Tip #9 to Get a Top SAT Math Score Page 2

Here are three more math vocab terms. Memorize them, practice using them, and remember to underline them in questions. That will avoid heaps of careless errors.

Factors—'numbers that divide into a number evenly (i.e., without a remainder).

Example: The factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.

More Math Vocab

When asked for the factors of a number, make a list of pairs like the ones shown above. This eliminates the possibility of missing any.

Prime factors—'the factors of a number that are also prime numbers. (Remember, a prime number is a number whose only factors are 1 and itself.)

Example: The prime factors of 48 are 2 and 3. These are the factors of 48 that also happen to be prime numbers.

Multiples—'all the numbers that are divisible by a certain number.

Example: The multiples of 3 are 3, 6, 9, 12, 15, 18, 21, etc.

Let's look at this question:

Solution: To complete this question, we factor 100 just as we factored 48 above.

More Math Vocab

Then we circle any prime numbers in the list. The numbers 2 and 5 are the only prime numbers in the list, so set P has 2 members.

Example Problems

Easy

Each of the following is a factor of 120 EXCEPT
1. 3
2. 4
3. 5
4. 6
5. 7
Which of the following is an odd number that is a factor of 114 ?
1. 2
2. 13
3. 55
4. 57
5. 61
How many integer factors does the number 48 have?
1. None
2. 1
3. 3
4. 8
5. 10

Medium

If P is the set of all different, real number, prime factors of 48 that are also factors of 100, how many members does set P contain?
1. None
2. 1
3. 2
4. 3
5. 52
What is the lowest number that is a multiple of 10, 12, and 15 ?
1. 1800
2. 900
3. 120
4. 60
5. 30

Hard

If a, b, and c are different prime numbers, how many factors does ab²c have?
1. 3
2. 5
3. 9
4. 10
5. 12

Answers

Answer to Brian's Math Magic Trick #1—Multiples: You said carrot. How did I guess your vegetable? Check out my website for answers:

www.BrianLeaf.com/carrot

E List the factors of 120, or better yet "Use the Answers." Just divide 120 by each answer choice. If the number goes into 120 evenly (without a remainder), then it is a factor. And 7 is not a factor, since 120 ÷ 7 = 17.14.
D This question tests if you know the terms "odd" and "factor." Once you do, this is a simple "Use the Answers" type question. Try each answer. Choice A does not work because 2 is not an odd number. Then to test each of the other choices, simply divide 114 by the number. If it goes in evenly, which means you get an integer (no decimal or fraction), then it is a factor. For example, 114 ÷ 55 = 2.07, and therefore 55 is not a factor of 114. Choice D is correct since 114 ÷ 57 = 2.
E This question simply asks you to factor 48 and count how many factors there are.

To factor a number, make a list of pairs, as shown below. When you reach 6 × 8, you know that you have them all since only 7, which does not go into 48, is between 6 and 8. Finding factors systematically like this is far better than randomly jotting down numbers. This is true for any SAT math question. Systematic and organized is better than scattered and random. It avoids careless errors, allows you to look back at your work, and helps you make the leap to the next step needed on a complicated question.

B To complete this question, we factor 48 just as we did above. However, this time we circle any prime numbers in the list. The numbers 2 and 3 are the only prime numbers in the list. Then we write the list of the factors for 100: 1, 2, 4, 5, 10, 20, 25, 50, 100. Of the prime factors of 48, only 2 is also a factor of 100. So set P has 1 member.
D This question could be difficult, but "Use the Answers" makes it easy! Just test each answer choice by dividing it by 10, 12, and 15. Now 60 is the lowest number on the list that is divisible by 10, 12, and 15. Notice that several other answers work also, but 60 is the lowest. Make sure to finish the question and try all choices!
E Just try numbers for a, b, and c. This is called "Make It Real" and is discussed more in Skill 16. Let's say a = 3, b = 5, and c = 7. Then ab²c = (3)(5)²(7) = 525. The long way to do this question is to list the factors of 525: 1, 3, 5, 7, 15, 21, 25, 35, 75, 105, 175, 525. The shortcut is to notice that since a, b, and c are prime, the factors will be 1, a, b, c, ab, ac, bc, b², ab², cb², abc, and ab2c.

Go to: Tip #10

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