### All SAT Math Resources

## Example Questions

### Example Question #3 : Integers

When 8 integers are multiplied their product is negative, then at most how many of the integers can be negative?

**Possible Answers:**

4

8

1

7

5

**Correct answer:**

7

When one multiplies two negative numbers (or any even multiple) the result is a positive number. However, when one multiplies three negative numbers (or any odd multiple) the product is negative. If the result of multiplying 8 negatives is odd, the largest number of negative integers will be the largest odd number, in this case 7.

### Example Question #4 : Integers

If n is an odd integer, all of the following must be odd integers EXCEPT:

**Possible Answers:**

n^{2} +2n + 1

n^{3} + 2

n^{2}

n^{2} + 4n + 4

n^{2} + 2n

**Correct answer:**

n^{2} +2n + 1

Let's examine the choice n^{2}. We can rewrite n^{2} as n * n, which would be mutiplying an odd number (because n is odd) by an odd number. Multiplying two odd numbers always produces another odd number. So this can't be the correct answer.

Next, let's look at n^{3} + 2. We can rewrite this as n^{2} * n + 2. We already established that n^{2} must be odd, so then n^{2} * n must also be odd. If we then add 2 to an odd number, we still get an odd number. So we can eliminate this choice as well.

Now, let's look at the choice n^{2} + 2n. Let's factor this as n(n + 2). We know that n must be odd, and we know that n+2 must be odd. Therefore, n(n + 2) is odd, because multiplying two odd numbers gives us an odd number.

Next, we can consider n^{2} + 4n + 4, which we can factor as (n + 2)(n + 2). Once again, n + 2 is an odd number, so multiplying n + 2 by itself will also be odd.

Finally, let's analyze n^{2} + 2n + 1. We can rewrite this as (n + 1)(n + 1). Since n is odd, n+1 must be an even number. When we multiply an even number by an even number, we get an even number, so (n + 1)(n + 1) must be even, and it cannot be odd.

The answer is n^{2} + 2n + 1.

### Example Question #5 : Even / Odd Numbers

If n and m are both positive even integers, which of the following must be odd?

I. (n + 1)(m + 1)

II. nm + 1

III. nm + m

**Possible Answers:**

II only

II and III only

I, II, and III

I and II only

I only

**Correct answer:**

I and II only

Let us analyze I, II, and III one at a time.

Because n and m are both even, if we increase either by 1, the result will be an odd number. Thus, n + 1 and m + 1 are both odd. When two odd numbers are multiplied together, the result is always an odd number. Thus (n + 1)(m + 1) must be an odd number.

Because n and m are even, when we multiply two even numbers together, we always get an even number. Thus nm is even. However, when we then add one to an even number, the result will be an odd number. Thus, nm + 1 is odd.

We just established that nm is even. If we subtract an even number from an even number, the result is always even. Thus, nm – m is an even number.

Only choice I and II will always produce odd numbers.

The answer is I and II only.

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

odd * odd * odd =

**Possible Answers:**

even * even

even * odd

odd * odd

odd * odd * even

even * even * even

**Correct answer:**

odd * odd

The even/odd number properties are good to know. If you forget them, however, it's easy to check with an example.

Odd * odd = odd. If you didn't remember that, a check such as 1 * 3 = 3 will give you the same answer. So if odd * odd = odd, (odd * odd) * odd = odd * odd = odd, just as 3 * 3 * 3 = 27, which is odd. This means we are looking for an answer choice that also produces an odd number. Let's go through them.

even * even = even (2 * 2 = 4)

even * odd = even (2 * 3 = 6)

odd * odd = odd (1 * 3 = 3) This is the correct answer! But just to double check, let's go through the last two.

even * even * even = even * even = even (2 * 2 * 2 = 8)

odd * odd * even = odd * even = even (1 * 3 * 2 = 6)

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

When is evaluated, what number is in the ones digit?

**Possible Answers:**

**Correct answer:**

Let us say we have two numbers, and whose ones digits are and B, respectively. If we want to know the ones digits of the product of and , all we need to do is to look at the ones digit of the product of and . For example, if we multiply 137 and 219, then the ones digit will be the same as the ones digit of . Since the ones digit of 63 is 3, the ones digit of 137 x 219 will also be 3. In short, we really only need to worry about the ones digits of the numbers we multiply when we try to find the ones digit of their product.

We want to find the ones digit of . An exponent is essentially just a short hand for repeated multiplication. Let us look at the ones digit of the first few exponents of 2013.

- the ones digit is 3.

