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.

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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?

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.”

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.

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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

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?

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!

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

thanks!

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 :).