GMAT Number Theory - Articles, Videos & Problems

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.

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.

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.

Number Theory