Divisibility Rules

By 2: last digit is even (0,2,4,6,8). By 3: digit sum divisible by 3. By 4: last two digits form a number divisible by 4. By 5: ends in 0 or 5. By 6: divisible by both 2 and 3. By 8: last three digits divisible by 8. By 9: digit sum divisible by 9. By 10: ends in 0. By 11: difference between sum of digits at odd places and even places is 0 or a multiple of 11. By 25: last two digits are 00, 25, 50 or 75. Memory trick: for 6, 12, 15, 18 break into co-prime factors and check each (e.g. 12 = 4 and 3).

Long division in an SSC CHSL exam hall costs you 60 seconds you do not have. The divisibility rule for 11 crunches the same check into 8 seconds, no calculator, no scratch — purely a digit-by-digit dance.

Definition: A number is divisible by 11 if the difference between the sum of its digits in odd positions and the sum of its digits in even positions (counted from the right) is either 0 or a multiple of 11 (such as ±11, ±22, ±33).

Definition: "Position from the right" means the units digit is position 1, tens digit is position 2, hundreds digit is position 3, and so on. Odd positions are 1, 3, 5, ... (units, hundreds, ten-thousands). Even positions are 2, 4, 6, ... (tens, thousands, hundred-thousands).

The mechanical drill — alternate + and − from the right

Take any number. Starting from the rightmost (units) digit, label the digits with alternating signs: +, −, +, −, +, ... Then add together all the digits with the + sign, add together all the digits with the − sign, and find the difference. If the difference is 0 or any multiple of 11, the original number is divisible by 11.

Worked example as in the source: 41809.

Step 1: Label from the right: 9(+), 0(−), 8(+), 1(−), 4(+).
Step 2: Plus-positions sum = 9 + 8 + 4 = 21.
Step 3: Minus-positions sum = 0 + 1 = 1.
Step 4: Difference = 21 − 1 = 20.
Step 5: Is 20 a multiple of 11? No — the multiples are 0, 11, 22, 33, ...
Conclusion: 41809 is not divisible by 11.

Now a second drill, this time with a number that is divisible: 482218.

Step 1: From the right: 8(+), 1(−), 2(+), 2(−), 8(+), 4(−).
Step 2: Plus-positions sum = 8 + 2 + 8 = 18.
Step 3: Minus-positions sum = 1 + 2 + 4 = 7.
Step 4: Difference = 18 − 7 = 11. Yes, 11 is a multiple of 11.
Conclusion: 482218 is divisible by 11. Verify: 482218 ÷ 11 = 43838. Clean.

Why the rule works — the algebra behind the trick

This is one of those magical-looking rules that has a perfectly logical explanation. Consider what each power of 10 does mod 11:

10^0 = 1 ≡ +1 (mod 11)
10^1 = 10 ≡ −1 (mod 11)
10^2 = 100 ≡ +1 (mod 11)
10^3 = 1000 ≡ −1 (mod 11)
10^4 = 10000 ≡ +1 (mod 11)

Notice the pattern: every power of 10 alternates between +1 and −1 (mod 11). So a number d_n d_{n−1} ... d_1 d_0, when reduced mod 11, becomes

(d_0)(+1) + (d_1)(−1) + (d_2)(+1) + (d_3)(−1) + ...

That is exactly the alternating sum from the right. The number is divisible by 11 if and only if this alternating sum is ≡ 0 (mod 11). The "rule" is just modular arithmetic in disguise.

You do not need to remember this proof for the exam, but understanding it once cements the rule for life. You stop fearing it because you know why it works.

A reverse problem — fill in the missing digit

Examiners love the variant: "What digit should replace * in 5*132 to make it divisible by 11?"

Question: Find the digit * in 5*132 such that the number is divisible by 11.

Solution:
Step 1: Label from the right: 2(+), 3(−), 1(+), *(−), 5(+).
Step 2: Plus-sum = 2 + 1 + 5 = 8.
Step 3: Minus-sum = 3 + *.
Step 4: Difference = 8 − (3 + *) = 5 − *.
Step 5: For divisibility, 5 − * must be 0 or ±11 or ±22, ...
Step 6: * is a single digit (0–9). 5 − * = 0 gives * = 5. 5 − * = −11 gives * = 16 (not allowed). 5 − * = 11 gives * = −6 (not allowed).
Conclusion: The missing digit is 5, giving 55132. Verify: 55132 ÷ 11 = 5012, exact.

A second variant — three-digit palindromes

A neat insight: every three-digit palindrome with the same first and last digit and any middle digit has a clean divisibility pattern with 11.

Take aba (e.g., 525). Label from the right: a(+), b(−), a(+). Plus-sum = 2a, minus-sum = b. Difference = 2a − b. So aba is divisible by 11 iff 2a − b ≡ 0 (mod 11). For 525: 2(5) − 2 = 8, not a multiple of 11. For 484: 2(4) − 8 = 0, yes — divisible. 484 ÷ 11 = 44. Clean.

Why it matters: SSC CHSL, SSC CGL, RRB NTPC, and bank PO/clerk papers almost always have a number-system question testing divisibility by 7, 11, or 13. The rule for 11 is the single most common one because the alternating-sum trick is elegant and unique. Add to that 2–3 marks from algebra/quant questions that secretly use divisibility (e.g., "Find the smallest multiple of 11 with...") and you are looking at 4–5 marks per Tier-I that hinge on this one tool.

Real-world parallel

Real-world example: ISBN-10 codes use a divisibility-by-11 checksum. If you weight the first 9 digits of an ISBN by their position and check that the total is divisible by 11 (with X representing 10 if needed), the ISBN is valid. The same modular-arithmetic logic underlies your everyday book barcodes.

