Every divisibility rule for SSC CGL, CHSL, CPO and other competitive exams — why each rule works, how to build a rule for any divisor, the missing-digit, counting and remainder question types, and the algebraic patterns that are always divisible. Every concept used across all chapter tests is explained here from zero.
Divisibility is the fastest-scoring topic in SSC Quantitative Aptitude. A divisibility question rarely needs any real calculation — it needs the right rule, read off the right digits, in under thirty seconds. Questions from this chapter appear in SSC CGL, SSC CHSL, SSC CPO, SSC MTS and nearly every other competitive exam, and they also hide inside HCF-LCM, remainder and number-pattern questions.
These notes explain every rule from the beginning — not just what the rule says but why it works, because the "why" is what lets you build a rule for a divisor you have never met. Each rule comes with a worked example, the exam shortcut, and the mistake that most often costs a mark.
Notes fix the method; only practice fixes the speed. Keep these open in one tab and a test in another — start with the chapter tests for this topic, then move to full papers once the rules feel automatic.
Graded across three levels, matched to these notes. Every question carries a step-by-step solution.
Start practising Previous yearReal shift-wise CGL papers with answer keys and full solutions. The best guide to what actually gets asked.
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Take a mock testHow to use these notes: read a section, cover the example with your hand, solve it yourself, then check. A rule you have only read is forgotten by evening. A rule you have used once stays.
A number N is divisible by d when dividing N by d leaves no remainder. We also say "d divides N", "N is a multiple of d", or "d is a factor of N" — all four phrases mean the same thing.
The division algorithm. For any division, Dividend = Divisor × Quotient + Remainder, and the remainder is always smaller than the divisor. Divisibility is the special case where the remainder is 0.
This one line answers a whole family of questions. If a division leaves 13 behind, and the divisor is 19 times the remainder and 13 times the quotient, then divisor = 247, quotient = 19, and the dividend is 247 × 19 + 13 = 4706. Always start from the quantity that is actually given.
Two positive numbers total 6318. Dividing the larger by the smaller gives 11 with 18 left over. Find the smaller.
Let the smaller be S. Then the larger is 11S + 18, and the sum is 11S + 18 + S = 12S + 18 = 6318.
So 12S = 6300 and S = 525. Check: 11 × 525 + 18 = 5793, and 5793 + 525 = 6318 ✔
The trap: if the question gives a difference instead of a sum, the coefficient is 11 − 1 = 10, not 11 + 1 = 12. One word decides which. The arithmetic usually will not warn you — 6300 ÷ 10 = 630 looks perfectly respectable.
A number divided by 8 and separately by 14 leaves no remainder either time, and the first quotient is 9 more than the second. Find the number.
N/8 − N/14 = 9. Over the LCM 56: 7N/56 − 4N/56 = 3N/56 = 9, so N = 168.
Check: 168 ÷ 8 = 21, 168 ÷ 14 = 12, and 21 − 12 = 9 ✔
"Divisible by 3 and 9 both" is not the same as "divisible by 27." 3 and 9 are not co-prime — they share the factor 3 — so passing both tests guarantees only 9. This single idea (co-prime parts) runs through the entire chapter; Section 7 makes it precise.
These rules all read the last few digits — one digit for each factor of 2 in the divisor.
| Divisor | Look at | Why | Example |
|---|---|---|---|
| 2 | last 1 digit — must be 0, 2, 4, 6, 8 | 10 = 2 × 5 | 4738 ✔ |
| 4 | last 2 digits form a multiple of 4 | 100 = 4 × 25 | 7316 → 16 ✔ |
| 8 | last 3 digits form a multiple of 8 | 1000 = 8 × 125 | 17512 → 512 ✔ |
| 16 | last 4 digits form a multiple of 16 | 10000 = 16 × 625 | 4173840 → 3840 ✔ |
| 32 | last 5 digits form a multiple of 32 | 100000 = 32 × 3125 | 57529696 → 29696 ✔ |
Everything above the last k digits is a multiple of 10k, and 10k = 2k × 5k. So those higher digits contribute nothing to the remainder under 2k. Only the last k digits matter.
Count the twos in the divisor before choosing how many digits to read. 112 = 24 × 7 needs four digits, not three — the 8-rule would let wrong numbers through.
