Achievable logoAchievable logo
CLT
Sign in
Sign up
Purchase
Textbook
Practice exams
Feedback
Community
How it works
Exam catalog
Mountain with a flag at the peak
Textbook
1. Introduction
2. CLT Quantitative Reasoning: Tools and Strategies
3. Math Reasoning
3.1 Number Properties
3.2 Factors and Multiples
3.3 Percents
3.4 Sequences
3.5 Logic
3.6 Average (Arithmetic Mean)
3.7 Two Conditions Questions
4. Geometry
5. Algebra
6. Grammar & Writing: Intro and Passage Types
7. Grammar & Writing: Question Types
8. Verbal Reasoning
9. Wrapping Up
Achievable logoAchievable logo
3.2 Factors and Multiples
Achievable CLT
3. Math Reasoning

Factors and Multiples

9 min read
Font
Discuss
Share
Feedback

Introduction

Understanding factors and multiples is essential for mastering the basics of number theory, which appears frequently in various mathematical contexts. These concepts are foundational in solving problems related to divisibility, prime numbers, and least common multiples, among others.

Approach Question

Consider the number 120. Which of the following numbers are factors of 120?

I. 5
II. 6
III. 7

A. I and II only
B. I and III only
C. II and III only
D. None of the above

Explanation

Factors are numbers that divide into another number completely without leaving a remainder. Multiples, on the other hand, are products obtained when a number is multiplied by an integer.

For the number 120:

  • 5 is a factor because 120 divided by 5 equals 24.
  • 6 is a factor because 120 divided by 6 equals 20.
  • 7 is not a factor because 120 divided by 7 leaves a remainder.

Thus, the correct answer is B. I and III only.

Definitions
Factor
A number that divides another number without leaving a remainder. Example: 2 is a factor of 8 because 8÷2=4.
Multiple
A product of a given number and any integer. Example: 20 is a multiple of 4 because 4×5=20.
Prime Number
A number greater than 1 that has no factors other than 1 and itself.
Composite Number
A number that has more factors than just 1 and itself.
Greatest Common Divisor (GCD)
The largest number that is a factor of two or more numbers. More commonly known as the Greatest Common Factor (GCF).
Least Common Multiple (LCM)
The smallest number that is a multiple of two or more numbers. When the LCM is used to add fractions together, it is referred to as the Least Common Denominator (LCD).

Topics for Cross-Reference

  • Number Properties

Variations

Questions might ask you to identify the LCM or GCD of a set of numbers, or determine if a number is prime or composite based on its factors.

Strategy Insights

  1. Knowing the factors of a number can quickly help determine its divisibility by other numbers. Use the factor tree to help you break a number into its prime factors.
  2. Using prime factorization can also simplify the process of finding the LCM and GCF (GCD) of numbers.

Flashcard Fodder

We recommend the following when it comes to memory work in this area:

  • Rememorize your times tables (yep, like grade school!) but do them up to 15×15.
  • Memorize the powers of 2 (perfect squares) up to 202 and the powers of 3 (perfect cubes) up to 103.
  • Memorize the powers with based 2 up to 210, the powers of 3 and 4 up to 35 and 45, and the powers of 5 up to 54.

Sample Questions

Difficulty 1

What is the greatest common divisor of 8 and 12?

A. 2
B. 3
C. 4
D. 6

(spoiler)

The answer is C. As noted in the Definitions section, both divisor and factor refer to a number that divides evenly into another number. That means, in turn, that the greatest common divisor (identical to the greatest common factor) must be the largest number that divides into two or more numbers. Choice A doesn’t work here because, although 2 is a divisor of both 8 and 12, it is not the largest divisor. Choice B is not right because 3 divides evenly into 12 but not into 8. Similarly, choice D gives us a number that divides into 12 but not into 8. Choice C proposes the largest number that is a divisor of both 8 and 12.

Difficulty 2

Which of the following is both a multiple of 6 and a factor of 90?

A. 15
B. 30
C. 45
D. 60

(spoiler)

The answer is B. Choice A doesn’t work because 15 is a factor of 90 but not a multiple of 6. Choice C has the same problem. Meanwhile, Choice D features the opposite conditions: 60 is a multiple of 6 but not a factor of 90.

Difficulty 3

If a number is a multiple of 6, which of the following must be true?

