, find it less helpful to think of individual multiplication rules for even numbers and more helpful to think in terms of: “The only way a product of integers is odd is if every integer in the set is odd.”

Because now, we can just think about our undesired outcome, the only outcome left:

4. x is odd AND y is odd, making the product xy odd

And how many such outcomes are there in this problem? Well at this point, we can treat it like a simple combinatorics problem. There are two odd numbers in the x group (1 and 3) and two odd numbers in the y group (5 and 7):

_2_ * _2_ = 4 possible odd xy products

And for the total, there are four possible x values and three possible y values:

_4_ * _3_ = 12 total possible xy products

That gives us a 4/12 or 1/3 probability of getting our undesired odd xy product. And as discussed in the previous post, we can now simply subtract that 1/3 from 1 to get:

1 – 1/3 = 2/3 probability that we get our desired even xy product.

For a little “homework,” try the following Official Guide GMAT problem. It has an underlying topic that we’ll discuss next time:

If x and y are integers, is xy even?
(1) x = y + 1.
(2) x/y is an even integer.

If you enjoyed this GMAT odds and evens article watch Mike solve this Number Theory problem with multiple solution paths.

Sep 2020

Data Sufficiency Problem

I’m here with a number theory data sufficiency problem. Like many of the other problems, we’re going to look at this problem over here, structurally, as well as mathematically. Taking a look at the stem the first thing we are struck by is the idea that we need to figure out this evenness and oddness.

1. What Do We Need?

When we ask ourselves what do we need: a few things should draw our attention: First, that one of these elements is squared. So if B is squared then no matter what it is its square will have the same identity: even squared is even, odd squared is odd. But also because we’re adding these two things together, for something to be odd one of them has to be odd, the other has to be even. But it could go either way so there’s a lot of moving pieces. The easiest thing to do is to say: “I need to know if each of them is even or odd.” But, of course, we know that the GMAT is not going to give us this information.

2. Start With Statement 1

Let’s take a look then at statement one and statement two. And because statement one is very straightforward we should begin there. So here we’re told very quickly succinctly: one is even one is odd. We can run a scenario and plug in some numbers, a two and a three for example, or deal with it at the identity level Either way, that gives us a straight answer that is sufficient. If that’s not visible to you I would suggest that you review your number properties. In general, this is enough and we say: “Okay, well, one’s sufficient.”

3. Statement Number 2

Then we get into number two and we have this “B plus C” is odd and immediately we might end up dismissing this and this would be a mistake. The reason we end up dismissing it is because one was so straightforward in addressing what we needed that two feels like because B and C aren’t extricated from each other that it’s almost too complex. So the GMAT may have lulled us into a sense of security with statement number one, which I think is one of the really neat structural features of this problem. If statement one were more complex we would actually spend more time looking at statement two.

Diving into statement two a little more deeply we can see that because B plus C is odd rather than even one must be even the other must be odd. And because it doesn’t matter which, something that we ascertained when we were looking at the question stem which is why that proactive thinking is really important, we can say well as long as one is each then that’s going to be sufficient as well. And so here the answer is D.

4. Further Information

I want to put up a third piece of information. And this is a really useful thing to do when you’re self-prepping is to look at data sufficiency and then postulate what other piece of information might have some subtlety, might the GMAT give us to induce us to an incorrect answer by modulating the complexity not in the question stem but in the introduced information. So here we have C equals B over 2. What this means is B must be even. Take a minute to think about that. We can’t know anything about C but B must be even because they’re integers and because you can slice B into two B is the even one. It’s tempting to move that 2 over and say 2C equals B and say: “Wait, C is even.”

But if you think about that a little more deeply it doesn’t add up because what we’re doing is multiplying C. An odd or an even number times 2 is going to result in an even. So this is a really great problem form because the same pattern of even/odd identities with different embedded equations and different ways of hiding whether B or C or M, N, or X and Y, or P and Q are odd or even is a very common trope especially as you get to the more challenging levels of the GMAT where you have these abstract DS questions, abstract inequalities that are really the bread and butter of 700 plus.

5. Examine This Problem Form Deeply

So as a more general problem form this is one to examine deeply and play around with in a whole bunch of different ways. You can introduce exponents, absolute value, inequalities as I mentioned, quadratic identities are a big one where, for example, you have a difference of two squares and then you’ve got one piece or the other, the X plus Y or X minus Y and they give you information on that. And so as you’re doing this, one of the most important pivot questions to look at is: “How do I convert this piece of information into the information they’re asking me about in the question stem?” or vice versa: “How do I relate this information in the question stem to this piece of information?” because almost certainly they’re going to be related and it’s in that relationship that you determine whether or not it’s sufficient.

And typically as the subtleties increase that relationship is what defines the entire problem. I realize that’s a little meta but these questions are a little meta. So I hope this helps! Wishing you guys a great day and like and subscribe below and we’ll see you real soon!

If you found this data sufficiency problem video helpful, try your hand at this percentage problem or this probability problem.