Reduce the fixed part first. For a blank in 7x7y16 under 8: the last three digits are y16 = 100y + 16. 16 is already a multiple of 8, and 100y leaves 4y. So you need 4y to be a multiple of 8, i.e. y even. Reducing 100 to its remainder 4 turns a three-digit test into a one-line condition.
Which digit z makes 8 divide 4718z6?
Last three digits: 8z6 = 806 + 10z. Now 806 leaves 6 under 8, and 10z leaves 2z. Need 6 + 2z to be a multiple of 8, i.e. 3 + z a multiple of 4, so z = 1, 5 or 9. Accepted digits sit 4 apart because 10 leaves remainder 2 under 8, and four steps of 10 add 40, a multiple of 8.
For the 4-rule the last two digits are a block. In 9080xy the digits x and y together must form a multiple of 4 — 04, 16, 28, 40, 52, 64, 76, 88 — not each on its own. Test them as one two-digit number.
The same last-digits idea, but now one digit per factor of 5.
| Divisor | Test | Example |
|---|---|---|
| 5 | Ends in 0 or 5 | 2350 ✔ |
| 10 | Ends in 0 | 8970 ✔ |
| 25 | Last two digits 00, 25, 50 or 75 | 3675 ✔ |
| 125 | Last three digits 000, 125, 250, 375, 500, 625, 750, 875 | 548625 ✔ |
| 625 | Last four digits form a multiple of 625 | 675000 ✔ |
These are the cheapest screens in the chapter. Whenever the divisor carries a 5, read the last digit first. In a "which of these is divisible by 45" question, three of four options usually die on the last digit alone. And when a blank sits in the last position, the 5-rule pins it to just two values (0 or 5) before any arithmetic starts.
Neither 25 nor 7 leaves anything behind in 10x6367y. Find x × y.
The last two digits are 7y. Of the four allowed endings, the only one starting with 7 is 75 — so y = 5 from two digits alone. With y fixed, the 7-test has a single unknown: 10x63675 = 10063675 + 100000x, which leaves 6 + 5x under 7, so x = 3. Answer 15. The 25-rule did the hard work in one glance.
The 10-rule pins a blank to exactly one value (0), the 5-rule to two, the 25-rule to the four endings. Run the rule that admits the fewest possibilities first — it hands every later rule a smaller problem.
A number is divisible by 3 when its digit sum is divisible by 3, and by 9 when its digit sum is divisible by 9. More than that: a number and its digit sum leave the same remainder under 3 and under 9. So the digit sum tells you the remainder, not just yes-or-no.
Why: 10 = 9 + 1, 100 = 99 + 1, 1000 = 999 + 1. Every power of 10 leaves remainder 1 under 9 (and under 3). So each digit contributes only itself to the remainder, and adding the digits captures everything.
What is left when 4728635 is divided by 9?
Digit sum 4+7+2+8+6+3+5 = 35, and 3+5 = 8. Remainder 8. Casting out nines: drop any 9 and any pair summing to 9 before adding — here 4+5 and 7+2 go, leaving 8+6+3 = 17 → 8. Same answer, half the work.
The gap of z4728 is at the front and 9 must divide the result. Find z.
Digit sum 21 + z. The only multiple of 9 in 21 to 30 is 27, so z = 6. Note the leading position forbids z = 0 — a leading zero would shorten the number. It did not matter here, but on another number it would.
Accepted digits sit 3 apart under the 3-rule, 9 apart under the 9-rule. So a blank under 3 usually has three or four answers (a chain like 1, 4, 7 or 0, 3, 6, 9), while a blank under 9 has exactly one. Read the remainder of the fixed digits and the whole list follows.
27 has no digit-sum rule. The digit sum settles 9 and no more. For 27 (or 81), use the digit sum as a screen to cut the field to one or two candidates, then confirm by division. Treating 9 as sufficient is the commonest error on this type — it leaves a false second answer standing. Example: for 660x332y under 135 = 27 × 5, the 9-screen gives (7,0) and (2,5); only 66073320 ÷ 27 = 2447160 is whole, so (7,0) is the answer.
Write the digits from the right. Add the odd-position digits, add the even-position digits, and subtract. The number is divisible by 11 when that difference is 0 or a multiple of 11 (positive or negative). The difference also gives the remainder once reduced into 0 to 10 — a difference of −2 means remainder 9.
