SSC CGL Divisibility Rules Practice Test 2026 — Free Chapter Wise Questions

Rules for 2 to 19, missing digit questions, composite divisors and divisibility of algebraic expressions — 50 questions per test, with full solutions and shortcut tricks in English or हिंदी.

2 Difficulty Levels 50 Questions per Test 30 min optional timer English & हिंदी

Rule of Divisibility Tests — Pick a Level and Start

Tap a level to see every test in it. Start at Level 1 the day you finish the rules, then move to Level 2 with the timer running to push accuracy under pressure.

Level 1 Basic Checking… View tests

Direct use of a single rule — is the number divisible by 4, 6, 8, 9 or 11, find the missing digit, and the smallest number to add or subtract. Attempt these without the clock while the rules are still fresh.

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Level 2 Tier I Standard Checking… View tests

Real SSC CGL Tier I difficulty — rules for 7, 13, 17 and 19, composite divisors split into co-prime factors, two unknown digits at once, and the first algebraic divisibility questions. This is the score that tells you the chapter is exam-ready.

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The rule of divisibility is the smallest chapter in the SSC CGL Quantitative Aptitude syllabus by direct weight and one of the most valuable by indirect weight. A shift may carry only one question that names divisibility outright — usually a missing digit or a smallest number to add — but the same rules are what let you cancel a fraction on sight, spot a factor without dividing, and shorten an LCM, remainder or simplification question that would otherwise eat a minute.

These practice tests put fifty divisibility questions in front of you in one sitting, which is what turns the rules from something you can recite into something you apply without thinking. Every question comes with a full solution and the shortcut, in English or Hindi.

How Many Divisibility Questions Are Asked in SSC CGL?

Divisibility rarely gets a question to itself. It sits underneath the whole arithmetic block, and that block together is worth four to six marks in a typical Tier I shift.

ChapterTier I (out of 25)Tier II Paper I (out of 30)
Rule of Divisibility0–10–1
Number System2–33–4
LCM & HCF1–21–2
Remainder Theorem0–11
Simplification1–22–3

Typical spread across recent SSC CGL shifts. SSC does not publish a chapter-wise breakup, so treat these as planning ranges. Tier I maths is always 25 questions in a locked 15-minute section.

What You Will Practice in the Rule of Divisibility Chapter

Core topics

  • Divisibility rules for 2, 3, 4, 5, 6, 8, 9, 10 and 11
  • Rules for 7, 13, 17 and 19 by the multiply-and-adjust method
  • Divisibility by 16, 25 and 125 from the last few digits
  • Composite divisors such as 12, 15, 18, 24, 36, 45, 72 and 88, split into co-prime factors
  • Missing digit questions — one unknown digit, and two unknown digits with two conditions
  • The smallest number that must be added to or subtracted from a number to make it divisible
  • Divisibility of an − bn and an + bn
  • Divisibility of expressions and of products of consecutive integers
  • The largest number that always divides a given expression
  • Using divisibility to test whether a given option can be the answer

Divisibility rules you must know cold

DivisorRule
2Last digit is even
3Sum of the digits is divisible by 3
4Last two digits form a number divisible by 4
5Last digit is 0 or 5
6Divisible by both 2 and 3
8Last three digits form a number divisible by 8
9Sum of the digits is divisible by 9
10Last digit is 0
11Difference between the sum of digits in odd and even places is 0 or a multiple of 11
16Last four digits form a number divisible by 16
25Last two digits are 00, 25, 50 or 75
125Last three digits form a number divisible by 125

The awkward ones — 7, 13, 17 and 19

Each of these has the same shape. Chop off the last digit, do one small operation with it, and repeat on the shorter number until you can judge it by sight.

DivisorWhat to do with the last digitExample
7Double it and subtract from the rest8232 → 823 − 4 = 819 → 81 − 18 = 63 ✔
13Multiply by 4 and add to the rest2197 → 219 + 28 = 247 → 24 + 28 = 52 ✔
17Multiply by 5 and subtract from the rest221 → 22 − 5 = 17 ✔
19Double it and add to the rest209 → 20 + 18 = 38 ✔

In the exam you will almost never need the rule for 7 more than twice on the same number. If it is still large after two rounds, the intended method was probably something else.

