GRE Math: Divisibility Rules

1. Which of the following is NOT divisible by 3?

(A)  231

(B)  246

(C)  285

(D) 326

(E)  411

One way to do this problem is to divide each answer choice by 3 and see which one leaves a remainder. A far better approach, however, is to apply the following rule of divisibility for 3. If the sum of the digits of any number is a multiple of 3, then that digit is divisible by 3.


Let’s try to apply this rule below:

111:  1 + 1 + 1 = 3.  Three is a multiple of three so 111 is divisible by three.

3,456: 3 + 4 + 5 + 6 = 18. Eighteen is a multiple of 3 so 3,456 is divisible by three.

2,789: 2 + 7 + 8 + 9 = 26. Twenty-six is NOT a multiple of three. Therefore, 2,789 is not divisible by three.


Now let’s return to the original question.

(A)  231: 2 + 3 + 1 = 6 (Multiple)

(B)  246: 2 + 4 + 6 = 12 (Multiple)

(C)  285: 2 + 8 + 5 = 15 (Multiple)

(D) 326: 3 + 2 + 6 = 11 (Not Multiple) Answer

(E)  411: 4 + 1 + 1 = 6 (Multiple)


The next important rule to know is the rule of divisibility by 4: If the last two digits of a number are a multiple of 4, then the number is divisible by 4.

724: the last two digits are 24. 24 is a multiple of 4. (Divisible)

470: 70 is not a multiple of 4 (70/4 = 17.5). (Not Divisible)

40,004: 04, or 4 is a multiple of 4 (Divisible)


Now let’s try the following problem in which you have to combine everything you learned today:

Which of the following integers is divisible by 12?

(A)  1,442

(B)  1,653

(C)  1,728

(D) 2,048

(E)  2,884


What, you gasp. Divisibility of 12. You never taught us that! Well, for a number to be divisible by 12, it has to be divisible by both 4 and 3, because 4 x 3 = 12.

(A)  1,442: 42 is not a multiple of 4.

(B)  1,653: 53 is not a multiple of 4 (though the sum of the digits is divisible by 3).

(C) 1,728: 28 is a multiple of 4; the sum of the digits, 18, is a multiple of 3.

(D) 2,048: sum of digits 14. Not a multiple of 3.

(E)  2,884 : sum of digits is 22. Not a multiple of 3.

Answer: C.

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  • Chris Lele

    Chris Lele is the Principal Curriculum Manager (and vocabulary wizard) at Magoosh. Chris graduated from UCLA with a BA in Psychology and has 20 years of experience in the test prep industry. He's been quoted as a subject expert in many publications, including US News, GMAC, and Business Because. In his time at Magoosh, Chris has taught countless students how to tackle the GRE, GMAT, SAT, ACT, MCAT (CARS), and LSAT exams with confidence. Some of his students have even gone on to get near-perfect scores. You can find Chris on YouTube, LinkedIn, Twitter and Facebook!

14 Responses to GRE Math: Divisibility Rules

  1. Sumanvitha Vommina May 23, 2019 at 1:12 am #

    Hi Chris. Could you let us know the divisibility rule for 7?

    • Magoosh Test Prep Expert
      Magoosh Test Prep Expert May 24, 2019 at 8:11 am #

      Hi Sumanvitha,

      You can read more about the divisibility rule for 7 here: 🙂

      • Tanzim October 23, 2019 at 10:50 am #

        That Okay but needs a huge calculation for large number. We can check recursively whether a number is divisible by 7 or not. Dividing the number until finished is the prettiest method i think.

        • Magoosh Test Prep Expert
          Magoosh Test Prep Expert December 6, 2019 at 8:14 pm #

          Hi Tanzim! Yes, you’re right that you can use the calculator to check for divisibility. I recommend learning both the mental math way and the calculator way, just to be safe. In this case, I would be sure to learn the divisibility rules for at least the first ten numbers. A good intuition is important!

  2. Lisanne October 6, 2015 at 3:35 am #

    I was wondering if this ’12’ trick and also the ‘6’ trick (number should be both divisible by 2 and by 3) works for other combinations, like 2 and 4 (numbers that are both divisible by 2 and 4 are also divisible by 8).