Why: 10 = 11 − 1, 100 = 99 + 1, 1000 = 1001 − 1. Powers of 10 alternate between remainder +1 and −1 under 11, so the digits get alternating signs.
Which digit z makes 11 divide 4z8153?
From the right: 3, 5, 1, 8, z, 4. Odd positions 3+1+z = 4+z; even positions 5+8+4 = 17. Difference z − 13, which for a digit lies in −13 to −4. The only multiple of 11 there is −11, so z = 2.
The position of the blank changes its sign. Had z sat one place over, the condition would read −4 − z and the answer would be 7, not 2. Count the places from the right before assigning the digit.
Two blanks under 11 usually give a total, not the digits. In 25x493y the 11-condition comes out as x + y − 1, so x + y = 1 or 12. Combined with the 3-rule (x + y leaves remainder 1) only x + y = 1 survives — two different numbers work, but their digit total is fixed. That is why the question asks for x + y.
Divided by 11, 7392648 leaves 10. What is the least to subtract so it leaves 3?
Subtracting k lowers the remainder by k: 10 − k = 3 gives k = 7. Check by the rule on 7392641: odd sum 23, even sum 9, difference 14 → 14 − 11 = 3 ✔. The rule worth keeping: to move from remainder r down to target t, subtract (r − t) mod 11.
These have no digit-sum or alternating-sum rule. Three tools cover them: the 1001 method, the 999 method, and osculators.
1001 = 7 × 11 × 13, so 1000 ≡ −1 under each of them. Break the number into three-digit blocks from the right and alternate the signs. The result leaves the same remainder as the original.
8471625 → 625 − 471 + 8 = 162 → 162 = 7 × 23 + 1, so the remainder under 7 is 1. One split, and the same 162 also gives the remainder under 11 and under 13.
999 = 27 × 37, so 1000 ≡ +1 under each. Break into three-digit blocks from the right and simply add them — no alternating signs. 6095084 → 84 + 95 + 6 = 185 = 37 × 5, so 37 divides it.
Do not mix the two: 1001 alternates (−1), 999 adds (+1). Confusing them gives wrong answers for both.
Strip the last digit, multiply it by the osculator, and add or subtract from what is left. Repeat until small. If the result is a multiple of the divisor, so was the original. The osculator comes from the nearest multiple of the divisor ending in 1 or 9:
| Divisor | Nearest multiple ending in 1 or 9 | Osculator | Action |
|---|---|---|---|
| 7 | 21 | 2 | subtract |
| 13 | 39 | 4 | add |
| 17 | 51 | 5 | subtract |
| 19 | 19 | 2 | add |
| 23 | 69 | 7 | add |
| 29 | 29 | 3 | add |
| 31 | 31 | 3 | subtract |
The sign rule: if the multiple ends in 9 (one below a ten), add; if it ends in 1 (one above a ten), subtract. 19 → add 2; 21 → subtract 2.
Which of 1769603, 2894811, 3442257, 4741485 is a multiple of 31?
Osculator −3. For 2894811: 289481 − 3 = 289478; 28947 − 24 = 28923; 2892 − 9 = 2883; 288 − 9 = 279 = 31 × 9. So 2894811. The other three end at 167, 323 and 448 — none a multiple of 31.
An osculator tests divisibility but does not report the remainder. If a question asks "what is left over under 17", use staged division (peel off 17 × 300000, then 17 × 70000, and so on) — the quotient assembles itself from the pieces and the remainder falls out at the end.
For a blank under 7 or 13, split by place value. 2z6134 = 206134 + 10000z. Reduce each piece: 206134 leaves 5 under 7, 10000 leaves 4. So the number leaves 5 + 4z, and z = 4 is the only digit making that a multiple of 7. Shorter and safer than the block method when the unknown sits inside a block.
To test a composite divisor, split it into parts that are co-prime (no common factor beyond 1) and test each part. A number passing all the part-tests is divisible by their product. 72 = 8 × 9; 45 = 5 × 9; 264 = 8 × 3 × 11; 132 = 4 × 3 × 11; 165 = 3 × 5 × 11.
