Welcome back to our series on number properties. The last two articles introduced GMAT quant problems that revolve around place value in multi-digit integers. This article returns to the “divisibility and factors” core of number properties. But now we are looking at the “flip side” of divisibility and factors: consecutive integers, multiples of integers and, particularly, their “spacing” on the number line. Let’s talk through the key properties:
Counting from 1, every nth integer (on a number line) is a multiple of n.
It’s not necessary to refer to a number line for this to be true, but I find the spatial aspect of number lines helpful.
Here’s the property at work:
1 2 3 4 5 6 7 8 9 10 11 12
We see that counting from 1, every third integer is a multiple of 3. This should make sense. Here is another, more general way to express this property:
Multiples of positive integer n occur every nth integer (on a positive number line).
This version of the property leaves out the “counting from 1” specification. When stated this way, we don’t have to know “where we are” on a number line. This leads to a helpful application of the principle:
Any series of n consecutive positive integers contains one multiple of integer n.
Regardless of “where we are” on a number line, a series of 3 consecutive integers contains a multiple of 3. Since multiples of 3 occur every third integer, we can’t list three consecutive integers without one of them being a multiple of 3.
Try listing four consecutive positive integers, none of which is a multiple of 4 – or five consecutive positive integers, none of which is a multiple of 5. It’s impossible!
Here’s a final extension of the principle:
The product of n consecutive positive integers is divisible by positive integer n.
When a series of 3 consecutive positive integers is multiplied together, the multiple of 3 in the series becomes a factor of the product. Therefore the product has at least one prime factor of 3 and is divisible by 3.
Let’s put this knowledge to work on some official GMAT quant problems:
How many integers between 10 and 20, inclusive, have a remainder of 1 or 2 when divided by 3?
I love this problem because it combines the property we’re learning in this article with our remainders article from this series!
Let’s think about the possible remainders when dividing by 3. The remainder when dividing an integer by 3 can only be 1, 2, or 0. This problem asks how many integers in the range 10 to 20, inclusive, have remainders of 1 or 2 when divided by 3. This is another way of asking how many integers in the range are NOT multiples of 3 (since the multiples of 3 are the numbers that, when divided by 3, have a remainder of 0).
The best way to go about answering the question is to take the full number of integers and subtract the number of multiples of 3. 10 to 20 inclusive is 11 integers, and the only multiples of 3 in the range are 12, 15, and 18. So the number of non-multiples of 3 (or the number of integers that leave a remainder of 1 or 2 when divided by 3) is 11 – 3 = 8. The correct answer is B.
Here’s a more challenging problem:
If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?
This problem serves as an introduction to a common way for the GMAT to encode the concept of consecutive integers. “The product of three consecutive integers” can be encoded as n(n + 1)(n + 2) or (n – 1)(n)(n + 1) or even n(n – 1)(n – 2).
Divisibility by 8 (the property at issue here) depends on possessing all the prime factors of 8. 8 = 23, so any product of three consecutive integers that has three prime factors of 2 will be divisible by 8.
Let’s look at a common case of the “spacing of multiples of integer n” property that we’re learning in this article. We can say that multiples of 2 (also known as even integers) occur every second integer. Odds and evens alternate! A series of three consecutive integers can only be even->odd→even or odd->even->odd.
The relevance of this property to the problem at hand is that whenever n is even, n + 2 is also even. Since half of the integers from 1 to 96 inclusive (half of the n values) are even, half of the three-consecutive-integer series in the range have at least two prime factors of 2. In such cases, n and (n + 2) are both even, each providing at least one prime factor of 2.
Here’s another helpful property for this problem: in the series of even integers, every other integer is a multiple of 4. We can think of the even-numbered members (the second, fourth, sixth, etc. members) of the series of even integers as “double even” numbers. They have at least two prime factors of 2. In other words, they are divisible by 4.
For this problem, this means that whenever n and (n + 2) are even (which is true in half of the cases), one of the values has at least two prime factors of 2. So together, when n and (n + 2) are even, they have a total of at least three prime factors of 3, and the product n(n + 1)(n + 2) is divisible by 8.
We can eliminate answers A and B (¼ and ⅜), because we know already that at least half of the n values in the range lead to a product that is divisible by 8. So is answer choice C (½) correct, or are some of the odd→even->odd products divisible by 8 too?
If n and (n + 2) are both odd, then the only way for the product (n)(n + 1)(n + 2) to be divisible by 8 is if the single even integer (n + 1) “carries the team” and contains all three required prime factors of 2. Or, perhaps more simply, whenever (n + 1) is a multiple of 8, the product n(n + 1)(n + 2) will be divisible by 8.
Given the property that multiples of integer 8 occur every 8th integer, (n + 1) will be a multiple of 8 in ⅛ of the cases. So the total fraction of cases where n(n + 1)(n + 2) is divisible by 8 is ½ + ⅛. (The ½ is from our work with the even->odd->even possibilities). ½ + ⅛ = 4/8 + ⅛ = ⅝. The correct answer is D.
That was challenging, so here’s a “cool down” official problem to close the article:
How many positive three-digit integers are divisible by both 3 and 4?
Remembering that divisibility is all about prime factors, we should translate this question like so: how many positive three-digit integers have at least one prime factor of 3 and two prime factors of 2?
We can go further by multiplying all these prime factors together to find the product 12. So for a final “translation” of the question: how many positive three-digit integers are divisible by – or multiples of – 12?
This is the product of the original integers 3 and 4, but this is only true because 3 and 4 share no prime factors. Be careful with cases like this. If the integers in the question shared any prime factors, it would be an erroneous oversimplification to simply multiply them together. Always think in terms of prime factors.
Now that we’re thinking in terms of multiples (via prime factors), we can apply the “spacing of multiples” property to easily solve this problem. There are about 900 (to be precise, 899) positive three-digit integers, and we need to know how many of these integers are multiples of 12.
Since multiples of 12 occur every 12th integer, about 1/12 of these 900ish integers are multiples of 12. I say “about” because it also depends on how the range of integers “lines up” with multiples of 12. As a small example, a series of thirteen consecutive integers may contain one or two multiples of 12. If the first integer in the series is a multiple of 12, the last one will be as well. But in every other case, the series will contain only one multiple of 12.
The answer choices for this problem differ enough from one to the next that we don’t need such precision. The quotient of 900/12 will get us close enough to determine the correct answer. In fact, it happens to work exactly. 900/12 = 75, so the correct answer is A.
This concludes our study of the spacing of multiples and series of consecutive integers. In the next and final article in this series, we’ll learn some special properties of even and odd integers.
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Contributor: Elijah Mize (Apex GMAT Instructor)