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Consecutive integers. This is one of the test’s favorite kind of math problems here. The word consecutive means in a row or one following another. And so, consecutive integers are integers that are in a row, 1, 2, 3, 76, 77, 78. Or negative 15, negative 14, negative 13. Those are examples of consecutive integers.
Here, for simplicity, I’ve just shown sets of three consecutive integers, but you could have five consecutive integers, ten consecutive integers, 50 consecutive integers. You could have any number of consecutive integers. And of course, they can be small numbers, large numbers, positive or negative. In fact, a set of three or more consecutive integers could contain both positive and negative numbers if the set contains zero.
So for example, the set negative 1, 0, 1 is a set of three consecutive integers that contains one positive number and one negative number. So let’s talk about some basic facts. A set of n consecutive integers will always contain one number divisible by n. So if we have three consecutive integers, one of them is divisible by 3. If you have 17 consecutive integers, one of them is divisible by 17.
So some of you may be wondering, that set that we just saw a moment ago, negative 1, 0, 1, that’s a set of three consecutive integers. Where’s the integer divisible by 3? Well, the trick here is that 0 is divisible by every integer. 0, in fact, is a multiple of every integer because you can take any integer and multiply it by 0 and you get 0.
So 0 is a multiple of every integer. Therefore, divisible by every integer, so that’s the number divisible by 3. Another fact. If n is odd, then the sum of a set of n consecutive integers will always be divisible by n. So this is interesting.
Take any three numbers in a row, add them up, you get something divisible by 3. Take any five numbers in a row, add them up, you get something divisible by 5, and so forth for any odd number. In a set of three consecutive integers, you could have two evens and one odd, or two odds and one even, depending on where you start. So in other words, if you are told that you have three consecutive integers, you don’t exactly know how many evens, how many odds without being given some other piece of information like the starting value or the middle value, something like that.
If you have a set of four consecutive integers, you know you absolutely have to have two evens and two odds. Even if you’re told nothing else, four consecutive integers automatically two evens and two odds. And in fact, that extends if you have an even number of consecutive integers. Then within that set, the evens and odds have to be evenly divided.
Now, gets interesting when we start talking about the algebraic representations of consecutive integers. It’s relatively easy to recognize consecutive integers when you’re given actual numbers. That’s not really a challenge. Test is not gonna print a bunch of consecutive numbers and then expect you to recognize that they’re consecutive integers.
That’s gonna be easy. What the test likes to do, they like to give you algebraic representations of consecutive integers. Well, what on earth does this mean? So, we’re gonna represent here consecutive integers using algebraic notation, using the variable n, where here n could be any integer at all.
And if n is any integer, all three of these sets are sets of consecutive integers. So for example, just to make things easy, let’s pick n equals 10. That first set would be 10,11,12,13. Four consecutive integers. The second set would be 8, 9, 10, 11, 12.
That would be 5 consecutive integers. And if n equals 10, the third set would be 25, 26 and 27. That would be three consecutive integers. So the test absolutely loves to give you algebraic representations of consecutive integers and expect you to recognize them in context. Here’s a very challenging problem.
I suggest pause the video and then we’ll talk about how to solve this. If n is an integer greater than 10, then the expression, that algebraic expression, must be divisible by which are the following? Well, this is a problem that can really drive people crazy because if they go down the road of trying to plug in numbers, well, let’s plug in n equals 11, n equals 12, n equals 13, you’re gonna get horribly large, complex numbers.
It’s gonna be a real disaster. So, there are many ways to make this problem an immensely difficult, virtually insolvable problem. It turns out that there’s a trick, as there often is with difficult math questions. And the trick is, let’s look at that algebraic expression and notice that first term, n squared minus two n.
We could factor out an n from that expression, in which case we’d have n times n minus 2, and then that would be multiplied by b plus 1 and then n minus 1. Now we have four different factors. Of course, you can multiply four factors in any order you want. So let’s rearrange the order. If we write them in this order, we see, aha, what we really have here in hidden form is a product of four consecutive integers.
