Since it’s Thanksgiving week, let’s think about pie for a second. No, not mathematical pi, just actual real edible pies. For Thanksgiving I’m in charge of making dessert, so I’ll be bringing two pies, one pumpkin and one apple. Let’s say that I sliced the apple pie into 12 pieces, and the pumpkin pie, since it held together better, into 18.

Fast forward to the end of the evening. My pies were a big hit, and I have almost none left. In fact, all I have is three pieces of apple and four pieces of pumpkin. I want to combine the remaining slices into a single pie pan, so that they take up less space in the fridge. How do I figure out if my remaining pie will fit in one pan?

Well, let’s start by writing down the remaining amounts of pie in the form of fractions. Remember, one of the definitions of a fraction is parts of a whole, so let’s apply that definition to figure out our starting fractions.

The apple pie was cut into 12 pieces, and we have three out of twelve left. So our apple pie fraction is

3

12

The pumpkin was cut into 18 pieces, and we have four left. So the pumpkin fraction is

4

18

To figure out how much pie we have left and whether it will fit into one pan, we’ll have to add these two fractions together.

3 + 4

12 18

Now, the first problem we run into is that our pies were cut into different amounts of pieces. We can’t accurately compare the two pies until we subdivide them into the same size pieces. So to do this, we’ll use a process that often confuses students when they first learn it: finding the Lowest Common Denominator.

I think this concept confuses students mostly because the terms used are long and complicated. So let’s talk about Lowest Common Denominator for a moment. We only specify Lowest because there are many different denominators that would work for this problem, but most of them are bigger than they need to be. We don’t want to work with numbers that are any larger than they need to be, so this process ensures that we don’t have to do any extra reducing after the fact.

We need to find a number that can divide evenly by both of our current denominators (though not necessarily at the same time). So all of the factors involved in either of our current denominators need to be present in our LCD. Let’s take the foolproof, long-way-round method and break each denominator down into its prime factors. You can do this with a factor tree:

12 18

3 x 4 3 x 6

3 x 2 x 2 3 x 3 x 2

So we see a lot of overlap here between the two denominators, right? Both contain a 3 and a 2. This is where the “lowest” part of the Lowest Common Denominator comes in. Worst case scenario, we could simply multiply each denominator by the other one and we’d come up with a number that both would divide by, right? But look at that overlap. That denominator might be common, but it certainly wouldn’t be the lowest – you’d be able to divide both top and bottom of each fraction by a 3 and a 2 (or by a 6, if you’re feeling efficient). We need all of those factors to be represented in the LCD, but we only need the duplicates to be represented once. What I like to do is rewrite the last lines of the factor trees one on top of the other, to make it easier to see the overlap:

12 = **3 x 2** x 2

18 = 3 x **3 x 2**

So that 3 x 2 of overlap we only need once. So let’s rewrite out the prime factors that we need in our LCD:

LCD = 3 x 3 x 2 x 2

There. We have two 3’s, so the 18 will be happy, and two 2’s, so the 12 will be happy. What does that come out to?

LCD = 3 x 3 x 2 x 2 = 36

So our target number is 36. Now, let’s go back to our original problem:

3 + 4

12 18

Now, remember that any number divided by itself is one? Good. And remember that one of the definitions of a fraction is a division problem you don’t want to do yet? Great. So we can multiply either of these fractions by a new fraction composed of a number over itself, and not change the actual value of the number. We’re simply rewriting it in a form that’s more useful to us. It’s basically the backwards application of the process for reducing a fraction. Only this time we’re not reducing; we’re making it more complex.

We’re going to do this in two separate cases. Just look at the first fraction for a moment. What number do we need to multiply the 12 by to get our target of 36?

If you’re stumped, just go back to the prime factor makeup of our target number and figure out what pieces are missing from the 12:

12 = 3 x 2 x 2

LCD = **3 x** 3 x 2 x 2

So we’re missing a 3. So multiply that first fraction by 3/3, which, remember, is just a fancy way of writing “1”, so we can do that without messing up the value of the fraction:

3 x 3 = 9

12 3 36

Okay, good sign. Our denominator is 36, which was our target. Let’s take the second case. Look at the second fraction:

4

18

18 = 3 x 3 x 2

LCD = 3 x 3 x 2** x 2**

So we’re missing a 2 here. Multiply by 2/2, which is just a fancy way of writing “1”:

4 x 2 = 8

18 2 36

Excellent! Hang on, we’re almost done. Now, what we’ve essentially found here are equivalent fractions for each of our originals that just happen to have the same denominator. Well, they don’t really “just happen” to have them; we did that on purpose to make our lives easier. So now we just plug these new forms of each fraction into the original problem:

9 + 8

36 36

And we have our pies subdivided into the same size pieces! Now we can add them normally to figure out if we have less than a whole pie’s worth of leftovers:

9 + 8 = 17

36 36 36

17 is less than 36, so we have less than a whole pie. We can safely put them both in the same pan without overflow.

Now if I could just figure out how to fit that turkey in the fridge…