Sarkari RiseLogin

Number Systems Questions for SSC CGL

Free, AI-curated practice for the Number Systems section of SSC CGL. We have 33+ verified questions in this bank. Below: 5 sample questions. Sign up free to unlock unlimited practice + AI explanations + per-topic analytics.

▶ Start free — SSC CGL mockAll SSC CGL resourcesAlready a user? Sign in →
Why this topic matters · 8 min read
Number Systems is one of the most heavily tested topics in SSC CGL Quant, appearing in both Tier 1 and Tier 2. Expect 3-5 direct questions per paper on divisibility, remainders, LCM/HCF, unit digits, and number properties. The questions are not very hard but are designed to waste time if you don't know the shortcuts. Mastering this topic is a guaranteed score booster.

Types of Numbers

Numbers are classified into Natural (1, 2, 3...), Whole (0, 1, 2, 3...), Integers (...-2, -1, 0, 1, 2...), Rational (p/q form where q is not zero), Irrational (non-terminating, non-repeating like root 2, pi), Real (all of the above). Prime numbers have exactly 2 factors: 1 and themselves. Composite numbers have more than 2 factors. Note: 1 is neither prime nor composite. 2 is the only even prime.

  • 1 is neither prime nor composite — this is a favourite trick question
  • 2 is the only even prime number
  • Every prime number greater than 3 is of the form 6n+1 or 6n-1
  • 0 is a whole number but NOT a natural number
  • Rational numbers include all terminating and repeating decimals
  • Irrational numbers: root 2, root 3, pi, e — cannot be expressed as p/q

Divisibility Rules

These rules let you check divisibility without actual division — critical for speed in SSC CGL. Know rules for 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 cold. The rule for 11 is a favourite: alternating sum of digits. For 7, double the last digit, subtract from remaining number, repeat.

  • Div by 2: last digit even
  • Div by 3: sum of digits divisible by 3
  • Div by 4: last two digits divisible by 4
  • Div by 8: last three digits divisible by 8
  • Div by 9: sum of digits divisible by 9
  • Div by 11: (sum of odd-position digits) minus (sum of even-position digits) = 0 or multiple of 11
Key formulas
Divisibility by 11
(D1 + D3 + D5 ...) - (D2 + D4 + D6 ...) = 0 or multiple of 11
When: Checking if a large number is divisible by 11 — very common in SSC CGL
Worked examples

Is 29,detector: 2 + 9 = 11, so 29 divisible by? No. Try 29348: (2+3+8)-(9+4) = 13-13 = 0, so yes divisible by 11.

Is 7896 divisible by 8? Last 3 digits = 896. 896 divided by 8 = 112. Yes, divisible.

LCM and HCF

HCF (Highest Common Factor) is the largest number that divides all given numbers. LCM (Lowest Common Multiple) is the smallest number divisible by all given numbers. A key relationship: HCF x LCM = Product of two numbers (only for TWO numbers, not three). SSC CGL loves word problems involving bells ringing, tiles, and ropes.

  • HCF x LCM = Product of the two numbers (valid only for exactly 2 numbers)
  • HCF of fractions = HCF of numerators / LCM of denominators
  • LCM of fractions = LCM of numerators / HCF of denominators
  • If HCF of two numbers is H, both numbers are multiples of H
  • For numbers in ratio a:b, the numbers can be written as aH and bH where H is HCF
Key formulas
HCF x LCM relation
HCF(a,b) x LCM(a,b) = a x b
When: When one of HCF, LCM, or one number is missing for a pair of numbers
HCF of fractions
HCF = HCF of numerators / LCM of denominators
When: When finding HCF of fractional values
LCM of fractions
LCM = LCM of numerators / HCF of denominators
When: When finding LCM of fractional values
Worked examples

HCF of 12 and 18 is 6. LCM = (12 x 18) / 6 = 36. Check: 36 is divisible by both 12 and 18. Correct.

Two numbers in ratio 3:4 have HCF 6. Numbers are 18 and 24. LCM = 18x24/6 = 72.

Remainders and Cyclicity

Remainder questions are very common in SSC CGL. The key trick is that remainder of a product equals product of individual remainders (mod concept without the name). For powers, find the pattern (cycle) of remainders — it usually repeats within 4 steps. Unit digit questions are a special case of remainder when divided by 10.

  • Remainder(a x b) / n = [Remainder(a/n) x Remainder(b/n)] / n
  • Unit digit of powers follows a cycle of at most 4
  • Unit digit cycle of 2: 2,4,8,6 — repeats every 4 powers
  • Unit digit cycle of 3: 3,9,7,1 — repeats every 4 powers
  • Numbers ending in 0,1,5,6 always end in 0,1,5,6 regardless of power
  • Numbers ending in 4: cycle is 4,6 (even power = 6, odd power = 4)
Key formulas
Remainder of product
Rem[(a x b) / n] = Rem[Rem(a/n) x Rem(b/n)] / n
When: Finding remainder of large products or powers without full calculation
Unit digit via cyclicity
Find (exponent mod 4): if 0 use 4th power, if 1 use 1st power, etc.
When: Finding unit digit of expressions like 7^95 or 3^47
Worked examples

Unit digit of 7^95: 95 mod 4 = 3. Cycle of 7 is 7,9,3,1. 3rd position = 3. Unit digit is 3.