Composite divisors — split into co-prime factors

DivisorTest these twoDo not use
123 and 42 and 6
153 and 5
182 and 93 and 6
243 and 84 and 6
364 and 96 and 6
455 and 93 and 15
888 and 114 and 22

The two factors must be co-prime, meaning their HCF is 1. This is the single most common trap in the chapter: 18 is divisible by 2 and by 6, yet it is not divisible by 12.

Algebraic divisibility — the four results

ExpressionDivisible byWhen
an − bna − bFor every n
an − bna + bOnly when n is even
an + bna + bOnly when n is odd
an + bna − bNever

Also worth memorising

  • The product of any n consecutive integers is divisible by n! — so n(n+1)(n+2) is always divisible by 6
  • n³ − n is always divisible by 6, and n⁵ − n by 30
  • A number is divisible by 99 if the sum of its digits taken in pairs from the right is divisible by 99
  • A number is divisible by 999 if the sum of its digits taken in threes from the right is divisible by 999
  • Smallest number to add = divisor − remainder; smallest number to subtract = the remainder itself

Solved Examples — SSC CGL Rule of Divisibility

Q1. What digit should replace x so that 34x5 is divisible by 9?
The digit sum must be a multiple of 9. Here it is 3 + 4 + x + 5 = 12 + x, and the next multiple of 9 after 12 is 18, so x = 6. Answer: 6
Q2. What digit should replace * in 7*2589 to make it divisible by 11?
Odd places give 7 + 2 + 8 = 17 and even places give * + 5 + 9 = * + 14. The difference is 17 − (* + 14) = 3 − *, which must be 0 or a multiple of 11, so * = 3. Answer: 3
Q3. Is 8232 divisible by 7?
Drop the last digit, double it, subtract: 823 − 4 = 819. Repeat: 81 − 18 = 63, and 63 = 7 × 9. Answer: Yes, 8232 = 7 × 1176
Q4. Show that 540 − 1 is divisible by 24.
Write the power in a useful form: 540 = (5²)20 = 2520. Since an − bn is always divisible by a − b, the expression 2520 − 120 is divisible by 25 − 1. Answer: divisible by 24
Q5. What is the smallest number that must be added to 1000 to make it divisible by 45?
1000 ÷ 45 leaves quotient 22 and remainder 10, because 45 × 22 = 990. The number to add is 45 − 10 = 35, which takes it to 1035 = 45 × 23. Answer: 35

How the Divisibility Chapter Tests Work

  • 50 questions per test, with a full question palette — answered, skipped, marked for review, not visited.
  • Optional 30-minute timer — with the clock to build exam speed, without it while you are still learning the rules.
  • Marking: +2 correct, −0.5 wrong, 0 unattempted — the same scheme as the real paper.
  • Solutions, shortcut tricks and concept notes open after you submit, question by question.
  • English and हिंदी, chosen at the start of every test.
  • Free with a TrickySSC account — sign in once with Google or your mobile number, and every test is open to you.

How to Prepare the Rule of Divisibility for SSC CGL

  1. Learn the rules for 2 to 11 first and test them on phone numbers. They cost nothing to practise and they are the ones that appear inside other chapters.
  2. Treat 7, 13, 17 and 19 as one method, not four rules. All four chop the last digit and adjust; only the multiplier and the sign change. Learning them as a family takes a fraction of the time.
  3. Always split a composite divisor into co-prime factors. Most wrong answers in this chapter come from testing 12 as 2 and 6, or 24 as 4 and 6.
  4. Do Level 1 untimed, the same day you finish the rules. The aim is correct recall, not speed.
  5. Move to Level 2 with the 30-minute clock on. A divisibility question that takes more than thirty seconds means you are still dividing instead of testing.
  6. Study it as a block. Do number system, LCM and HCF and remainder theorem right after — they reuse these rules constantly.
  7. Verify against a real paper. Once Level 2 is consistent, attempt a previous year paper and then a full mock test under locked sectional timing.

