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.