https://focustutoring.com/gmat-blog/gmat-odds-and-evens/By: Rich Zwelling, Apex GMAT Instructor

Date: 18th March, 2021

Last time, we signed off with an Official Guide GMAT problem that provided a nice segue into Number Theory, specifically today’s topic of GMAT Odds and Evens. Now we’ll discuss the solution. Here’s the problem, in case you missed it and want to try it now:

*If x is to be chosen at random from the set {1, 2, 3, 4} and y is to be chosen at random from the set {5, 6, 7}, what is the probability that xy will be even?*

*A) 1/6
*

*B) 1/3*

*C) 1/2*

*D) 2/3*

*E) 5/6*

**Method #1 (Certainly passable, but not preferable)**

Since there are so few numbers involved here, you could certainly take a brute-force approach if pressed for time and unsure of a faster strategy. It doesn’t take long to map out each individual product of *x*y* systematically and then tally up which ones are even. Here I’ll use red to indicate **not even** and green to indicate **even**:

**1*5 = 5
**

**1*6 = 6**

**1*7 = 7**

**2*5 = 10
**

**2*6 = 12**

**2*7 = 14**

**3*5 = 15
**

**3*6 = 18**

**3*7 = 21**

**4*5 = 20
**

**4*6 = 24**

**4*7 = 28**

Since we know that all probability is Desired Outcomes / Total Possible Outcomes, and since we have 8 even results out of 12 total possible outcomes, our final answer would be 8/12 or **2/3**.

However, this is an opportune time to introduce something about odd and even number properties and combine it with the method from our “undesired” probability post…

**Method #2 (Far preferable)**

First, some number theory to help explain:

You might have seen that there are rules governing how even and odd numbers behave when added, subtracted, or multiplied (they get a little weirder with division). They are as follows for addition and subtraction:

Even ± Odd = Odd

Odd ± Odd = Even

And with multiplication, the operative thing is that, when multiplying integers, just *a single even number* will make the entire product even. So for example, the following is true:

Odd * Odd * Odd * Odd * Odd * Odd * Odd * Odd * Odd * … * Odd = Odd

But introduce just a single even number into the above product, and the entire product becomes even:

**Even** * Odd * Odd * Odd * Odd * Odd * Odd * Odd * Odd * Odd * … * Odd = **Even**

This makes sense when you think about it, because you are *introducing a factor of 2 to the product*, so the entire product must be even.

When considering the above rules, you could memorize them, or you could turn to SCENARIO examples with simple numbers to illustrate the general pattern. For example, if you forget what Odd * Odd is, just multiply 3 * 5 to get 15, which is odd, and that will help you remember.

This can also help you see that introducing just a single even number makes an entire product even:

3 * 3 * 3 = 27

2 * 3 * 3 * 3 = 54

So how does all of that help us get the answer to this problem faster?

Well, the question asks for the *desired* outcome of *xy being even*. That presents us with three possibilities:

1. *x* could be even AND *y* could be odd

2. *x* could be odd AND *y* could be even

3*. x* could be even AND *y* could be even

This is why I, personally, find it less helpful to think of individual multiplication rules for even numbers and more helpful to think in terms of: **“The only way a product of integers is odd is if every integer in the set is odd.”**

Because now, we can just think about our *undesired* outcome, the only outcome left:

*4. x* is odd AND *y* is odd, making the product *xy* odd

And how many such outcomes are there in this problem? Well at this point, we can treat it like a simple combinatorics problem. There are two odd numbers in the *x *group (1 and 3) and two odd numbers in the *y* group (5 and 7):

_2_ * _2_ = 4 possible odd *xy* products

And for the total, there are four possible *x* values and three possible *y* values:

_4_ * _3_ = 12 total possible *xy* products

That gives us a 4/12 or 1/3 probability of getting our *undesired* odd *xy* product. And as discussed in the previous post, we can now simply subtract that 1/3 from 1 to get:

1 – 1/3 = **2/3** probability that we get our *desired* even *xy* product.

For a little “homework,” try the following Official Guide GMAT problem. It has an underlying topic that we’ll discuss next time:

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

*(1) x = y + 1.*

*(2) x/y is an even integer.*

If you enjoyed this GMAT odds and evens article watch Mike solve this Number Theory problem with multiple solution paths.

Find other GMAT Number Theory topics here:

Odds and Ends (…or Evens)

Consecutive Integers (plus more on Odds and Evens)

Consecutive Integers and Data Sufficiency (Avoiding Algebra)

GMAT Prime Factorization (Anatomy of a Problem)

A Primer on Primes