Remainder when 2^10 is divided by 7: 2^1=2, 2^2=4, 2^3=1 (mod 7), cycle length 3. 10 mod 3 = 1. Remainder = 2.

Factors and Number of Factors

If a number N = p^a x q^b x r^c (prime factorisation), then total number of factors = (a+1)(b+1)(c+1). Sum of factors and product of factors can also be found from prime factorisation. SSC CGL tests this in the form: how many factors does 360 have, or find numbers with exactly 5 factors.

  • Number of factors of N = p^a x q^b x r^c is (a+1)(b+1)(c+1)
  • A perfect square has an odd number of factors
  • A number has exactly 2 factors only if it is prime
  • A number has exactly 3 factors only if it is square of a prime
  • Sum of all factors of p^a = (p^(a+1) - 1) / (p - 1)
Key formulas
Total factors
If N = p^a x q^b x r^c, then Total Factors = (a+1)(b+1)(c+1)
When: Counting total divisors of a number — direct SSC question type
Worked examples

How many factors does 360 have? 360 = 2^3 x 3^2 x 5^1. Factors = (3+1)(2+1)(1+1) = 4x3x2 = 24.

Which number less than 50 has exactly 3 factors? Must be square of prime: 4(2^2), 9(3^2), 25(5^2), 49(7^2). All qualify.

⚠ Common mistakes to avoid
  • Using HCF x LCM = product formula for THREE numbers — it only works for exactly two numbers. For three numbers there is no such direct formula.
  • Forgetting that 1 is neither prime nor composite — questions often ask to count primes up to a limit and aspirants include or exclude 1 incorrectly.
  • In unit digit cyclicity, dividing exponent by 4 and using remainder 0 as the 0th power. When remainder is 0, always use the 4th position in the cycle, not the 0th.
  • Confusing HCF of fractions with LCM of fractions — the formula flips the denominator operation. Mnemonic: HCF goes small so it divides by the bigger (LCM) in denominator.
  • In divisibility by 11, starting the count of odd and even positions from the wrong end — always start from the rightmost digit as position 1 (odd position).
🧠 Memory aids
  • HOLE for fraction LCM/HCF: H-over-L, L-over-H. HCF of fractions = H on top over L on bottom. LCM of fractions = L on top over H on bottom.
  • Unit digit cycles — LAST DIGIT GANG: 0,1,5,6 are stubborn (never change). 4 and 9 have 2-step cycles. 2,3,7,8 have 4-step cycles.
  • For prime check of number N: test divisibility only by primes up to root of N. Example for 97: root 97 is about 9.8, so only test 2,3,5,7.
  • 6n plus or minus 1 rule: All primes above 3 fit 6n+1 or 6n-1. Use to quickly eliminate non-primes: if a number is NOT of this form, it is composite.
🎯 SSC CGL exam tips
  • SSC CGL Tier 1 typically has 2-3 questions on LCM/HCF word problems (bells, pipes, tiles), 1 on unit digits, and 1 on remainders or divisibility. Tier 2 goes deeper with remainder theorems and factor-based questions.
  • Unit digit questions appear almost every year and can be solved in under 30 seconds with cyclicity — never waste time computing actual powers.
  • Remainder questions in recent papers (2023-2024) have used the pattern: find remainder when a large power of a small number is divided by a single digit. Always find the cycle first.
  • LCM/HCF word problems follow 3 standard templates: (1) when do two events coincide again — take LCM, (2) largest tile/rope/plank that fits — take HCF, (3) minimum quantity that fills containers exactly — take LCM.
  • Spend no more than 60-75 seconds per Number Systems question in Tier 1. If a remainder question looks long, use the cycle method — if you find yourself doing long division for large numbers, you are doing it wrong.

Sample questions

Q1 · medium · AI-verified
When a number is divided by 56, the remainder is 29. What will be the remainder when the same number is divided by 8?
  1. 3
  2. 7
  3. 1
  4. 5
Q2 · medium · AI-verified
How many numbers between 1 and 200 are divisible by both 4 and 6?
  1. 8
  2. 25
  3. 16
  4. 33
Q3 · hard · AI-verified
The ratio of two numbers is 3:4 and their LCM is 180. What is the sum of the two numbers?
  1. 84
  2. 120
  3. 96
  4. 105
Q4 · hard · AI-verified
If the 7-digit number 4567x75 is divisible by 9, what is the value of x?
  1. 6
  2. 8
  3. 2
  4. 4
Q5 · medium · AI-verified
The sum of the digits of a two-digit number is 9. If the digits are reversed, the new number is 27 more than the original number. What is the original number?
  1. 45
  2. 27
  3. 18
  4. 36
💡 Want answers + explanations + 28+ more Number Systems questions? Sign up free →
⭐ Recommended for SSC CGL aspirants

Pro 6-month

all your target exams · 6 months · unlimited mocks + AI
₹799~₹4.4/day
Sign up free, then unlockSee all plans →

More SSC CGL topics

Current Affairs
203+ practice questions
Geography
130+ practice questions
One Word Substitution
114+ practice questions
History
105+ practice questions
Cloze Test
103+ practice questions
Antonyms
101+ practice questions

Free practice, AI explanations, 24 exams — all in one app

Daily 10-Q quiz · AI doubt solver in Hindi + English · adaptive mocks · 4,150+ verified PYQs.

Sign up freePricingTry Daily 10-Q