### All SAT Math Resources

## Example Questions

### Example Question #1 : How To Multiply Even Numbers

If x is an even integer and y is an odd integer. Which of these expressions represents an odd integer?

I. xy

II. x-y

III. 3x+2y

**Possible Answers:**

II only

I and II only

I and III only

II and III only

I, II, and III only

**Correct answer:**

II only

I)xy is Even*Odd is Even. II) x-y is Even+/-Odd is Odd. III) 3x is Odd*Even =Even, 2y is Even*Odd=Even, Even + Even = Even. Therefore only II is Odd.

### Example Question #2 : How To Multiply Even Numbers

If x is an even number, y is an odd number, and z is an even number, which of the following will always give an even number?

I. xyz

II. 2x+3y

III. z^{2 }– y

**Possible Answers:**

II only

I only

I, II and III

II and III only

I and II only

**Correct answer:**

I only

I. xyz = even * odd * even = even

II. 2x + 3y = even*even + odd*odd = even + odd = odd

III. z^{2 }– y = even * even – odd = even – odd = odd

Therefore only I will give an even number.

### Example Question #3 : How To Multiply Even Numbers

If *x* and *y *are integers and at least one of them is even, which of the following MUST be true?

**Possible Answers:**

*x* + *y* is odd

*xy* is odd

*x* + *y *is even

*xy* is even

Nothing can be determined based on the given information

**Correct answer:**

*xy* is even

Since we are only told that "at least" one of the numbers is even, we could have one even and one odd integer OR we could have two even integers.

Even plus odd is odd, but even plus even is even, so *x* + *y* could be either even or odd.

Even times odd is even, and even times even is even, so *xy* must be even.

### Example Question #26 : Even / Odd Numbers

Let *a* and *b* be positive integers such that *ab*^{2} is an even number. Which of the following must be true?

I. *a*^{2 }is even

II. *a*^{2}*b* is even

III. *ab* is even

**Possible Answers:**

II and III only

I, II, and III

I only

I and II only

II only

**Correct answer:**

II and III only

In order to solve this problem, it will help us to find all of the possible scenarios of *a*, *b*, *a*^{2}, and *b*^{2}. We need to make use of the following rules:

1. The product of two even numbers is an even number.

2. The product of two odd numbers is an odd number.

3. The product of an even and an odd number is an even number.

The information that we are given is that *ab*^{2} is an even number. Let's think of *ab*^{2} as the product of two integers: *a* and *b*^{2}.

In order for the product of *a* and *b*^{2} to be even, at least one of them must be even, according to the rules that we discussed above. Thus, the following scenarios are possible:

Scenario 1: *a* is even and *b*^{2} is even

Scenario 2: *a* is even and *b*^{2} is odd

Scenario 3: *a* is odd and *b*^{2} is even

Next, let's consider what possible values are possible for *b*. If *b*^{2} is even, then this means *b* must be even, because the product of two even numbers is even. If *b* were odd, then we would have the product of two odd numbers, which would mean that *b*^{2} would be odd. Thus, if *b*^{2} is even, then *b* must be even, and if *b*^{2} is odd, then *b*^{ }must be odd. Let's add this information to the possible scenarios:

Scenario 1: *a* is even, *b*^{2} is even, and *b* is even

Scenario 2: *a* is even, *b*^{2} is odd, and *b* is odd

Scenario 3: *a* is odd, *b*^{2} is even, and *b* is even

Lastly, let's see what is possible for *a*^{2}. If *a* is even, then *a*^{2 }must be even, and if *a* is odd, then *a*^{2} must also be odd. We can add this information to the three possible scenarios:

Scenario 1: *a* is even, *b*^{2} is even, and *b* is even, and *a*^{2} is even

Scenario 2: *a* is even, *b*^{2} is odd, and *b* is odd, and *a*^{2} is even

Scenario 3: *a* is odd, *b*^{2} is even, and *b* is even, and *a*^{2} is odd

Now, we can use this information to examine choices I, II, and III.

Choice I asks us to determine if *a*^{2} must be even. If we look at the third scenario, in which *a* is odd, we see that *a*^{2} would also have to be odd. Thus it is possible for *a*^{2} to be odd.

