*Number properties* questions are about how numbers “behave.” What happens when we multiply an even number by an odd number? What about when we divide a positive number by a negative? These are the kinds of questions raised by number properties questions. In answering these questions, some calculation will typically be involved. The better you understand the categories of numbers and how they interact, the quicker and more confident you’ll be in answering number properties questions.

If *n* is an integer, which of the following is not necessarily true about $n(n−1)(n−2)$?

A. It is even.

B. It is divisible by 3.

C. It is divisible by $n_{2}−n$.

D. It is a perfect cube.

We begin by defining *integer*: a number that appears on the real number line, such as $−7$, $0$, or $15$.

If *n* is an integer, then $n−1$ is the integer that comes immediately before $n$ and $n−2$ is the integer that comes before $n−1$. As an example, we could be talking about the integers $4$, $3$, and $2$. Since the order doesn’t matter when multiplying integers (multiplication is *commutative*), we might as well list them in ascending order, so $2$, $3$, $4$.

There is no further information we need to consider in the question itself, so we can begin evaluating the answer choices. Choice A is necessarily true because three consecutive integers are bound to contain at least one even number, and an even number always results in an even product. (Just one even number multiplied by a trillion odd numbers would still yield an even product!) We can see this in that $2×3×4=24$, but even if we try a set with only one even, the result is still even: $3×4×5=60$.

Choice B is a little trickier, but still necessarily true. Think of the factor tree: as long as you can find at least one three among the factors of a number, that number must be divisible by three. The products we already showed, $24$ and $60$, bear this out. Another way to say this is that given three consecutive integers, exactly one of them must be divisible by $3$. This makes their product divisible by $3$; interestingly, it also makes their sum always divisible by $3$; we can see this by labeling the integers $x$, $x+1$, and $x+2$ and adding them together to make $3x+3$. This expression can be factored as $3(x+1)$, which must be divisible by $3$.

To understand choice C, a little algebra is required. Because of the commutative and associative properties of multiplication (*note*: it’s not necessary to know these terms, but we mention them here in case you’ve heard of them), we are permitted to multiply any two of the three factors in this question together to understand an aspect of the overall product. You may have recognized that choice C’s $n_{2}−n$ is the same as $n(n−1)$. If the overall product is divisible by both $n$ and $n−1$, then it must be divisible by $n_{2}−n$.

This leaves us with choice D as the apparently correct answer. As logic teaches us (and as the CLT will test us!), we can disprove a universal (or “must”) statement by finding just one *counterexample*. But first, let’s define *perfect cube*: an integer that is the result of cubing (raising to the third power) an integer). Examples of perfect cubes include $1$, $8$, and $125$. The two products we’ve already generated–-$24$ and $60$–-are not perfect cubes (the closest perfect cubes to them are $27$ and $64$, respectively). So we have more than enough to show that choice D is *not* necessarily true, and therefore the answer!

Most number properties on the CLT will contain something about *odd* and *even* numbers, *positive* and *negative* numbers, or both. See **Flashcard Fodder** for what you need to know about these types of numbers.

We recommend quizzing yourself on the list below and adding flashcards for any of the facts you don’t know automatically:

$even+eveneven+oddodd+oddeven×eveneven×oddodd×oddpositive×negativenegative×negativepositive÷negativenegative÷negative =even=odd=even=even=even=odd=negative=positive=negative=positive $

If *p* is an integer and $p_{2}$ is odd, which of the following must be true?

A. $p$ is odd.

B. $p$ is even.

C. $p$ is prime.

D. $p$ is negative.

(spoiler)

The answer is **A**. Understanding that exponents tell us to multiply by the same factor, we can resort to our even and add rules to confirm that it doesn’t matter how many times you multiply an odd number by itself, it will remain odd. If $p_{2}$ is odd, that means $p$ must be odd, and any power of $p$ will also be odd. Answer choices C and D reflect possible realities, but not all odd numbers are prime and not all odd numbers are negative.

If *x* and *y* are integers such that $x+y$ is odd, which of the following statements must be true?

A. Both $x$ and $y$ are odd.

B. Both $x$ and $y$ are even.

C. One of $x$ and $y$ is even and the other is odd.

D. Neither $x$ nor $y$ is odd.

(spoiler)

The answer is **C**. Reviewing our even and odd rules, we see that there is only one way to make an odd sum: one even and one odd number. (The order doesn’t matter, since addition is commutative.) All three wrong answers would result in an even sum. Only choice C gives up the mix of opposites that we need. Note that this is even true is zero is one of the numbers; a little-known fact is that zero is an even number. Using $0$ in a sum with an odd number still fits the pattern.

If $m$ and $n$ are integers and *mn* is even, which of the following must be true?

A. $m$ is even.

B. $n$ is even.

C. At least one of $m$ or $n$ is even.

D. Both $m$ and $n$ are even.

(spoiler)

The answer is **C**. At least one of *m* or *n* must be even to make the product even. Remember, though: only one even number is required to make an even product, even if it’s paired with thousands of odds. An even number in a multiplication problem is like a drop of food coloring in a large bowl of water; even one “drop” (even number) is enough to affect the whole “bowl” (product).

If $r$ and $s$ are consecutive integers, which of the following is FALSE?

A. $r$ + $s$ is odd.

B. $rs$ is even.

C. $r$-$s$_ is odd…

D. $r_{2}−s_{2}$ is even.

(spoiler)

The answer is **D**. There are two ways to approach the problem. One way is to reason mathematically, observing that of two consecutive integers, one must be even and one must be odd. Working through the choices, we apply our even and odd rules. We should get an odd sum but an even product, so choices A and B are both true (and therefore wrong answers). Subtraction works the same as addition when it comes to even/odd rules, so choice C is just as true as A. But with choice D, we know by multiplication that we will still have one even and one odd integer even after squaring both. Subtracting even from odd (or vice versa) makes odd, so choice D is false and therefore the correct answer.
Alternatively, we can plug in numbers here. We could use 1 and 2, 2 and 3, or the like. If you try any pair of consecutive integers, you’ll find that choices A, B, and C are true every time, while choice D is false.

If *m* and *n* are integers such that $m_{2}=n_{3}$, which of the following statements must be true?

A. $m$ is a factor of

n.

B. $m$ is a multiple ofn.

C. $m$_ is a perfect square.

D. $n$ is a perfect cube.

(spoiler)

The answer is **B**. To understand the set of possible integers that can apply to this question, we should plug in numbers by trial and error. Interestingly, we can make the equation $m_{2}=n_{3}$ work if we use either 0 or 1. However, both of these are “weird” numbers (they behave differently than most numbers), and unfortunately, using either zero or one for both numbers makes all four answer choices true, so that doesn’t help us.

Math reasoning further reveals that we need to use positive numbers for $m$ and $n$, because a negative number squared is positive but a negative number cubed is negative. It would make sense to start with $n$, plugging in numbers until $n_{3}$ is also a perfect square. 2 and 3 don’t yield perfect squares when cubed, but 4 does: $4x4x4=64$, which is $8_{2}$. Excellent! We’ve found numbers that can work: 8 for $m$ and 4 for $n$. Let’s test the choices with these values.

Choice A doesn’t work; 8 is not a factor of 4, but rather a multiple. But because because that’s true, choice B appears to work well. Testing the others, we see that choice C is out (8 is not a perfect square), nor do choice D (4 is not a perfect cube).

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