That unlocks the whole problem. We know if we have four consecutive integers, we have two even and two odd. Well, if we’re multiplying two even numbers, each number is divisible by 2, so that means the product would have to be divisible by 4. So the product is definitely divisible by 4. If we have three consecutive integers in a row, one of them is divisible by 3.
So if we have four in a row, we definitely have at least one divisible by 3, so we have a factor of 3. We have even numbers, so we have a factor of 2. So, all we need is a factor of 3 and a factor 2 to give us a factor of 6. So, this is definitely divisible by 6. 18, this is a little trickier.
18 is 2 times 3 times 3. Okay. Well, we know that we have one factor of 2 at least. We have more than one factor 2 actually. We know we have one factor 3, do we have a second factor 3? Are we guaranteed that we have a second factor of 3?
It would be possible. Let’s think about just four consecutive integers, 3, 4, 5, 6. It’s possible to get two numbers that are divisible by 3 in a set of four consecutive integers. But it’s also possible, 4, 5, 6, 7, that we could only have one number divisible by 3.
And since this is a must be true kind of question, we can’t assume that we have two consecutive integers. We could, that we have two numbers divisible by 3. We could have two numbers divisible by 3, but we don’t necessarily have two numbers divisible by 3. So we cannot conclude with absolute certainty that this is gonna be divisible by 18.
So it has to be divisible by 4 and 6, and therefore, the answer is answer choice C. Sometimes, the test may also ask about consecutive multiples of a number. What does this mean exactly? Here’s some examples, 25, 30, 35, 40. Notice that is 5 times 5, 6, 7, and 8.
In other words, if you take any number and multiply it by a set of consecutive integers, you get consecutive multiples of that number. Another way to say this is 25 is a multiple of 5, and then all we are doing is we are simply adding 5 after that, so plus 5, 30, plus 5, 35, plus 5, 40. 35, 42 and 49 are consecutive multiples of 7, all divisible by seven.
And in fact, we start with 35, we just add 7 and add 7, we get the other two numbers. 48, 51 and 54 are consecutive multiples of 3. That’s 3 times 16, 17 and 18. And of course, what we’re doing is,we’re just adding 3 to get from one number to the next. 65, 78, 91 and 104 are consecutive multiples of 13.
That is, 13 times 5, 6, 7 and 8. Then we’re started, we can start with 65,. You just add 13 and you get the other numbers. So the test likes asking questions about consecutive multiples. Here’s an example of a question. So pause the video here and then I’ll talk about this.
N equals 135 is the lowest number of a set of 11 consecutive multiples of 5. What is the difference between the lowest and the highest number in the set? So, piece of strategy number one. If we’re just comparing the difference between the highest and the lowest, it doesn’t matter where it is on the number line.
Sso we have our low, we have our high, and then there’s gonna be some difference between them. Now, the test is telling us explicitly that this low number’s 135 and then it goes up to some other number. We could figure out that big number. But the point is we don’t have to because we could put this anywhere on the number line.
Anywhere on the number line that we have a set of 11 consecutive multiples of 5, there’s gonna be the same difference between the highest and lowest. So I’m gonna just make it easier. I’m gonna ignore that 135. I’m just gonna say the lowest number is 5. And of course, that is 5 times 1.
Of course, the next number would 10. That’s 5 times 2. And we go up 11 multiples, so the 11th multiple, of course, would have to be 5 times 11, which is 55. And so really, we want the difference between the high and the low. 55 minus 5 is just 50.
That’s the difference. So that’s a very easy way to solve that question. In summary, we talked about consecutive integers, the examples and basic rules. We talked about the very important topic of the algebraic representation of the consecutive integers. That is, ordinarily how consecutive integers will appear on the test.
And we also talked about the issue of consecutive multiples of a number.
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