Banking sector also uses similar mod-checks. The MICR code on Indian cheques does not specifically use mod-11, but the same family of weighted-sum checks (mod 7 in some cases) catches data-entry errors before they reach the clearing house.

Common misconceptions, cleared

Common misconception: "Start counting from the left." Wrong. The rule pairs each digit with the right power of 10, and 10^0 lives at the rightmost digit. Counting from the left flips the sign pattern and gives you the negative of the correct difference — which, fortunately, is still ±0 or ±11k, so the test still works. But to avoid confusion, always start from the right.

Common misconception: "If the alternating sum is positive, the number is divisible by 11." Sign does not matter — only whether the absolute value is 0 or a multiple of 11. So both +11 and −11 confirm divisibility.

Common misconception: "I have to compute the difference exactly." You only need to know it mod 11. If you spot 22, you can stop and declare "divisible." If you spot 18, you can stop and declare "not divisible." You never need long division.

Common misconception: "The rule works only for small numbers." It works for any number, however large. Try 1331 (which is 11^3): 1(−) + 3(+) + 3(−) + 1(+) (from right) = (1+3) − (3+1) = 0. Divisible. Try 14641 (which is 11^4): from right 1(+), 4(−), 6(+), 4(−), 1(+) = (1+6+1) − (4+4) = 8 − 8 = 0. Divisible. The rule scales effortlessly.

A 30-second drill template

In your scratch sheet, write the number once, draw an arrow over each digit with alternating + and − starting from the right, compute the two sums, subtract, and check against 0, ±11, ±22, ±33. That is the entire workflow. With practice you can do it mentally for 5-digit numbers and on paper for 8–9 digit numbers in under 20 seconds.

Q: For what value of x is the number 34x5 divisible by 9? Solution: A number is divisible by 9 if its digit sum is divisible by 9. Digit sum = 3 + 4 + x + 5 = 12 + x. We need 12 + x to be a multiple of 9. The next multiple of 9 after 12 is 18, so 12 + x = 18, giving x = 6. Check: 3465, digit sum = 18, divisible by 9. Answer: x = 6.

Divisibility rules are the fastest tools in SSC CHSL Quant. They crack factor questions, "find the missing digit", HCF/LCM and simplification problems without long division. Expect these to appear directly (1 question) and indirectly in many others. Learning them cold saves 30–40 seconds per problem.

The rule table

Divisor	Test
2	last digit even
3	digit sum ÷ 3
4	last 2 digits ÷ 4
5	ends in 0 or 5
6	passes 2 and 3
8	last 3 digits ÷ 8
9	digit sum ÷ 9
11	(sum of odd places − sum of even places) ÷ 11
12	passes 3 and 4
25	last 2 digits are 00, 25, 50, 75

The tricky ones

• 7: drop the last digit, subtract twice it from the rest; repeat. 672 → 67 − 2×2 = 63 → divisible by 7. ✔
• 11: for 4,58,061, odd-place sum − even-place sum must be a multiple of 11 (including 0).
• Composite divisors: for 72 = 8 × 9, the number must pass both 8 and 9 (co-prime factors).

Exam Tricks & Tips

• 🎯 For "divisible by 12", never test 12 directly — check 3 and 4 separately; far faster.
• 🎯 For a missing digit with divisibility by 9, make the digit sum a multiple of 9 and solve for the digit.
• 🎯 Divisibility by 11: for even-digit palindromes and alternating patterns the difference is often exactly 0 → divisible.
• 🎯 To test 8, look only at the last 3 digits — ignore the rest entirely.
• 🎯 Break any divisor into co-prime factors (72=8×9, 36=4×9, 15=3×5) and test each; don't use non-co-prime pairs like 4×6.
• ❌ Don't test divisibility by 4 using the last one digit — always use the last two digits (and 8 uses the last three).

Expected exam pattern

"Which digit makes 34_78 divisible by 3?", "The largest number that divides…", or "By which of these is 4620 divisible?". One-step logic, ~20–25 seconds.

Quick recap

Memorise rules for 2–11; for composite divisors split into co-prime factors and test each. Remember 4→last 2 digits, 8→last 3, 11→alternating-place difference.

Cover the answer, recall, then check. 11 cards on divisibility rules for SSC CHSL.

Q1. Divisibility rule for 4?
A1. The number formed by the last two digits must be divisible by 4. (e.g. 1732 → 32 ÷ 4 ✔)

Q2. Divisibility rule for 8?
A2. The last three digits must form a multiple of 8. (e.g. 51,048 → 048 ÷ 8 ✔)

Q3. Divisibility rule for 11?
A3. (Sum of digits in odd places) − (sum in even places) is 0 or a multiple of 11.

Q4. How do you test divisibility by 12?
A4. Check 3 and 4 separately (12 = 3×4, co-prime). Pass both ⇒ divisible by 12.

Q5. Is 61,809 divisible by 11?
A5. Odd-place digits (from right): 9+8+6 = 23; even-place: 0+1 = 1; difference 23−1 = 22, a multiple of 11 → Yes.

Q6. Quick rule for divisibility by 7?
A6. Drop the last digit, subtract twice it from the remaining number; if the result is a multiple of 7, so is the original. (672 → 67−4=63 ✔)

Q7. Rule for 6?
A7. Must be divisible by both 2 and 3.

Q8. Rule for 9?
A8. The digit sum must be divisible by 9.

Q9. Find digit x so that 4x3 is divisible by 3.
A9. 4+x+3 = 7+x must be a multiple of 3 → x = 2, 5, or 8.

Q10. Divisibility rule for 25?
A10. Last two digits are 00, 25, 50 or 75.

Q11. Why split 72 into 8×9 rather than 6×12 when testing divisibility?
A11. 8 and 9 are co-prime; 6 and 12 share factors, so passing both would not guarantee divisibility by 72.