6th Grade Math Textbook, Fundamentals

ends in 5; divisible by 5

9  9 18; divisible by 3

3  3 6; divisible by 3

11 divisible by 11

ends in 5; divisible by 5

• 9  9 18; divisible by 9

•• 9 is divisible by 3.

••• all prime numbers

495

5 99

5 9 11

53311

9  9 18; divisible by 9

• divisible by 3

•• divisible by 7

••• divisible by 11

••••^7331113 all prime numbers

143

1001

9

9009

1001

(^33)
337
Divisibility Rules
A number is divisible by:
2 if it is an even number
(ends in 0, 2, 4, 6, or 8).
6 if it is divisible by both 2 and 3.
3 if the sum of the digits is divisible by 3. 8 if the last three digits form a number
divisible by 8.
4 if the last two digits form a number
divisible by 4.
9 if the sum of the digits is divisible by 9.
5 if the ones-place digit is 5 or 0. 10 if the ones-place digit is 0.
Lesson 5-1 for exercise sets. &KDSWHU 
3UDFWLFH $FWLYLWLHV
Tell whether each number is primeor composite.

1. 41 2. 300 3. 264 4. 51 5. 67

Write the prime factorization of each number in exponential form.

6. 30 7. 80 8. 63 9. 52 10. 160

11.Discuss and Write Can the product of two prime numbers be a prime number?

Explain. Give examples to support your explanation.

So the prime factorization of 495 is 3^2 • 5 • 11.

When the divisibility rules for 2, 3, 5, or 9 do not work, try dividing by other

prime numbers. To find the prime factorization of 9009, start by trying 7, 11, 13, 17, and 19.

Find the prime factorization of 495.

Method 1 Make a Factor Tree Method 2 Use Division

1

5

3

3

11

495

99

33

11

Method 1 Make a Factor Tree Method 2 Use Division

1

9009

3003

1001

143

13

3

3

7

11

13

So the prime factorization of 9009 is 3^2 • 7 • 11 • 13.

9  9 18; divisible by 3

3  3 6; divisible by 3

divisible by 7

divisible by 11

divisible by 13