Why co-prime matters: if the parts share a factor, passing both guarantees only their LCM, which is smaller than the product. 24 split as 4 × 6 is wrong — 12 passes both and is not a multiple of 24. Use 8 × 3.
| Divisor | Split | Divisor | Split |
|---|---|---|---|
| 6 | 2 × 3 | 72 | 8 × 9 |
| 12 | 4 × 3 | 88 | 8 × 11 |
| 15 | 3 × 5 | 99 | 9 × 11 |
| 18 | 2 × 9 | 104 | 8 × 13 |
| 22 | 2 × 11 | 112 | 16 × 7 |
| 24 | 8 × 3 | 117 | 9 × 13 |
| 33 | 3 × 11 | 126 | 2 × 9 × 7 |
| 36 | 4 × 9 | 144 | 16 × 9 |
| 44 | 4 × 11 | 168 | 8 × 3 × 7 |
| 45 | 5 × 9 | 176 | 16 × 11 |
| 48 | 16 × 3 | 189 | 27 × 7 |
| 55 | 5 × 11 | 198 | 2 × 9 × 11 |
| 56 | 8 × 7 | 208 | 16 × 13 |
| 63 | 9 × 7 | 216 | 8 × 27 |
| 66 | 2 × 3 × 11 | 224 | 32 × 7 |
Order the tests by how many candidates they kill. In a "which of these four" question: parity first (free), then the last digits, then the digit sum, and only then the expensive 7 or 13 test — on the one survivor. Three of four candidates usually die before any real arithmetic.
Which of 393831, 399500, 815472, 872704 does 168 divide?
168 = 8 × 3 × 7. Parity: 393831 is odd, out. Digit sums: 399500 → 26, 872704 → 28, both fail 3; 815472 → 27 ✔. Only 815472 needs the 7-test: 472 − 815 = −343 = 7 × (−49) ✔. Answer 815472. Note 872704 clears 8 easily and fails only on 3 — passing the hardest-looking rule says nothing about the easiest.
"Divisible by 6 twice over" is not 36. 6 × 6 shares everything, so a number can pass the 6-test and fail 36 — 18 does. Split 36 as 4 × 9. In general, never split a divisor into two copies of the same factor.
A single blank under a single rule. Apply the rule with the blank as an unknown, and read off which digits 0 to 9 satisfy it. The question then asks for the unique digit, the largest, the smallest, the sum of all accepted digits, or how many there are.
Which digit lets 9 divide 7z3821? Digit sum 21 + z; only 27 is reachable, so z = 6.
Several digits let 4 divide 3z4. Largest minus smallest? Last two digits z4 = 10z + 4, need z even: 0, 2, 4, 6, 8. So 8 − 0 = 8. The trap is forgetting that 0 is accepted — the gap is not the leading position.
Several digits let 3 divide 62z471. Which is the second-smallest? Digit sum 20 + z, need z leaving remainder 1: 1, 4, 7. Second-smallest is 4.
Zero. Inside the number, 0 is a perfectly good digit and is often the only answer — 94z27615 under 11 gives z = 0 and nothing else. At the front of the number, 0 is forbidden because it would shorten the number. Check where the blank is before ruling anything in or out.
Read the remainder of the fixed part, then subtract from the divisor. For 38z14 under 9, the fixed digits sum to 16, which leaves 7; so z must supply 9 − 7 = 2. One subtraction instead of a list. But watch the reflex: the answer is the gap to the next multiple, not the remainder itself.
68x338y is a multiple of 72. Find x + y.
72 = 8 × 9. Last three digits 38y = 380 + y; 380 leaves 4 under 8, so y = 4. Digit sum 28 + x + 4 = 32 + x; only 36 is reachable, so x = 4. Answer 8. Each rule pinned one digit.
82x298y answers to 7 and to 12. Largest x × y?
4-rule: y = 0, 4 or 8. 3-rule: x + y leaves remainder 1. 7-test: 82x298y = 8202980 + 10000x + y, and 8202980 leaves 2, 10000 leaves 4, so the number leaves 2 + 4x + y. Branch by y: y = 0 gives x = 3 but fails 3; y = 4 gives x = 2 or 9, only (9,4) passes 3; y = 8 gives x = 1 or 8, only (8,8) passes 3. Products 36 and 64 → 64. Five candidates from 7, cut to two by the cheap 3-rule.