Next, we can analyze *a*^{2}*b*. In the first scenario, we see that *a*^{2} is even and *b* is even. This means that *a*^{2}*b* would be even. In the second scenario, we see that *a*^{2} is even, and *b* is odd, which would still mean that *a*^{2}*b* is even. And in the third scenario, *a*^{2} is odd and *b* is even, which also means that *a*^{2}*b* would be even. In short, *a*^{2}*b* is even in each of the possible scenarios, so it must always be even. Thus, choice II must be true.

We can now look at *ab*. In scenario 1, *a* is even and *b* is even, which means that *ab* would also be even. In scenario *2*, *a* is even and *b* is odd, which means that *ab* is even again. And in scenario 3, *a* is odd and *b* is even, which again means that *ab* is even. Therefore, *ab* must be even, and choice III must be true.

The answer is II and III only.

### Example Question #4 : How To Multiply Even Numbers

Let equal the product of two numbers. If , then the two numbers COULD be which of the following?

**Possible Answers:**

8 and 8

20 and 4

32 and 2

0 and 16

2 and 8

**Correct answer:**

2 and 8

The word "product" refers to the answer of a multiplication problem. Since 2 times 8 equals 16, it is a valid pair of numbers.

### Example Question #5 : How To Multiply Even Numbers

If and is an odd integer, which of the following could be divisible by?

**Possible Answers:**

**Correct answer:**

If is an odd integer then we can plug 1 into and solve for yielding 13. 13 is prime, meaning it is only divisible by 1 and itself.

### Example Question #173 : Arithmetic

is even

is even

Therefore, which of the following must be true about ?

**Possible Answers:**

It could be either even or odd.

It must be even.

It must be odd.

**Correct answer:**

It could be either even or odd.

Recall that when you multiply by an even number, you get an even product.

Therefore, we know the following from the first statement:

is even or is even or both and are even.

For the second, we know this:

Since is even, therefore, can be either even or odd. (Regardless of what it is, we can get an even value for .)

Based on all this data, we can tell nothing necessarily about . If is even, then is even, even if is odd. However, if is odd while is even, then will be even.

### Example Question #31 : Integers

In a group of philosophers, are followers of Durandus. Twice that number are followers of Ockham. Four times the number of followers of Ockham are followers of Aquinas. One sixth of the number of followers of Aquinas are followers of Scotus. How many total philosophers are in the group?

**Possible Answers:**

**Correct answer:**

In a group of philosophers, are followers of Durandus. Twice that number are followers of Ockham. Four times the number of followers of Ockham are followers of Aquinas. One sixth of the number of followers of Aquinas are followers of Scotus. How many total philosophers are in the group?

To start, let's calculate the total philosophers:

**Ockham:** * <Number following Durandus>, or

**Aquinas: *** <Number following Ockham>, or

**Scotus: ** divided by , or

Therefore, the total number is:

### Example Question #32 : Integers

If n is an integer that is not equal to 0, which of the following must be greater than or equal to n?

I. 7n

II. n + 5

III. n^{2}

**Possible Answers:**

I, II, and III

I and II only

II only

II and III only

I and III only

**Correct answer:**

II and III only

I is not always true because a negative number multiplied by 7 will give a number that is more negative than the original. II is true because adding 5 to any number will increase the value. III is true because squaring any number will increase the magnitude of the value, and squaring a negative number will make it positive.

### Example Question #33 : Integers

Which of the following integers has an even integer value for all positive integers and ?

**Possible Answers:**

**Correct answer:**

There are certain patterns that can be used to predict whether the product or sum of numbers will be odd or even. The sum of two odd numbers is always even, as is the sum of two even numbers. The sum of an odd number and an even number is always odd. In multiplication the product of two odd numbers is always odd. While the product of even numbers, as well as the product of odd numbers multiplied by even numbers is always even. So for this problem we need to find scenarios where the only possibile answers are even. can only result in even numbers no matter what positive integers are used for and , because must can only result in even products; the same can be said for . The rules provide that the sum of two even numbers is even, so is the answer.

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