Divisibility Rules Practice Quiz

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Janaisa Harris, BA (Mathematics) |
High School Math Teacher
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Janaisa Harris, an experienced educator, has devoted 4 years to teaching high school math and 6 years to tutoring. She holds a bachelor's degree in Mathematics (Secondary Education, and Teaching) from the University of North Carolina at Greensboro and is currently employed at Wilson County School (NC) as a mathematics teacher.
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  • 1/10 Questions

    Which of the following numbers is 36 divisible by?

    • 10
    • 5
    • 2
    • 7
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About This Quiz

You must have read about the divisibility rules in your maths textbooks. Do you still remember those rules? If yes, play it out this practice quiz and test your knowledge for the same. A divisibility rule is a short and useful method to identify whether any given number is divisible by a fixed divisor without performing the actual division. The See morebelow quiz contains ten questions. Let's see how easily you solve it. Shall we begin it now? Good luck!

Divisibility Rules Practice Quiz - Quiz

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  • 2. 

    Which number is divisible by 2?

    • 123

    • 236

    • 237

    • 345

    Correct Answer
    A. 236
    Explanation
    The number 236 is divisible by 2 because it ends in an even digit, which means it is an even number. In contrast, the other numbers listed (123, 237, and 345) do not end in even digits and are therefore not divisible by 2.

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  • 3. 

    1000 can be divisible by which number here?

    • 5

    • 3

    • 2

    • 2 and 5 both

    Correct Answer
    A. 2 and 5 both
    Explanation
    1000 can be divided evenly by both 2 and 5. When a number is divisible by 2, it means that it can be divided into two equal parts without leaving any remainder. Similarly, when a number is divisible by 5, it means that it can be divided into five equal parts without leaving any remainder. In the case of 1000, it can be divided by 2 to give 500 and by 5 to give 200. Therefore, the correct answer is that 1000 is divisible by both 2 and 5.

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  • 4. 

    The number 203 is divisible by 3 and 7

    • True

    • False

    Correct Answer
    A. False
    Explanation
    The number 203 is not divisible by 3 because the sum of its digits (2+0+3) is not divisible by 3. Additionally, 203 is divisible only by 7. 

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  • 5. 

    Which number is divisible by both 2 and 3?

    • 126

    • 100

    • 123

    • 133

    Correct Answer
    A. 126
    Explanation
    For a number to be divisible by both 2 and 3, it must meet the following conditions:
    Divisible by 2: The number must be even (i.e., its last digit must be 0, 2, 4, 6, or 8).
    Divisible by 3: The sum of its digits must be divisible by 3.
    Let’s check each number:
    126:
    Last digit is 6 (even), so it is divisible by 2.
    Sum of digits = 1 + 2 + 6 = 9, which is divisible by 3.
    Therefore, 126 is divisible by both 2 and 3.
    100:
    Last digit is 0 (even), so it is divisible by 2.
    Sum of digits = 1 + 0 + 0 = 1, which is not divisible by 3.
    Therefore, 100 is not divisible by both 2 and 3.
    123:
    Last digit is 3 (odd), so it is not divisible by 2.
    Therefore, 123 is not divisible by both 2 and 3.
    133:
    Last digit is 3 (odd), so it is not divisible by 2.
    Therefore, 133 is not divisible by both 2 and 3.
    Thus, the correct answer is 126.

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  • 6. 

    15000 can be divisible by which number here?

    • 2

    • 3

    • 5

    • All of the above

    Correct Answer
    A. All of the above
    Explanation
    The number 15000 can be divided evenly by 2 because it is an even number. It can also be divided evenly by 3 because the sum of its digits (1+5+0+0+0) is divisible by 3. Additionally, it can be divided evenly by 5 because it ends in 0 or 5. Therefore, all of the given numbers (2, 3, and 5) can divide 15000 evenly.

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  • 7. 

    Which number is divisible by 9?

    • 78500

    • 78531

    • 78532

    • 78534

    Correct Answer
    A. 78534
    Explanation
    A number is divisible by 9 if the sum of its digits is divisible by 9.
    7 + 8 + 5 + 3 + 1 = 24. 24 is not divisible by 9.
    7 + 8 + 5 + 3 + 2 = 25. 25 is not divisible by 9.
    7 + 8 + 5 + 3 + 4 = 27. 27 is divisible by 9.
    Therefore, 78534 is divisible by 9.

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  • 8. 

    Which number is divisible by both 3 and 9?

    • 78531

    • 78534

    • 3333

    • 1234

    Correct Answer
    A. 78534
    Explanation
    For a number to be divisible by both 3 and 9, the sum of its digits must be divisible by 9. Let's check each number:
    78531:
    Sum of digits = 7 + 8 + 5 + 3 + 1 = 24
    24 is divisible by 3, but not by 9.
    Therefore, 78531 is divisible by 3 but not by 9.
    78534:
    Sum of digits = 7 + 8 + 5 + 3 + 4 = 27
    27 is divisible by both 3 and 9.
    Therefore, 78534 is divisible by both 3 and 9.
    3333:
    Sum of digits = 3 + 3 + 3 + 3 = 12
    12 is divisible by 3, but not by 9.
    Therefore, 3333 is divisible by 3, but not by 9.
    1234:
    Sum of digits = 1 + 2 + 3 + 4 = 10
    10 is not divisible by either 3 or 9.
    Thus, the correct answer is 78534.

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  • 9. 

    If the sum of the digits of a number is divisible by 3, then the number will be divisible by which of the following?

    • 3

    • 9

    • 2

    • 7

    Correct Answer
    A. 3
    Explanation
    If the sum of the digits of a number is divisible by 3, then the number will be divisible by 3 as well. This is because when a number is divisible by 3, the sum of its digits is also divisible by 3. Therefore, if the sum of the digits of a number is divisible by 3, it implies that the number itself is divisible by 3.

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  • 10. 

    What numbers are divisible by 2, 3, and 10 at the same time?

    • 12

    • 30

    • 80

    • 120

    • 50

    Correct Answer(s)
    A. 30
    A. 120
    Explanation
    The numbers that are divisible by 2, 3, and 10 at the same time are called common multiples. In this case, 30 and 120 are the common multiples. Both 30 and 120 can be divided evenly by 2, 3, and 10, making them the correct answers.

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Janaisa Harris |BA (Mathematics) |
High School Math Teacher
Janaisa Harris, an experienced educator, has devoted 4 years to teaching high school math and 6 years to tutoring. She holds a bachelor's degree in Mathematics (Secondary Education, and Teaching) from the University of North Carolina at Greensboro and is currently employed at Wilson County School (NC) as a mathematics teacher.

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  • Apr 03, 2025
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