Finding Consecutive Multiples Using the Sum-Product Pattern
“Two consecutive positive multiples of three has a product of 54. What is their sum?”
Knowing our math vocabulary is always extremely important. Let’s look at what sort of terms we would need to know to solve this type of problem.
They need two “consecutive multiples” of 3 that are positive. That’s a lot at once, so let’s break that down. What do all these terms mean?
“Consecutive” you should have a fairly good clue about. If I wanted to show you consecutive numbers as multiples of 1 — in other words consecutive multiples of 1 — well, you do that all the time! It’s just basic counting. Example: 1, 2, 3, 4, 5, 6, …
Now here we are with consecutive multiples of 3. That means that we’re going in multiples of 3, just as you first learned your times tables back in elementary school. And these are consecutive, meaning they follow one after the other. Example: 3, 6, 9, 12, 15, 18, ….
What we’re doing therefore is looking for just two terms (they could be 15 and 18 or 9 and 12) that are consecutive (they follow one right after the other). I couldn’t choose, say, 3 and 18 because those are not consecutive; they don’t come one right after the other.
What’s also important in this problem? It’s important that they are asking for positive multiples of 3. Here’s something you won’t see for example: -3, -6, -9, … We can’t give consideration to these because they are not going to be possible answers. Even if this is going in the positive direction, such as -3, 0, 3, 6, 9 — these can’t work either (or at least -3 can’t, and it’s debatable whether we could even include 0). So that’s definitely not within the scope of our problem. Furthermore, if the product of the two numbers comes to 54, 0 is definitely out of the picture right away (since no other number multiplied by 0 could give us 54).
So 0 is out of the picture and negatives are out of the picture. We can only work with positive numbers and we’ll only need two of them.
That really breaks down all the terms that we have to worry about, but the question is still, how do we go about figuring out the product of two numbers that we don’t know (and are only given the clue of 54 being their product)? We know what it means now that they are consecutive positive multiples of 3, so the hint to a solution must be buried somewhere in that.
Here’s what we can do. I’m going to think of any unknown number as “\(x\)”. So the first term (the first of the two numbers), let’s just say, is \(x\):
1st number: \(x\)
Now, what is the second number going to be? Let’s think about it: we have to count in multiples of 3 — and it doesn’t necessarily have to start at 3, e.g., it could have started at 1 or 5 and then gone up by 3’s from there. Regardless, it’s going to start from our unknown number (or \(x\)) and we’re going to have to add 3 to it. So the first number is \(x\) and the second number is \(x+3\).
1st number: \(x\) 2nd number: \(x+3\)
We’ve just set up our problem, only instead of doing it as a word problem (ugh!), now we’ve set it up in mathematical language (awesome!):
\(x * (x+3) = 54\)
Next, we know our first number times our second number equals 54, and that’s exactly the hint they gave us. Now all we have to do is find their sum!
Here’s how we can do that (using the “sum-product” pattern or method):
\(C_{factor 1}+C_{factor 2}=B\)
This “formula” says that the first factor of \(C\) plus the second factor of \(C\) equals \(B\). This is the rule that I’m going to use to solve this. And you’re probably wondering, “What on earth does that mean!?”
I’ll be you will recognize it once we’ve used it. First, let’s just distribute the \(x\):
\(x^2+3x=54\)
And then let’s simply subtract 54 from both sides so that everything is equal to 0:
\(x^2+3x-54=0\)
What does this look like? Well, it looks a lot like the sort of quadratic equations that we’re constantly asked to solve in algebra 1 all the way through algebra 2! Factoring this will give us the two factors we need to look at to find \(B\).
What is \(B\) again, you ask?
Recall that the way a quadratic equation is typically expressed in standard form is
\(Ax^2+Bx+C=0\)
So the sum-product pattern is what we’ll use to find the factors pattern is what we will use to find the factors of this particular equation because \(A=1\). (In other words, when you have the equation in the form \(x^2+Bx+C=0\), we can use the sum-product pattern/method.)
What we do is we take -54 (our \(C\) term) and look for two factors that can go into -54 that also will add up to give us our \(B\) term (3):
some first number \(×\) some second number \(=-54\) , and
same first number \(+\) same second number \(=3\)
For a first try: 1 and 54 are two factors of 54. But these aren’t going to be helpful at all. There’s no way that I could add or subtract 1 and 54 that would give me 3, so we’ve got to keep going. If we try 2 and 27, that doesn’t look like that’s going to bring us close to 3 either…
Here’s a hint: we could spend a lot of time doing it this way (54 will also be divisible by 3 and so forth), but if we’re trying to find two terms that we can either add or subtract together that’ll give us a such a small number as 3, then these two numbers had better be pretty close together.
Let’s try coming from the opposite direction (from bigger factors closer together to small one farther apart). The closest factors to each other that are still factors of 54 are 6 and 9. And that’s going to work! But clearly these two factors being multiplied together requires one of them to be negative to come up with -54, so the question is, if I were to combine them would a positive 6 and a negative 9 give us a (positive) 3? Or would a negative 6 and a positive 9 give us 3? Hopefully the answer is pretty clear:
\(-6×9=-54\) , and
\(-6+9=3\)
It would have to be a negative 6 and a positive 9 if we want a positive 3. So converting these into factors for our quadratic expression now, we get:
\((x-6)(x+9)=0\)
This makes sense since what we’ve essentially done (through the sum-product pattern) is “de-FOIL” (or “un-FOIL”?) the expression into these two binomials. Were we to expand them we would get back our original quadratic equation, so this is correctly factored.
Now what? Let’s think about what this result means: when we have solved for x-intercepts before using quadratic equations, we’ve known what the next step for solving for \(x\) looks like. We have to split up the two binomials such that either one of these could equal 0 — so we make them both equal to 0:
\((x-6)(x+9)=0\) \(x-6=0\) and \(x+9=0\)
Equating them each to zero and finding \(x\) for each of them is now just the simplest of algebra: we just carry the number over to the other side!
\(x-6=0\) and \(x+9=0\) \(x=6\) \(x=-9\)
(This is precisely how we would normally find the intercepts of a parabola on the x-axis — or just solve for \(x\) in general, as you’ve been taught to do.)
What we’ve solved just now by finding these values of \(x\) is the value of the first of those two multiples of 3 — those very first two numbers in other words. Our first number is actually equal to one of these solutions for x! But which one?
1st number: \(6\)? or \(-9\)?
2nd number: \(x+3\)
The problem asks for consecutive positive multiples of three, so immediately we should see that -9 is going to be extraneous to what we need. So the only possible value that \(x\) could be for this number is 6.
1st number: \(6\) 2nd number: \(6+3=9\)
There you have it! The first number is 6 and the second number must be 6+3 = 9. Almost done! So to finally answer the question, what is the sum of these two numbers (drumroll, please):
1st multiple of 3 \(+\) 2nd multiple of 3 \(=\) ?
\(6+9=15\)
That’s our answer!
