GRE Quant: Divisibility Rules and Shortcuts

Divisibility rules and shortcuts support specific moments in GRE Quant problem solving where quick checks on numbers add clarity and efficiency. These ideas stay simple in structure and help you test numerical relationships smoothly, recognize useful signals inside a problem, and move forward with better direction on select Quant questions without unnecessary effort.

The following video, part of our GRE preparation online course, covers divisibility rules for important values such as 2, 3, 4, 5, 9, 11, 20, 25, and 50, along with higher powers of 2, higher powers of 5, and their common combinations. It explains how these rules apply within expressions and conditions and shows how they guide sound numerical decisions. Spend time with this lesson and apply these ideas thoughtfully during your GRE practice questions, GRE sectional tests, and GRE practice tests.

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Divisibility Rule for 2 and 5 and Their Powers and Combinations…

Divisibility tests on GRE Quant

Divisibility rules for various numbers can be determined by examining a specific number of ending digits. The core logic is based on whether the last n digits are divisible by the divisor in question.

Divisibility by 2, 5, and 10

To find the remainder when dividing by 2, 5, or 10, look only at the last digit of the number.

Example (876345): * For 2, the remainder is 5 % 2 = 1.
- For 5, the remainder is 5 % 5 = 0.
- For 10, the remainder is 5 % 10 = 5.

Divisibility by 4, 20, 25, 50, and 100

To find the remainder for these divisors, look at the last two digits of the number.

Example (9887963450): * For 4, the remainder is 50 % 4 = 2.
- For 20, the remainder is 50 % 20 = 10.
- For 25, the remainder is 50 % 25 = 0.
- For 50, the remainder is 50 % 50 = 0.
- For 100, the remainder is 50 % 100 = 50.

The Mother Rule: Divisibility by 2^n or 5^n

This general rule states that for any divisor that is a power of 2 or 5, you must divide the last n digits by that divisor to find the remainder.

Rule for 2^n: Divide the last n digits by 2^n.
Rule for 5^n: Divide the last n digits by 5^n.

Example: 8377334346380

For 8 (which is 2^3): Divide the last 3 digits (380) by 8. Remainder: 380 % 8 = 4.
For 125 (which is 5^3): Divide the last 3 digits (380) by 125. Remainder: 380 % 125 = 5.
For 16 (which is 2^4): Divide the last 4 digits (6380) by 16. Remainder: 6380 % 16 = 12.

Divisibility Rule for 3 and 9…

Divisibility tests on GRE Quant

Divisibility tests for 3 and 9 are determined by calculating the sum of the digits of a number and then dividing that sum by 3 or 9. The remainder of this division matches the remainder of the original number.

Rule for 3 and 9

Add all the digits of the number together. Divide that sum by 3 (for divisibility by 3) or by 9 (for divisibility by 9).

Example 1: 43457

Sum of digits: 4 + 3 + 4 + 5 + 7 = 23
Dividing by 3: 23 % 3 = 2 (The remainder is 2)
Dividing by 9: 23 % 9 = 5 (The remainder is 5)

Example 2: 876351

Sum of digits: 8 + 7 + 6 + 3 + 5 + 1 = 30
Dividing by 3: 30 % 3 = 0 (The remainder is 0, meaning it is perfectly divisible by 3)
Dividing by 9: 30 % 9 = 3 (The remainder is 3)

Divisibility Rule for 11…

Divisibility tests on GRE Quant

To determine if a number is divisible by 11, or to find its remainder when divided by 11, calculate the difference between the sum of the odd-numbered digits and the sum of the even-numbered digits. If that difference is then divided by 11, the resulting remainder is the same as the remainder of the original number.

Example 1: 54321

Sum of odd-numbered digits (1st, 3rd, 5th): 1 + 3 + 5 = 9
Sum of even-numbered digits (2nd, 4th): 2 + 4 = 6
Difference: 9 – 6 = 3
Remainder with 11: 3 % 11 = 3
The remainder of 54321 when divided by 11 is 3.

Example 2: 738695

Sum of odd-numbered digits: 5 + 6 + 3 = 14
Sum of even-numbered digits: 9 + 8 + 7 = 24
Difference: 14 – 24 = -10
Remainder with 11: -10 % 11 = -10
Special rule: Whenever the remainder is negative, add the divisor (11) to it.
Final calculation: -10 + 11 = 1
The remainder of 738695 when divided by 11 is 1.

Divisibility Rule for 6, 12, 15, 18, 22…

Divisibility tests on GRE Quant

A number is divisible by a composite number if it is divisible by its co-prime factors. This allows you to determine divisibility without memorizing a unique rule for every large number.

Divisibility by 6

The number must be divisible by both 2 and 3.

Divisibility by 12

The number must be divisible by both 3 and 4.

Divisibility by 15

The number must be divisible by both 3 and 5.

Divisibility by 18

The number must be divisible by both 2 and 9.

Divisibility by 22

The number must be divisible by both 2 and 11.

Note

Rather than memorizing an exhaustive list of individual rules, focus on the broad concept: if a number satisfies the divisibility requirements of two co-prime factors that multiply to the target number, it is divisible by that target.