The digits of 4A3B4 multiply to 12 and 36 divides the number. Find A + B.
A × B = 12 allows only (2,6), (3,4), (4,3), (6,2). The 4-rule needs B even, leaving (2,6), (3,4), (6,2). The 9-rule needs 11 + A + B a multiple of 9, i.e. A + B = 7 → only (3,4). Answer 7. The product condition cut a hundred cases to four in one line.
Cap the range a relation allows before hunting for multiples. If A = B + 3, then B ≤ 6; if B = 2A, then A ≤ 4. Without the cap a "solution" like B = 10 slips through and looks fine until the end.
When a sum is fixed and a product is asked, the most balanced pair wins. If x + y = 13, then (7,6) beats (9,4) and (5,8). When squares are asked, the pair with the single largest digit usually wins, but compute every legal pair — (3,8) gives 73 while (8,2) gives only 68.
"In how many ways can the blanks be filled" is the two-blank method with counting at the end instead of picking. Branch on the blank pinned by the last-digits rule, count the values of the other blank in each branch, and add.
How many pairs (x, y) make 24 divide 72x4y6?
8-rule on 4y6 = 406 + 10y: need y = 1, 5 or 9. 3-rule: 19 + x + y, so x + y leaves remainder 2.
y = 1 → x = 1, 4, 7 (3 pairs). y = 5 → x = 0, 3, 6, 9 (4 pairs). y = 9 → x = 2, 5, 8 (3 pairs). Total 10.
One branch gives four because its chain of x-values starts at 0.
Pairs with a fixed total t. For t ≤ 9 there are t + 1 pairs (0..t); for t ≥ 10 there are 19 − t. So a total of 6 gives 7 pairs, 10 gives 9, 15 gives 4. Totals near the middle are the richest. Under 9 the totals sit 9 apart, so only two are ever reachable and the counts stay small.
(0,0) counts when neither blank leads the number. Discarding zeros by habit gives 6 instead of 7 for 393xy under 15 — and 6 is offered as a distractor.
Multiples of k from 1 to n: ⌊n/k⌋ (throw away the decimal part).
Multiples of k from a to b: ⌊b/k⌋ − ⌊(a−1)/k⌋ — the second term strips
everything below the start. Use a − 1, not a, so that a itself is counted if it qualifies.
| Question | Count | Example |
|---|---|---|
| Divisible by a and b | multiples of LCM(a,b) | 1 to 2400 by 16 and 20: LCM 80 → 30 |
| Divisible by a or b | n(a) + n(b) − n(LCM) | 500 to 2000 by 9 or 12: 167 + 125 − 42 = 250 |
| Divisible by a but not b | n(a) − n(LCM) | 1 to 900 by 7 not 21: 128 − 42 = 86 |
| By a, not b, not c | n(a) − n(ab) − n(ac) + n(abc) | 1 to 1200 by 5, not 3, not 4: 240 − 80 − 60 + 20 = 120 |
The overlap divisor is the LCM, not the product. For 9 or 12 the overlap is 36, not 108. For 7 not 21, the overlap is 21 itself (one divisor already contains the other). Using the product undercounts the overlap and inflates the answer.
A tally of multiples of 31, kept from 1 up to some ceiling N, reads 96. How high can N be?
⌊N/31⌋ = 96 means 31 × 96 ≤ N < 31 × 97, i.e. 2976 ≤ N < 3007. Highest is 3006. The trap is 2976 — that is the lowest ceiling giving 96, not the highest. The tally does not change between one multiple and the next.
Check with a small block. The pattern repeats every LCM. For "by 5, not 3, not 4" the block 1 to 60 contains six qualifying numbers (5, 10, 25, 35, 50, 55), and 1200 = 20 blocks → 120. When the range is a whole number of blocks, this confirms the inclusion–exclusion arithmetic in one step.
Divide once: N = qk + r. Then
658341 is below a multiple of 173. How far must it climb?
173 × 3805 = 658265, remainder 76. Climb = 173 − 76 = 97. The remainder 76 answers "how far down", and is offered on purpose. Read which direction the question asks before writing the remainder down.
24, 45 and 16 will none of them go into 63295. Least to add?
