Why this topic matters · 8 min read
Number System and Divisibility is a foundational topic in CSAT Paper 2, appearing almost every year either directly (divisibility rules, remainders, LCM/HCF) or embedded in word problems. Typically 2-4 questions per paper. CSAT is qualifying (33% cutoff) but many aspirants fail it — so mastering this topic gives a safe buffer. Questions range from easy rule-based to tricky remainder theorem problems.
Types of Numbers
Numbers are classified in layers. Natural numbers start from 1. Whole numbers include 0. Integers include negatives. Rational numbers can be written as p/q (q not zero). Irrational numbers cannot (like root 2, pi). Real numbers are all of these combined. For UPSC CSAT, you mainly deal with integers and natural numbers.
- Natural numbers: 1, 2, 3, ... (no zero, no negatives)
- Whole numbers: 0, 1, 2, 3, ... (zero included)
- Integers: ...-2, -1, 0, 1, 2, ... (negatives included)
- Prime numbers: exactly 2 factors — 1 and itself. Example: 2, 3, 5, 7, 11, 13
- 1 is neither prime nor composite — this is a classic trap
- 2 is the only even prime number
Divisibility Rules (The Core Shortcuts)
Divisibility rules let you check if a number divides another without actual division. In CSAT, these save 30-60 seconds per question. Memorise rules for 2 through 11 — beyond that, just divide. The rule for 7 is ugly; skip it and just divide directly.
- Div by 2: last digit is even (0,2,4,6,8)
- Div by 3: sum of digits divisible by 3. Example: 123 — sum=6, yes
- Div by 4: last two digits divisible by 4. Example: 1732 — 32/4=8, yes
- Div by 5: last digit is 0 or 5
- Div by 6: divisible by BOTH 2 and 3
- Div by 8: last three digits divisible by 8
- Div by 9: sum of digits divisible by 9. Example: 729 — sum=18, yes
- Div by 10: last digit is 0
- Div by 11: (sum of digits at odd positions) minus (sum at even positions) = 0 or multiple of 11
LCM and HCF
HCF (Highest Common Factor) is the largest number that divides two or more numbers. LCM (Lowest Common Multiple) is the smallest number that is a multiple of two or more numbers. Think of HCF as the biggest shared piece and LCM as the smallest common meeting point. A key relationship ties them together: for any two numbers a and b, HCF x LCM = a x b. This formula is your fastest tool when one of the four values is unknown.
- HCF divides, LCM is divisible — never mix these up
- HCF is always less than or equal to the smaller number
- LCM is always greater than or equal to the larger number
- For fractions: HCF of fractions = HCF of numerators / LCM of denominators
- For fractions: LCM of fractions = LCM of numerators / HCF of denominators
- Word problem signal for LCM: bells ringing together, tiles fitting, events repeating
Key formulas
HCF x LCM relation
HCF(a,b) x LCM(a,b) = a x b
When: Use when 3 of the 4 values are given and one must be found
HCF by division
Divide larger by smaller, then divide divisor by remainder repeatedly until remainder = 0. Last divisor = HCF
When: Use for large numbers where prime factorisation is slow
Worked examples
HCF(12,18): 18 = 12x1 + 6, then 12 = 6x2 + 0. HCF = 6. LCM = (12x18)/6 = 36
Three bells ring at 6, 8, 10 min intervals. They ring together at LCM(6,8,10) = 120 min = after 2 hours
Remainders and the Remainder Theorem
Remainder problems are the trickiest CSAT number system questions. The key idea: when a number N is divided by d, N = d x quotient + remainder. The remainder is always less than d. For large powers (like 7 raised to 99 divided by 5), find the cyclicity of remainders — the pattern repeats every few steps. This avoids impossible calculations.