SSC CGL विभाज्यता के नियम (Rule of Divisibility) चैप्टर टेस्ट — हिंदी में

विभाज्यता के नियम SSC CGL गणित का सबसे छोटा अध्याय है, लेकिन इसका उपयोग सबसे ज़्यादा होता है। सीधे तौर पर हर शिफ्ट में 0–1 प्रश्न आता है — जैसे लुप्त अंक ज्ञात करना या सबसे छोटी संख्या जोड़ना — पर यही नियम संख्या पद्धति, लघुत्तम-महत्तम, शेषफल और सरलीकरण के प्रश्नों में बार-बार काम आते हैं।

TrickySSC पर इस अध्याय के सभी टेस्ट हिंदी में उपलब्ध हैं — प्रश्न, विकल्प, विस्तृत हल और शॉर्टकट ट्रिक सब हिंदी में। दो स्तर हैं: लेवल 1 (आधारभूत) और लेवल 2 (टियर I स्तर)। प्रत्येक टेस्ट में 50 प्रश्न, वैकल्पिक 30 मिनट का टाइमर और +2 / −0.5 की वही मार्किंग जो असली परीक्षा में होती है।

ध्यान रखने योग्य बात: 12, 24, 36 जैसी भाज्य संख्याओं के लिए हमेशा सह-अभाज्य गुणनखंडों से जाँच करें — 12 के लिए 3 और 4, 24 के लिए 3 और 8। 12 को 2 और 6 से जाँचना ग़लत है, क्योंकि 18 दोनों से विभाजित होती है फिर भी 12 से नहीं।

SSC CGL Rule of Divisibility — Frequently Asked Questions

What is the divisibility rule of 7?
Remove the last digit, double it, and subtract it from the number that is left. Repeat until the number is small enough to judge. If the final result is 0 or a multiple of 7, the original number is divisible by 7. For example, 8232 gives 823 − 4 = 819, then 81 − 18 = 63, and 63 is a multiple of 7.
What is the divisibility rule of 11?
Add the digits in the odd places and the digits in the even places separately, then take the difference. If the difference is 0 or a multiple of 11, the number is divisible by 11. This is the rule SSC uses most often in missing digit questions.
What is the divisibility rule of 13?
Remove the last digit, multiply it by 4, and add it to the number that is left. Repeat until the number is small. For 2197 this gives 219 + 28 = 247, then 24 + 28 = 52, and 52 is 13 × 4, so 2197 is divisible by 13.
How do I check divisibility by a composite number like 12 or 36?
Break the divisor into two co-prime factors and test both. For 12 use 3 and 4, for 36 use 4 and 9, for 45 use 5 and 9. The factors must be co-prime. Testing 12 as 2 and 6 is wrong, because 18 passes both of those and is still not divisible by 12.
How many divisibility questions are asked in SSC CGL?
Zero to one question comes directly from this chapter in a Tier I shift, usually a missing digit or a smallest number to add. Its real value is indirect — divisibility is the tool used inside number system, LCM and HCF, remainder theorem and simplification, which together are worth four to six marks.
Is an − bn always divisible by a − b?
Yes, for every n. It is also divisible by a + b when n is even. Separately, an + bn is divisible by a + b only when n is odd, and it is never divisible by a − b. These three results settle most of the algebraic divisibility questions SSC asks.
How many questions are in each divisibility practice test?
Fifty questions per test, with an optional 30-minute timer and the real exam marking scheme of +2 for a correct answer, −0.5 for a wrong one and 0 for an unattempted question.
Are the divisibility tests available in Hindi and are they free?
Yes to both. Questions, options, detailed solutions and shortcut tricks are available in हिंदी, and you pick English or Hindi before the test starts. Every test is free — you only need a free TrickySSC account, and the sign-in box opens on the page itself when you start a test.

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