LCM = 24 × 32 × 5 = 720 (the 24 from 16, the 32 from 45). 63295 = 720 × 87 + 655. Add 720 − 655 = 65. Missing the 24 and using 23 from 24 gives LCM 360 and a wrong answer.
Largest 5-digit multiple of 307: 99999 = 307 × 325 + 224, so 99999 − 224 = 99775.
Smallest 6-digit multiple of 239: 100000 = 239 × 418 + 98, so 100000 + (239 − 98) = 100141.
Come down from the ceiling; climb up from the floor. Going the wrong way gives a genuine multiple with the wrong number of digits.
"Which multiple is closer" needs the comparison done, not assumed. 43917 under 158 leaves 151, far more than half of 158, so the multiple above (43924, just 7 away) is nearer. Reflexively rounding down picks 43766, which is 151 away.
Adding k raises the remainder by k; subtracting k lowers it, wrapping round by the divisor. To move from remainder r to target t:
A wrap-round is needed exactly when the target lies on the wrong side of the current remainder.
Under 13, 4826395 leaves 2. Least to subtract so it leaves 6?
Target 6 is above the current 2, and subtracting only lowers — so go round: 2 − k = −7, k = 9. Formula: (2 − 6) mod 13 = 9. The reflex answers 2 (aim for zero) and 4 (6 − 2) are both wrong.
Aiming for remainder 0 by reflex. The question says "leaves 5 instead" — the target is 5, not 0. Read the target before computing anything.
Every prime factor of a product must come from one of the two factors. To make 8721 × 3n5 a multiple of 165 = 3 × 5 × 11: 8721 supplies the 3 (digit sum 18) but not the 5 or the 11. The 5 comes free from the final digit of 3n5; the 11 needs 3 − n + 5 = 8 − n to be 0 or 11, so n = 8. Testing 3n5 for 3 would have been wasted work.
6545 × 4n8 is a multiple of 112. Find n.
112 = 16 × 7. 6545 is odd (no twos) and 6545 = 7 × 935 (the 7 is covered). So 4n8 must supply all four twos — it must be a multiple of 16. The multiples of 16 in the 400s are 400, 416, 432, 448, 464, 480, 496; only 448 fits 4n8. n = 4. Testing for 8 instead of 16 would let 408 and 488 through as well.
"Divisible by 12 only / 25 only / both / neither": complete both tests first. 638712 clears 12 (last two digits 12, digit sum 27) and fails 25 (last two digits not 00/25/50/75) → 12 only. Reading the options early tempts you to stop after one test.
When both pass and the divisors are co-prime, the number is also divisible by their product; when they share a factor, only by their LCM — 24 and 45 together guarantee 360, not 1080.
Order: parity → last digits → digit sum → alternating sum → 7/13 blocks → osculator. Each step is cheaper than the next and usually kills at least one option. For 17, 19, 23, 29, 31 no cheap screen exists, so every candidate must be worked — but the osculator keeps each to four or five short steps.
| Expression | Factorises as | Guaranteed factor |
|---|---|---|
| an − 1 | (a − 1)(an−1 + … + 1) | a − 1 |
| an + 1 (n odd) | (a + 1)(an−1 − … + 1) | a + 1 |
| an − an−1 | an−1(a − 1) | a − 1 and the primes of a |
| an + an−1 | an−1(a + 1) | a + 1 and the primes of a |
| a3 − b3 | (a − b)(a2 + ab + b2) | a − b |
| a3 + b3 | (a + b)(a2 − ab + b2) | a + b |
| a4 − b4 | (a − b)(a + b)(a2 + b2) | all three brackets |
| an + an+1 + an+2 + an+3 | an(1 + a + a2 + a3) = an(1+a)(1+a2) | the bracket's primes |
Which of 11, 13, 25, 45 divides 615 − 614?
= 614(6 − 1) = 5 × 214 × 314. The primes are 2, 3 and one 5. 45 = 9 × 5 ✔; 25 needs two fives ✗; 11 and 13 are absent. Answer 45.
Largest single-digit divisor of 230 + 231 + 232?
= 230(1 + 2 + 4) = 230 × 7. Single digits that divide: 1, 2, 4, 7, 8. Largest is 8, not the tempting 7. The question asks for the largest, not the interesting prime.