  3. Ghadeer October 5, 2015 at 10:32 am #

    In one of the practice questions the questions asks the divisibility of 36, the answer is found by the following,

    The number 36 is 4 times 9, so if a number is divisible by 36, it must be divisible by both 4 and 9. To tell if a number is divisible by 4, we look at the final two digits, the tens and the ones places: if the last two digits form a two-digit number divisible by 4, then the whole number is divisible by 4. To tell is a number is divisible by 9, we add the digits: if the sum of the digits is divisible by 9, then the number is divisible by 9.

    My Question is why did he use the factors 4,9 rather than 6^2, I used 6 breaking it down to 2,3 and checking the divisibly for those led me to wrong answer!

    I’ve been thinking about it for some time now and I haven’t found an answer!! 36 can be 4X9 or 6^2 or any other factors are true ! why did we break it to 4 and 9 !!

    Thank you !

  4. Ghadeer September 27, 2015 at 1:57 am #

    Hi chris,

    Am little confused about finding the factors so we can check for divisibility, the example above you used 12 = 4X3 and not 6X2, that I understand because 6 can be factored further more into 2X3, but what if the Q is asking for factors of 36! the correct answer used 36 = 4X9 rather than 6X6 !

    I used the factor 6 and ended up checking divisibility of 2 and 3 which led me to a wrong answer!

    how can we determine how to factorize the number we are asked to check it’s divisibility?

    Thank you,

  5. Brian Aiken August 17, 2015 at 5:18 pm #

    I thought of some more tricks for divisibility by 3.

    It is possible to add the sum of the sum. For example: 7,346,892.

    7+3+4+6+8+9+2 = 39 and 3+9 = 12 (divisible by 3) and 1+2=3 (divisible by 3)

    But I prefer casting out. Remove all numbers that are multiples of 3.

    7,346,892 –> 7,×46,8×2 –> since 7+8=15 (divisible by 3) –> x,x46,xx2 = 4+6+2=12

  6. Fatima September 4, 2013 at 1:24 am #

    Hye Chris

    We know the rule that if “x” is divisible by “y” and “y” is in turn divisible by “z” , than “x” will also be divisible by “z”. Using that rule, can we calculate the prime factors of 12 i.e 2x2x3 and see if any of the numbers mentioned above contain the same factors in their Prime Factorization.
    If they do contain 2x2x3 hidden somewhere than they are divisible by 12.
    Is this an authentic approach for GRE questions or just luck that all my answers have been correct up till now using this rule?

    • Chris Lele
      Chris Lele September 4, 2013 at 1:14 pm #

      Hi Fatima,

      This is a great approach as well. All you have to do is divide by ‘2’ twice and then see if the remaining number is divisible by ‘3’. You are essentially doing the same thing as I did above, except for one extra step. It might take you a little longer, but you will get the same (correct) answer as if you were using the “12 rule.”

      Hope that helps!

  7. christina March 4, 2013 at 11:14 pm #

    for the last question, do the divisibility rules for 2 and 6 factor in as well? since 6 x 2= 12…


    • Chris Lele
      Chris Lele March 5, 2013 at 12:00 pm #

      Hi Christina,

      Good question!

      Actually, you don’t want to use to use 6 and 2 to test divisibility by 12, because the ‘2’ is redundant in the ‘6’ (3 x 2 = 6). To illustrate, 18 is divisible by 6 and 2, but it is not divisible by 12.

      Hope that makes sense :).

      • Candace June 17, 2014 at 2:20 pm #

        Also, it is important to note that just because a number is divisible by 4 and 2 (4*2=8) does not mean it is divisible by 8… for example… 7620 is divisible by 4 and 2 but it is not divisible by 8, even though 8 is divisible by both 4 and 2.

        • Kalyan September 1, 2017 at 11:44 pm #

          Just take 12, it is divisible by 2 and 4, but not 8. So, 3*4 = 12, rule does not apply here.
          I believe, it can be applicable when the base numbers ( 2,4, and 2,6) when one of them are not factors of the other number.
          Like, 3,4 and 3,5..etc
          A number divisible by 3 and 4, means its also divisible by 12.
          A number divisible by 3 and 5, the its divisible by 15.

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