Mar 2020

Remainder Number Theory Problem

Today we’re going to be looking at the Remainder Number Theory problem:

If 23 is divided by some integer n, the remainder is 3.

I. Is n even?
II. Is n a multiple of 5?
III. Is n a factor of 120?

(A) I and II only
(B) III only
(C) I, II, and III
(D) II and III only
(E) II only

Our big question is that originally we’re given this unknown number N and we know we just have a remainder 3. So the problem is presenting us information in a way that we’re not used to seeing and what we need to do is work backwards from this to drive the core insights.

Sorting Through the Information

So if we have a remainder of 3 on 23 this means that the chunk that isn’t remainder is 20. So what can our n be in those cases that will allow us to divide out by 20 and leave this remainder 3?

Well, first we know that n has to be greater than 3 because in order to have a remainder the amount we’re dividing by has to be something greater. The moment the remainder equalizes the thing we’re dividing by of course we get one more tick in the dividing by box and the remainder goes back down to zero.

Solving

So with 23 and a remainder of 3, our key number to look at is 20. Our factors of 20, that is the things that divide evenly into 20, are 1, 2, 4, 5, 10, and 20. Of course, 1 and 2 are below 3 and so they’re not contenders. So we end up with n being 4, 5, 10, or 20.

Check Against the Statements

So for number 1: Is N even? If N can be 4 but can also be 5 then we’re not assured that it’s even. Notice the data sufficiency problem type embedded here. So N is not necessarily even.

Is N a multiple of 5? Once again N is not because N could be 4 or 5. Finally, is N a factor of 20? And in this case, it is because 4, 5, 10, and 20 as we just said are all factors of that 20 that we’re looking for. So our answer here is 3 alone, answer choice A.

More Practice

Now here’s a more challenging problem at the same form, see if you can do it and we’re going to come back and in the next video talk about the solution and give you another problem.

So if 67 is divided by some integer N the remainder is 7. Our three things to look at are whether:

- N is even?
- If N is a multiple of 10?
- Or N is a factor of 120?

So give this one a try and see if you can use the principles from the easier problem on this more challenging one to make sure that you actually understand what’s going on. If not, re-watch this video and see if a review might allow you to answer this question.

If you enjoyed using this video for practice, try this one next: Wedding Guest Problem.

Jun 2019

Number Theory Problem Form – Wedding Guest

Today, we’re going to be looking at what at first seems to be an allocation problem. But on further reflection, actually turns out to be a much simpler number theory problem. Let’s take a look at it:

At a wedding, the bride’s side has 143 guests and the groom’s side has 77 guests. What is the largest number of identical tables that can be created if each table has to have an equal number of guests from the bride’s side and the groom’s side? An identical table is one where the number of guests from the bride’s side is the same at every table and the number of guests from the groom’s side is the same at every table.

A. 3
B. 5
C. 7
D. 11
E. 13

If we take a look at the problem stem these numbers 143 and 77. They stick out to us and they stick out not just because they don’t seem to have any relative association but also because they’re sort of odd-looking numbers, they don’t look like most the numbers were used to seeing. Say, 48 or 24 or 36 – something easily divisible clearly breakable into factors. Here, we’re given these two disparate numbers and we’re being asked to formulate not what the tables are made up of but how many tables there are.

Solving the Problem

So, we look at these two numbers and we examine first the 77 because it’s a simpler lower number. 77 breaks into factors of 7 and 11. This clues us in as to what to look for out of the 143. 143 must have a factor of 7 or 11. And in fact, 143 is evenly divisible by 11 and it gives us 13.

This means that the maximum number of tables is 11. Each one has 13 people from the bride’s side and 7 people from the groom’s side. 13 plus 7 there are 20 people at each table. Times the 11 tables is 220.

And, we can back check our math, 143 plus 77 is 220. We don’t need to go that far but that might help deliver some comfort to this method. So in reality this is a very creative clever way the GMAT is asking us for the greatest common factor.

Graphical Solution Path

Another way to think about this is that we need an equal number of groups from the bride and groom side. The number of people on the bride’s side doesn’t have to equal the number of people on the groom’s side. We just need them broken in into the same number of equal groups. Graphically, the illustration shows us how a certain number of different sized groups combined into this common table. So 13 and 7 and we have 11 groups of each.

Number Theory Problem Forming

This is a great problem to problem form. It will give you some additional mental math or common result experience by forcing you to figure out numbers that you can present that at first don’t look like they match but in fact do have a common factor. You’ll notice that if they had given you 16 people and 36 people finding that common factor might be easier.

So as you problem form this try and do it in a way that sort of obscures the common factor. Either try it maybe where they have multiple common factors and you tweak things like what is the greatest number what is the fewest number of tables. Or even do a perspective shift and take a look at say how many guests are represented at each table. Or on the bride’s side or on the groom’s side. Of course, there are conceptual shifts to this and you can make this story about anything. Once you control the story or rather the structure of the story this problem becomes very very straightforward.

Hope this helped, I look forward to seeing you guys soon.