I. It is a multiple of 2.
II. It is a multiple of 3.
III. It is a multiple of 4.

A. I and II only
B. II and III only
C. I, II, and III
D. None of the above

(spoiler)

The answer is A. We can approach this problem by recognizing that the prime factors of 6 are 2 and 3. This means that any multiple of 6 must also be multiples of 2 and 3. In one step, we’ve confirmed that statements I and II are true. What about statement III? For a number to be a multiple of 4, it must contain two 2’s as factors. But we only know for sure that this number contains one 2. So statement III goes too far for us to know for certain.

If you prefer to plug in numbers, you could try 12, 18, and 24. All three of those numbers are multiples of 2 and 3, so this exercise appears to confirm statements I and II. Regarding statement III, although 12 and 24 and multiples of 4, 18 is not, so we have proved that statement III does not work.

Difficulty 4

How many integers between 30 and 40 (inclusive) are both composite (non-prime) and odd?

A. 1
B. 2
C. 3
D. 4

(spoiler)

The answer is C. Let’s begin by removing the even numbers and leaving ourselves with only 31, 33, 35, 37, and 39. Which of these appear to be composite? Certainly 35, since its last digits tells us it’s divisible by 5. Let’s use our divisibility trick for 3 with the other numbers; which of them have digits that add up to a multiple of 3? It looks like 33 and 39 fit the bill while 31 and 37 do not.

A handy way to speed up your process with primes is to memorize the perfect squares through at least 102 (you should definitely do this for its own benefit!). If a number is smaller than a perfect square and is not prime, it must be divisible by the prime numbers smaller than the square root of that perfect square. For example, if a number is less than 72, which is 49, if it’s not prime it must be divisible by the prime numbers less than 7: 2, 3, or 5. Using this strategy, we can confidently say that 31 and 37 are prime because they are divisible by none of these factors.

It looks like we have three composite, odd numbers. But can we be sure that 31 and 37 are not composite? A good thing to remember is that the smallest composite, odd number not divisible by 3 or 5 is 49. So with smaller numbers, if an odd number is not divisible by 3 or 5, you can be confident that it is prime.

Difficulty 5

Which of the following integers is NOT a factor of 1,001?

A. 7
B. 9
C. 11
D. 13

(spoiler)

The answer is B. This is a difficult problem to approach without a calculator. Remember the UnCLES Method: look at the answers! The way the question is phrased, three of the answers must evenly divide into 1,001, so we can begin anywhere. Why not start with 11 since 10×11=1,100, which is reasonably close to 1,001? Since 1,100−1,001=99 and 99 is a multiple of 11, we can see that 11 works. But let’s use that to our advantage by finding the actual quotient; if 11 goes into 1,100 100 times and we then subtracted a total of nine 11’s, then the quotient must be 100−9=91.

(There is also a cool divisibility trick for 11, but this will apply only rarely, so feel free to skip to the next paragraph if you’d like to save time. If the sum of the odd-ordered (first, third, fifth, etc.) digits of a number is equal to the sum of the even-ordered digits of the number, or if the sums differ by 11 or a multiple of 11, then 11 is a factor of that number. We can see at a glance that the year 2024 is divisible by 11, since (2+2)=(0+4), and that 1969 is also divisible by 11 since (1+6) is exactly 11 less than (9+9). One can see at a glance that 1,001 must be divisible by 11.)

We now have a much smaller number to work with, but the number is still difficult. Does 91 have any factors besides 1 and itself? It might not seem so at first because, since its digits don’t add up to a multiple of 3, 91 is not divisible by 3. It clearly doesn’t divide by 5, so let’s try 7. If you realize that 7 goes into both 70 and 21, then 7 must go evenly into 70+21, which s 91. (There is also a cool trick involving two numbers equidistant from a multiple of 10: in this case 7×13=(10−3)(10+3)=102−32. This always works with numbers on opposite sides of 10-multiples: so, 98×102=1002−22=9996). And 7 goes in 10+3=13 times, so have our other two factors of the original number: 7 and 13!

Before we leave this question, we would be remiss not to mention that there is a faster way to arrive at the solution here. You are looking for the answer choice that is NOT a factor of 1,001. As a reminder, the “adding the digits” trick works not just with 3 but with 9 also. If a number’s digits don’t add to a multiple of 9, then the number itself is not divisible by 9. We can test this with 1,001 and see right away that it’s not divisible by 9 … which gives us the answer right away!

For Reflection

  1. How will understanding factors and multiples help you in solving real-world problems?
  2. What strategies will you employ to quickly find factors and multiples during tests?
  3. How comfortable are you with the prime factorization (factor tree) method? Look it up it feels unfamiliar. What resources could help you improve?

Sign up for free to take 5 quiz questions on this topic

All rights reserved ©2016 - 2025 Achievable, Inc.