Divisibility Rules (2, 3, 4, 5, 6, 9, 10) Practice Problems & Conceptual Mastery
Quickly determine whether multi-digit integers are divisible by 2, 3, 4, 5, 6, 9, and 10 using digit-sum, last-digit, and last-two-digit divisibility shortcuts without long division.
๐ Worked Step-by-Step Example
Add the digits: 1 + 3 + 8 = 12. Since 12 is divisible by 3 (12 รท 3 = 4), 138 is divisible by 3.
โ Frequently Asked Questions
Why does the digit-sum rule work for 3 and 9?
Every place value power of 10 can be expressed as 1 more than a multiple of 9 (e.g., 10 = 9 + 1, 100 = 99 + 1, 1000 = 999 + 1). When you expand a number algebraically, the powers of 9 drop out under modulo 9 and 3, leaving only the sum of the digits.
How do you test if a number is divisible by 6?
Since 6 = 2 ร 3 and gcd(2, 3) = 1, a number is divisible by 6 if and only if it is divisible by both 2 and 3. Check that the last digit is even (0, 2, 4, 6, 8) AND the sum of all digits is a multiple of 3.
What is the fastest way to check divisibility by 4?
Because 100 is divisible by 4, any multiple of 100 is also divisible by 4. Therefore, you only need to inspect the number formed by the last two digits. If that 2-digit number is divisible by 4 (or half of it is even), the entire number is divisible by 4.
Can a number be divisible by 9 but not by 3?
No. Any number divisible by 9 is automatically divisible by 3 because 9 is a multiple of 3. However, the reverse is not true: 12 is divisible by 3 but not by 9.
๐บ๏ธ Pedagogical Learning Roadmap
2-Digit Addition without Regroupingโ
Master core 2-Digit Addition without Regrouping concepts to build computational fluency.
Prerequisite2-Digit Addition with Regroupingโ
Master core 2-Digit Addition with Regrouping concepts to build computational fluency.
Next Step2-Digit by 2-Digit Multiplicationโ
Advance to 2-Digit by 2-Digit Multiplication practice drills and timed challenges.
Next Step2-Digit Subtraction without Borrowingโ
Advance to 2-Digit Subtraction without Borrowing practice drills and timed challenges.
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