To find the ones digit of the 2013 to the second power, we need to think of it as the product of 2013 and 2013. As discussed previously, if we want the ones digit of two numbers multiplied together, we just need to multiply their ones digits. Thus, if we multiply 2013 by 2013, then the ones digit will be the same as .

- ones digit is 9.

Next, we want to find the ones digit of 2013 to the third power. In order to do this, we will multiply the square of 2013 by 2013. It does not matter that we do not know exactly what 2013 squared equals, beacuse we only need to worry about the ones digit, which is 9. In other words, 2013 to the third power will have a ones digit that is equal to the ones digit of the product of 9 (which was the ones digit of 2013 squared) and 3 (which is the ones digit of 2013). When we multiply 9 and 3, we get 27, so the ones digit of 2013 to the third power is 7.

- ones digit of 7.

To find the ones digit of 2013 to the fourth power, we only need to worry about multiplying the ones digit of 2013 to the third power (which is 7) by the ones digit of 2013. When we mulitply 7 and 3, we get 21, which means that the ones digit of 2013 to the fourth power is 1.

- ones digit of 1.

To find the ones digit of 2013 to the fifth power, we will multiply 1 by 3, which gives us 3.

- ones digit of 3.

Notice that we are back to a ones digit with 3. If we multiply this by 2013, we will end up with a ones digit of 9. In other words, the ones digits repeat every fourth power.

The value of the ones digits of the powers of 2013 is as follows (starting with 2013 to the first power):

3, 9, 7, 1, 3, 9, 7, 1,....

We essentially want to find the 2013th term of the sequence above. Notice that every fourth term is 1, i.e. the sequence repeats every four terms. If a terms position in the sequence is a multiple of 4, then the term will be 1. In short, the 4th, 8th, 12th, 16th terms, and so on, will be 1. Because 2012 is a multiple of 4, the 2012th term in the sequence will be 1. (We can determine if a number is a multiple of 4 by looking at its last two digits.) This means that that 2013th term will be 3. Thus, 2013 to the power of 2013 has a ones digit of 3.

The answer is 3.

### Example Question #4 : Integers

If and are both odd integers, which of the following is not necessarily odd?

**Possible Answers:**

**Correct answer:**

With many questions like this, it might be easier to plug in numbers rather than dealing with theoretical variables. However, given that this question asks for the expression that is not always even or odd but only not necessarily odd, the theoretical route might be our only choice.

Therefore, our best approach is to simply analyze each answer choice.

: Since is odd, is also odd, since and odd number multiplied by an odd number yields an odd product. Since is also odd, multiplying it by will again yield an odd product, so this expression is always odd.

: Since is odd, multiplying it by 2 will yield an even number. Subtracting this number from will also give an odd result, since an odd number minus an even number gives an odd number. Therefore, this answer is also always odd.

: Since both numbers are odd, their product will also always be odd.

: Since is odd, multiplying it by 2 will give an even number. Since is odd, subtracting it from our even number will give an odd number, since an even number minus and odd number is always odd. Therefore, this answer will always be odd.

: Since both numbers are odd, there sum will be even. However, dividing an even number by another even number (2 in our case) does not always produce an even or an odd number. For example, 5 and 7 are both odd. Their sum, 12, is even. Dividing by 2 gives 6, an even number. However, 5 and 9 are also both odd. Their sum, 14, is even, but dividing by 2 gives 7, an odd number. Therefore, this expression isn't necessarily always odd or always even, and is therefore our answer.

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

Theodore has jelly beans. Portia has three times that amount. Harvey has five times as many as she does. What is the total count of jelly beans in the whole group?

**Possible Answers:**

**Correct answer:**

To find the answer to this question, calculate the total jelly beans for each person:

**Portia:** * <Theodore's count of jelly beans>, which is or

**Harvey:** * <Portia's count of jelly beans>, which is or

So, the total is:

(Do not forget that you need those original for Theodore!)

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

If you multiply an even number by and odd number, what type of number will the answer be?

**Possible Answers:**

A non-zero integer

Zero

Odd

Even

**Correct answer:**

Even

**If you multiply an even number by and odd number, what type of number will the answer be?**

For this question, let's try out a pair of numbers

Even: 6

Odd: 3

Now, if we multiply them, we get:

18, an even number.

Any number multiplied by an even number will yield an even number.

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

If two odd numbers are multiplied, what will the product be?

**Possible Answers:**

Not enough information to solve

Zero

Odd

Even

**Correct answer:**

Odd

**If two odd numbers are multiplied, what will the product be?**

Let's begin by making up two numbers and seeing what we get.

So, we get 15, an odd number. Regardless of the two odd numbers we choose, we will always get an odd product.

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