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Test divisibility by 2-12 with rules and explanations
The number must be even (last digit is 0, 2, 4, 6, or 8)
Example: 148 ends in 8 (even), so it's divisible by 2. 148 ÷ 2 = 74
The sum of all digits must be divisible by 3
Example: 123 → 1+2+3 = 6, and 6 ÷ 3 = 2, so 123 is divisible by 3
The last two digits form a number divisible by 4
Example: 1,316 → last two digits are 16, and 16 ÷ 4 = 4, so it's divisible by 4
The last digit must be 0 or 5
Example: 235 ends in 5, so it's divisible by 5. 235 ÷ 5 = 47
Must be divisible by both 2 AND 3
Example: 42 is even (÷2) and 4+2=6 (÷3), so 42 is divisible by 6
Double the last digit and subtract from the rest. Result divisible by 7
The last three digits form a number divisible by 8
The sum of all digits must be divisible by 9
Example: 729 → 7+2+9 = 18, and 18 ÷ 9 = 2, so 729 is divisible by 9
The last digit must be 0
Example: 450 ends in 0, so it's divisible by 10. 450 ÷ 10 = 45
Alternating sum of digits is divisible by 11
Must be divisible by both 3 AND 4
By 2, 5, and 10: These rules work because our number system is base-10. The last digit represents ones, so it determines divisibility by factors of 10.
By 3 and 9: These work because 10 ≡ 1 (mod 3) and 10 ≡ 1 (mod 9). This means each digit contributes its face value to the sum modulo 3 or 9.
By 4 and 8: Since 100 = 4 × 25 and 1000 = 8 × 125, higher place values are already divisible by 4 and 8. Only the last 2 or 3 digits matter.
By 6 and 12: These are composite numbers, so we check their prime factors. 6 = 2 × 3, and 12 = 3 × 4, so we test both conditions.
7 doesn't divide evenly into powers of 10 (our number system base), so there's no simple pattern in the digits. The rule (double the last digit and subtract) works but is more complex. For 7, it's often easier to just divide directly.
Yes! If a number is divisible by both m and n, and m and n are coprime (share no common factors), then the number is divisible by m × n. For example, if divisible by 3 and 4, it's divisible by 12. If divisible by 2 and 3, it's divisible by 6.
For composite numbers, break them into prime factors. 15 = 3 × 5 (check both), 25 = 5² (last two digits divisible by 25), 100 = 4 × 25 (last two digits are 00). For powers of 2 or 5, check the last n digits where 2ⁿ or 5ⁿ equals the divisor.
Because 10 ≡ -1 (mod 11). This means 10² ≡ 1, 10³ ≡ -1, and so on. Each place value alternates between adding and subtracting its digit. If the alternating sum is divisible by 11, so is the number.
Yes, but it's complex: Add 4 times the last digit to the rest of the number. Repeat until you get a recognizable multiple of 13. For example, 143 → 14 + (4×3) = 26, which is 2×13. Alternatively, just divide by 13 directly!
Divisibility rules help with mental math, simplifying fractions, finding GCF and LCM, factoring, and solving problems faster without a calculator. They're foundational for algebra, number theory, and even computer science (like hash functions and modular arithmetic).