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Welcome back to our series on number properties. This final article in the series will introduce some properties of even and odd integers that appear very frequently on GMAT quant. It’s not an overstatement to say that you should be looking for properties of evens and odds on almost every GMAT quant problem. We speak of the evenness or oddness of an integer as its identity. At the very least, a problem may give information from which you can deduce the identity of the correct answer choice, which can help you eliminate choices.

## Properties of even and odd integers

Even and odd numbers have certain properties when they are being multiplied, added, or subtracted. For our purposes, addition and subtraction are the same thing – it doesn’t matter which is occurring. Here are the properties for multiplication:

• In multiplication, the presence of a single even factor leads to an even product.

• An even product has at least one even factor.

• An odd product has only odd factors.

• A set of odd factors has an odd product.

To consolidate all of these rules into one, think about an even product like a drop of food coloring in a bottle of water – it colors all the liquid. It doesn’t matter how many odd factors there are – if even one factor is even, the product will be even as well.

Conversely, when you see an odd integer, you know that all of its factors or odd. This is another instance of the “contrapositive” relationship from the fifth article in this series. An even factor yields an even product. So if a product is odd, it must not have any even factors.

In terms of prime factors, this makes sense. Every even number (except for 0) has a prime factor of 2. So now matter how many odd factors exist in multiplication, the even factor will always smuggle that prime factor of 2 into the product, making the product even.

### Addition/Subtraction properties of even and odd integers

Now for the addition/subtraction properties of even and odd integers. In addition and subtraction, odd integers behave like negative numbers in multiplication or division.

When multiplication or division involves negative numbers, an odd number of negatives leads to a negative product or quotient, while an even number of negatives leads to a positive product or quotient. Every two negative members “cancel” each other. In the same way, every two odd addends “cancel” each other to render an even sum.

• If addition/subtraction involves an odd number of odd members, the sum is odd.
• If addition/subtraction involves an even number of odd members, the sum is even.

## Official GMAT problem for practice

Let’s get into some official GMAT problems:

If x and y are positive integers, is xy even?

1. x2 + y2 – 1 is divisible by 4.
2. x + y is odd.

(A) Statement (1) alone is sufficient, but statement (2) alone is not sufficient.

(B) Statement (2) alone is sufficient, but statement (1) alone is not sufficient.

(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

(D) EACH statement ALONE is sufficient.

(E) Statements (1) and (2) TOGETHER are not sufficient.

Let’s take statement 2 first since it’s much simpler. If x + y is odd, then one of the integers is odd and the other is even. Since one of them is even, their product xy is even. Statement 2 is sufficient.

Statement 1 tells us that x2 + y2 – 1 is divisible by 4. Let’s just say that x2 + y2 – 1 is even. If this is true, then x2 + y2 is odd. So one of the addends, x2 or y2, is odd, and the other is even. x2 is the product of the factors x * x, and y2 is the product of the factors y * y. Since every odd product has only odd factors, either x or y must be odd, with the other being even. Again, with an even factor, the product xy is even. Statement 1 is also sufficient, and the correct answer is D.

Here’s another:

If a  and b are integers, is a + b + 3 an odd integer?

1. ab is an odd integer.
2. ab is an even integer.

(A) Statement (1) alone is sufficient, but statement (2) alone is not sufficient.

(B) Statement (2) alone is sufficient, but statement (1) alone is not sufficient.

(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

(D) EACH statement ALONE is sufficient.

(E) Statements (1) and (2) TOGETHER are not sufficient.

For statement 1, if the product ab is odd, then each of the factors a and b is odd. Thus a + b + 3 can be represented as ODD + ODD + ODD, which has an odd sum (because the number of odd addends is odd). Statement 1 is sufficient.

Statement 2 tells us that a and b differ by an even integer. This means that a and b have the same identity. Any two odds differ by an even integer, and any two evens differ by an even integer. Alternatively, if two integers differ by an odd integer, one of the integers is odd and the other is even. In this case, we can’t determine whether a and b are both odd or both even, but the statement is still sufficient!

Remember that we were asked about the identity of the sum a + b + 3. If a and b are both odd, the sum is odd, as observed before. But if a and b are both even, the sum is still odd, because 3 is the only odd addend! So statement 2 is sufficient as well, and the correct answer is D.

### Official problem involving prime addends for practice

If the sum of a set of ten different positive prime numbers is an even number, which of the following prime numbers CANNOT be in the set?

(A) 2

(B) 3

(C) 5

(D) 7

(E) 11

This problem can be solved easily with a property of primes that we haven’t mentioned since the first article in this series: 2 is the only even prime number. This problem involves an even number of addends with an even sum. Since the sum is even, the number of odd addends must be even. Including the prime number 2 as one of the addends would reduce the number off odd addends from ten to nine, breaking the rule about the even sum. Therefore 2 cannot be included in the set, and the correct answer is A.

### Related data sufficiency problem for practice

Here’s a related data sufficiency problem:

Is the sum of four different positive prime integers an odd integer?

1. One of the integers is 11.
2. One of the integers is 2.

(A) Statement (1) alone is sufficient, but statement (2) alone is not sufficient.

(B) Statement (2) alone is sufficient, but statement (1) alone is not sufficient.

(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

(D) EACH statement ALONE is sufficient.

(E) Statements (1) and (2) TOGETHER are not sufficient.

Since all of these addends are prime, the only uncertainty regarding even/odd identity depends on whether 2, the only even prime number, is one of the addends. Statement 1 doesn’t tell us this, but statement 2 does. The correct answer is B.

### Number properties problem for practice

Here’s a final problem for the article and for our long series on number properties. A bittersweet moment indeed:

Is the positive integer n odd?

1. n2 + (n + 1)2 + (n + 2)2 is even.
2. n2 – (n + 1)2 – (n + 2)2 is even.

(A) Statement (1) alone is sufficient, but statement (2) alone is not sufficient.

(B) Statement (2) alone is sufficient, but statement (1) alone is not sufficient.

(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

(D) EACH statement ALONE is sufficient.

(E) Statements (1) and (2) TOGETHER are not sufficient.

First of all, the difference between the statements is nothing but whether these three consecutive integers are being added or subtracted. This is irrelevant to the question of even/odd identity. So if one statement by itself is sufficient, the other statement is as well. And if one statement by itself is insufficient, the other statement is as well, and combining the statements won’t help since they are essentially redundant. The correct answer can only be D or E.

In either statement, we have three addends with an even sum. This means that the number of odd addends is even. Since these are three consecutive integers, represented as n and (n + 1) and (n + 2), they must be odd->even->odd or even->odd->even. Only the odd->even->odd scenario is allowed by the “even sum” constraint, so n2 is odd, (n + 1)2 is even, and (n + 2)2 is odd again. Finally, the odd integer n2 is the product of the factors n * n. Since every odd product has only odd factors, n must be odd. Either statement is sufficient, and the correct answer is D.

This concludes our study of evens and odds, and here ends our series of articles on number properties. We’ve learned a lot, so be sure to put your knowledge to work in continued practice!

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