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.
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?
- 3
- 7
- 1
- 5
Q2 · medium · AI-verified
How many numbers between 1 and 200 are divisible by both 4 and 6?
- 8
- 25
- 16
- 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?
- 84
- 120
- 96
- 105
Q4 · hard · AI-verified
If the 7-digit number 4567x75 is divisible by 9, what is the value of x?
- 6
- 8
- 2
- 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?
- 45
- 27
- 18
- 36