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.

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.