GMAT Prime numbers article with questions
Posted on
29
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) 108
C) 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.

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

 

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Consecutive Integers and Data Sufficiency (Avoiding Algebra) Article
Posted on
25
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…

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

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gmat consecutive integers article
Posted on
23
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=35/3 (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:

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.

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

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odds and evens on the gmat
Posted on
18
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.

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

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Posted on
03
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.

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.

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.”

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.

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.

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.

 

 

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Posted on
14
Mar 2020

Remainder Number Theory Problem

Today we’re going to be looking at this problem and 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 it 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 in 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.

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Posted on
21
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.

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 were 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.

Which 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’s 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 a 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 broke 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.

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.

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