The answer is a − 1, and the smallest case n = 1 shows nothing larger is guaranteed. 4n − 1 → 3; 6n − 1 → 5; 8n − 1 → 7; 52n − 1 = 25n − 1 → 24; 32n − 1 = 9n − 1 → 8. Watch the exponent: the 2 in 52n makes the base 25, not 5.
"Which fails to divide" inverts the usual question — three options do divide and one does not. For 320 + 321 + 322 = 320 × 13, the total is odd, so 26 fails while 13, 39 and 117 all work. Read the word "fails".
The product of k consecutive integers is always divisible by k! — 3 in a row give 6, 4 give 24, 5 give 120. Among any 3 in a row one is a multiple of 3 and one is even.
| Expression | Why | Always divisible by |
|---|---|---|
| n3 − n = (n−1)n(n+1) | 3 consecutive | 6 |
| n5 − n = (n−1)n(n+1)(n2+1) | 3 consecutive, and n2+1 supplies the 5 | 30 |
| n7 − n | 3 consecutive, and 7 by Fermat's little theorem | 42 |
| n3 + 2n = (n3 − n) + 3n | both parts multiples of 3 | 3 |
| Sum of 5 consecutive = 5(n+2) | 5 times the middle one | 5 |
| Sum of cubes of 3 consecutive | 3n3+9n2+15n+9, and the bracket carries another 3 | 9 |
| Product of 3 consecutive evens = 8n(n+1)(n+2) | 8 from the evens, 6 from the run | 48 |
The smallest case (n = 1 or n = 2) always shows why nothing larger is guaranteed: 25 − 2 = 30, so 60 fails.
| Pattern | Equals | Guaranteed divisor |
|---|---|---|
| abab (2-digit block twice) | 101 × ab | 101 |
| abcabc (3-digit block twice) | 1001 × abc | 1001 = 7 × 11 × 13 |
| abcdabcd (4-digit block twice) | 10001 × abcd | 10001 = 73 × 137 |
| ababab (2-digit block thrice) | 10101 × ab | 10101 — not 101 |
| abccba (3-digit block then its reverse) | 100001a + 10010b + 1100c | 11 — not 1001 |
| abc + bca + cab (rotations) | 111(a + b + c) | 111 |
| abab − ab | 100 × ab | 100 |
| abab − baba | 909(a − b) | 909 |
Writing a k-digit block twice multiplies by 10k + 1. Mirroring it instead gives only 11. Read whether the second copy is repeated or reversed.
In a 3-digit number the hundreds digit is twice the tens digit and the units digit twice the hundreds. Largest number certain to divide it?
With tens digit b: number = 100(2b) + 10b + 4b = 214b. Only b = 1, 2 are legal (214, 428), and 214 caps the guarantee. Convert the relations to place value and the coefficient is the answer.
Downward is always safe: divisible by 132 ⇒ divisible by 44, because 44 is a factor of 132.
Upward needs co-prime parts: divisible by a and by b ⇒ divisible by ab only if a and b share no prime. Otherwise the
guarantee is the LCM, which is smaller — and the product overshoots the LCM by exactly the HCF.
| Claim | Verdict | Because |
|---|---|---|
| 8 and 9 ⇒ 72 | true | co-prime |
| 4 and 25 ⇒ 100 | true | co-prime |
| 9 and 20 ⇒ 180 | true | co-prime |
| 10 and 15 ⇒ 150 | false | share 5; LCM 30. Counter-example 30 |
| 6 and 14 ⇒ 84 | false | share 2; LCM 42. Counter-example 42 |
| 8 and 12 ⇒ 96 | false | share 4; LCM 24. Counter-example 24 |
| 4 and 22 ⇒ 88 | false | share 2; LCM 44. Counter-example 44 |
A single counter-example — the LCM itself — convicts every false claim.
Three separate verdicts, in order: is A true, is R true, does R explain A. Never read the options first. A true Reason can still fail to explain, and a true Reason never rescues a false Assertion — if 6317424 fails the 7-test, it is not a multiple of 168 however correct the rule "24 and 7 ⇒ 168" may be.
Fix the pairing you are surest of first — usually the 125 or 625 property, which reads off the last digits — and throw away every option that disagrees. Leave properties with no digit rule (81, 243, 343, prime squares like 121) until the field has narrowed to one candidate, then confirm by division.
