Divisibility Rules and Factors

In campus tests (TCS NQT, Infosys, Wipro, Cognizant, Accenture) the Number System is the highest-frequency Quant area — usually 2 to 4 questions per paper. Divisibility rules are the foundation: they let you factor, cancel and eliminate options in seconds instead of doing long division.

Core divisibility rules

Divisor	Test
2	Last digit even
3	Digit sum divisible by 3
4	Last two digits divisible by 4
5	Last digit 0 or 5
6	Divisible by 2 AND 3
8	Last three digits divisible by 8
9	Digit sum divisible by 9
11	Alternating digit sum (+ − + −) is 0 or a multiple of 11
7	Double the last digit, subtract from the rest; repeat

Composite divisors: split N into COPRIME factors. For 12 test 3 and 4; for 15 test 3 and 5; for 72 test 8 and 9. Do not use 2 and 6 for 12 — they share the factor 2.

Factors: every integer has a unique prime form, e.g. 360 = 2³ × 3² × 5. From that form you derive factor count, HCF, LCM and remainders.

Exam Tricks & Tips

• 🎯 For 11, sum odd-position and even-position digits separately; the difference must be 0 or a multiple of 11.
• 🎯 Split any composite divisor into COPRIME parts only (12 = 3×4, never 2×6).
• 🎯 The digit-sum test also gives the remainder: 3421 has digit sum 10, so 3421 ÷ 9 leaves remainder 1.
• 🎯 Divisible by 25 if the last two digits are 00, 25, 50 or 75.
• 🎯 For "find the missing digit so N is divisible by 9", make the digit sum a multiple of 9.
• ❌ Don't test 8 with only the last two digits — you need the last THREE (2,300 ends in "00" but 300 ÷ 8 is not exact, so it fails).
• ❌ Never call a number divisible by 6 just because it is even — it must also pass the 3-test.

Expected exam pattern: direct "which is divisible by X", find-the-digit blanks, and hidden steps inside remainder or LCM problems. 1 to 2 marks, ~40 seconds each.

Quick recap: Learn the 2–11 tests cold; split composite divisors into coprime factors; digit sum gives both divisibility and the remainder for 3 and 9.

Q1. Divisibility rule for 3?
A1. Sum of the digits is divisible by 3.

Q2. Rule for 4?
A2. The number formed by the last two digits is divisible by 4.

Q3. Rule for 8?
A3. The number formed by the last three digits is divisible by 8.

Q4. Rule for 11?
A4. Difference between the sum of odd-position and even-position digits is 0 or a multiple of 11.

Q5. Rule for 7?
A5. Double the last digit, subtract it from the remaining number, and repeat until small; if the result is divisible by 7, so is the number.

Q6. How do you test divisibility by 12?
A6. Test 3 and 4 (coprime factors); both must pass.

Q7. Is 918082 divisible by 11?
A7. Alternating sum = 2 − 8 + 0 − 8 + 1 − 9 = −22, a multiple of 11, so yes.

Q8. Remainder of 3421 ÷ 9?
A8. Digit sum = 10, and 10 mod 9 = 1, so remainder 1.

Q9. Prime factorisation of 360?
A9. 2³ × 3² × 5.

Q10. Test for divisibility by 25?
A10. Last two digits are 00, 25, 50 or 75.

Q11. Why can't you test 12 using 2 and 6?
A11. 2 and 6 are not coprime (share factor 2); a number can pass both yet not be divisible by 12.

Q12. Smallest digit x making 4x2 divisible by 3?
A12. 6 + x must be a multiple of 3, so x = 0.

Every campus aptitude test opens with number-system questions, and the fastest way to lose time is to actually divide big numbers. Divisibility rules let you decide "does 7 divide 3,58,024?" in seconds — and they quietly power half of the remainder, factor and simplification questions you will meet later.

What this topic tests: whether you can check divisibility by 2–11 instantly, use rules in reverse (find a missing digit), and recognise that a number's prime factorisation is the master key to everything about its factors.