The night before the exam, read only this page.
| Divisor | Rule | Because |
|---|---|---|
| 2 / 4 / 8 / 16 / 32 | last 1 / 2 / 3 / 4 / 5 digits | 10k = 2k × 5k |
| 5 / 25 / 125 / 625 | last 1 / 2 / 3 / 4 digits | same |
| 3, 9 | digit sum (gives the remainder too) | 10 ≡ 1 |
| 11 | alternating sum from the right | 10 ≡ −1 |
| 7, 11, 13 | alternate 3-digit blocks | 1001 = 7 × 11 × 13 |
| 27, 37 | add 3-digit blocks | 999 = 27 × 37 |
| 7, 17, 31 | osculator: subtract 2, 5, 3 × last digit | 21, 51, 31 end in 1 |
| 13, 19, 23, 29 | osculator: add 4, 2, 7, 3 × last digit | 39, 19, 69, 29 end in 9 |
| Composite | split into co-prime parts | 24 = 8 × 3, never 4 × 6 |
| Situation | Formula |
|---|---|
| Division algorithm | Dividend = Divisor × Quotient + Remainder, r < divisor |
| Two numbers, sum S, quotient q, remainder r | smaller = (S − r)/(q + 1) |
| Two numbers, difference D | smaller = (D − r)/(q − 1) |
| Multiples of k in 1..n | ⌊n/k⌋ |
| Multiples of k in a..b | ⌊b/k⌋ − ⌊(a−1)/k⌋ |
| a and b / a or b / a not b | n(LCM) / n(a)+n(b)−n(LCM) / n(a)−n(LCM) |
| Exactly one of a, b | n(a) + n(b) − 2 n(LCM) |
| Least to subtract / add | r / k − r (they sum to k) |
| Nearest multiple | below if r < k/2, above if r > k/2 |
| Shift remainder r → t | add (t−r) mod k | subtract (r−t) mod k |
| Largest n-digit multiple | (largest n-digit) − remainder |
| Smallest n-digit multiple | (smallest n-digit) + (k − remainder) |
| an − 1 | always divisible by a − 1 |
| a3 ± b3 | always divisible by a ± b |
| k consecutive integers | product divisible by k! |
| n3−n / n5−n / n7−n | 6 / 30 / 42 |
| 2-digit reversal | sum 11(a+b); difference 9(a−b) |
| 3-digit reversal | difference 99(a−c); minus digit sum → 9 |
| 4-digit outer swap | difference 999(a−d) |
| Block written twice | × 101 / 1001 / 10001 for 2 / 3 / 4 digits |
Answers: 1) 2 2) 8 3) 10 4) 250 5) 97 6) 9 7) 45 8) No — LCM is 42; counter-example 42
They are quick tests that tell whether one number divides another without doing the division. The last-digits rules cover 2, 4, 8, 5, 25 and 125; the digit-sum rule covers 3 and 9; the alternating-sum rule covers 11; and block methods and osculators cover 7, 13, 17, 19 and other primes. Composite divisors are tested by splitting them into co-prime parts.
Every power of 10 leaves remainder 1 when divided by 9, so each digit contributes only itself to the remainder. Adding the digits therefore gives the same remainder as the whole number. The same reasoning gives the rule for 3, and the rule for 11 comes from powers of 10 alternating between +1 and −1.
Two ways. The osculator: strip the last digit, double it, subtract from the rest, and repeat until the number is small. Or the 1001 method: split into three-digit blocks from the right, alternate the signs and add — the result leaves the same remainder as the original, and it works for 11 and 13 at the same time.
Split the divisor into co-prime parts and test each: 72 = 8 × 9, 88 = 8 × 11. The parts must share no factor — 24 must be split as 8 × 3, not 4 × 6, because 4 and 6 both carry a 2 and the test would pass numbers that fail 24.
Divide and take the remainder r. The least to subtract is r; the least to add is divisor minus r. The two always sum to the divisor, which is a quick check. For three divisors, use their LCM as the divisor.
After finishing a section here, attempt the matching Rules of Divisibility practice test on TrickySSC. The tests are graded across three levels, and every question carries a step-by-step solution.
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