The method

Beginner — the core rules

Memorise these; they are non-negotiable in placements.

Divisor	Test
2	last digit even
3	digit sum divisible by 3
4	last two digits divisible by 4
5	last digit 0 or 5
6	passes 2 and 3
8	last three digits divisible by 8
9	digit sum divisible by 9
11	(sum of odd-place digits − sum of even-place digits) divisible by 11

Intermediate — combine and factor

For divisors that are not on the list, split into coprime parts. To test divisibility by 12, check 3 and 4 (not 2 and 6 — those share the factor 2 and give false positives). By 72, check 8 and 9. By 99, check 9 and 11. The rule: break the divisor into coprime factors and apply each rule.

To find the factors of a number, write its prime factorisation: 360 = 2³ × 3² × 5. Every factor is built by choosing each prime to some allowed power. That single idea drives factor-counting, HCF/LCM and remainders.

Advanced — the fast approach

• 7, 13 rarely need a rule; if forced, use the "×3 subtract" trick for 7: drop the last digit, subtract twice it from the rest (e.g. 3-5-8 → 35 − 2·8 = 19, not divisible, so 358 isn't by 7).
• For missing-digit problems, apply the rule as an equation. "For which x is 4x2 divisible by 3?" → 4 + x + 2 = 6 + x divisible by 3 → x ∈ {0,3,6,9}.

Worked example

Is 61,71,49,203 divisible by 11? Mark alternate digits from the right: positions (odd) 3,2,4,7,6 and (even) 0,9,1,1... let us just sum odd-place = 3+2+9+7+... instead compute cleanly. Digits right-to-left: 3,0,2,9,4,1,7,1,6. Odd places (1st,3rd,…): 3+2+4+7+6 = 22. Even places: 0+9+1+1 = 11. Difference = 22 − 11 = 11, divisible by 11. Yes. No long division needed.

Where it appears

Number-system MCQs in TCS NQT, Cognizant, Capgemini, Infosys and Wipro aptitude rounds; also as a hidden step inside HCF/LCM, remainder and "largest number that divides" questions across every company test.

Speed tricks and shortcuts

• Digit-sum for 3 and 9 also gives the remainder: 3,58,024 has digit sum 22 → remainder 22 mod 9 = 4, so it leaves 4 on division by 9.
• Mnemonic for the coprime rule: "Split until they share nothing."
• For 11, remember "odds minus evens."

Key formulas

• 2: last digit even; 4: last two digits ÷4; 8: last three digits ÷8.
• 3: digit sum ÷3; 9: digit sum ÷9.
• 5: ends in 0 or 5; 10: ends in 0; 25: last two digits ÷25.
• 11: (sum of odd-place digits − sum of even-place digits) is 0 or ÷11.
• 6: divisible by 2 and 3; 12: by 3 and 4; composite rule = coprime factors.
• 7: subtract twice the last digit from the rest, repeat.

Usage: for a composite divisor, split into coprime factors and test each rule.

Worked Example

Problem: For which single digit x is the 6-digit number 24x8x6 divisible by 11?

Solution: The rule of 11 says a number is divisible by 11 when the difference between the sum of digits in odd positions and the sum in even positions is 0 or a multiple of 11.

Label the digits from the right (position 1 = units):

Position	6	5	4	3	2	1
Digit	2	4	x	8	x	6

• Odd positions (1, 3, 5): 6 + 8 + 4 = 18
• Even positions (2, 4, 6): x + x + 2 = 2x + 2

Difference = 18 − (2x + 2) = 16 − 2x.

Set this equal to a multiple of 11 (…, −11, 0, 11, …):

• 16 − 2x = 0 → x = 8 ✓ (valid digit)
• 16 − 2x = 11 → x = 2.5 ✗
• 16 − 2x = −11 → x = 13.5 ✗

Only x = 8 works. The number becomes 248886, and indeed 248886 ÷ 11 = 22626 exactly.

Answer: x = 8.