- Remainder is always non-negative and less than the divisor
- If N gives remainder r when divided by d, then N-r is exactly divisible by d
- Cyclicity trick: find remainders of base^1, base^2, base^3 ... until it repeats, then use the cycle to find any large power
- Divisibility by 9 or 3 is same as remainder of digit sum when divided by 9 or 3
- If two numbers give same remainder when divided by d, their difference is divisible by d
Key formulas
Basic division identity
N = Divisor x Quotient + Remainder, where 0 <= Remainder < Divisor
When: Always — it is the definition of remainder
Cyclicity of powers
Find pattern: a^1 mod d, a^2 mod d, a^3 mod d ... identify cycle length C, then a^n mod d = a^(n mod C) mod d
When: When asked: what is the remainder when a large power is divided by some number
Worked examples
What is remainder when 7^100 is divided by 5? Cycle: 7^1=7(rem 2), 7^2=49(rem 4), 7^3(rem 3), 7^4(rem 1), repeats every 4. 100 mod 4 = 0, so use 7^4 — remainder is 1.
A number leaves remainder 3 when divided by 7. What is remainder when double that number is divided by 7? Double remainder = 6. Answer: 6
Factors, Multiples and Perfect Numbers
A factor of N divides N exactly. Total number of factors is found from prime factorisation. If N = p^a x q^b x r^c, then total factors = (a+1)(b+1)(c+1). This formula appears directly in CSAT. Sum of factors and product of factors formulas are less common but good to know. Perfect numbers (like 6, 28) equal the sum of their proper factors — rarely tested but mentioned in statements.
- Number of factors of N = (a+1)(b+1)(c+1) if N = p^a x q^b x r^c
- To find prime factorisation: divide by 2, then 3, then 5, and so on
- A perfect square always has an odd number of total factors
- Number of even factors: reduce power of 2 by 1, then apply formula
- Co-prime numbers: HCF = 1. Example: 8 and 9 are co-prime
Key formulas
Total number of factors
If N = p^a x q^b x r^c, then Factors(N) = (a+1)(b+1)(c+1)
When: Direct CSAT questions: how many factors does 360 have, or find numbers with exactly N factors
Worked example
How many factors does 360 have? 360 = 2^3 x 3^2 x 5^1. Factors = (3+1)(2+1)(1+1) = 4x3x2 = 24
⚠ Common mistakes to avoid
- Treating 1 as a prime number — 1 is neither prime nor composite, always remember this
- Confusing HCF and LCM rules for fractions — many aspirants invert the formula. Memorise: HCF of fractions uses LCM in denominator
- In remainder problems, assuming remainder can equal the divisor — remainder is always strictly less than divisor
- Forgetting that LCM x HCF = a x b works ONLY for two numbers, not three or more
- In cyclicity problems, when the exponent is exactly divisible by cycle length, use the last value in the cycle (not the first)
🧠 Memory aids
- Divisibility by 6 = 2 AND 3 together — think SIX = TWO x THREE, both parents must approve
- HCF is the Humble Common Factor (smaller, modest), LCM is the Lofty Common Multiple (big, ambitious)
- For fractions: HCF goes cross, LCM goes straight — HCF numerator/LCM denominator, LCM numerator/HCF denominator
- Remainder cycle trick acronym: FIND-SPOT — Find the cycle, Spot where exponent lands
- Factor formula memory: N = p^a q^b r^c gives (a+1)(b+1)(c+1) — add one to each power, then multiply
🎯 UPSC CSE exam tips
- CSAT 2023 and 2022 had remainder and LCM-based word problems — these appear more often than pure theory questions, so focus on applications
- Divisibility rules for 11 are tested in statement-based MCQs like: which of the following is divisible by 11 — practice the alternating digit sum rule until it is automatic
- LCM bell-ringing and tiling problems are the most predictable word problems — solve them in under 60 seconds using direct LCM calculation
- For power-remainder questions, always write out 4-5 terms of the cycle first before computing — aspirants who skip this step make errors
- CSAT qualifying cutoff is only 33 percent (about 27-28 marks out of 80). Number system questions are among the most solvable — never leave these blank even with guessing after elimination
Q1 · medium · AI-verified
How many numbers between 1 and 200 (inclusive) are divisible by 3 but not by 9?
- 44
- 22
- 45
- 66
Q2 · medium · AI-verified
How many three-digit numbers are divisible by both 4 and 6?
- 48
- 75
- 25
- 50
Q3 · medium · AI-verified
What is the value of the expression: (1/1×2) + (1/2×3) + (1/3×4) + ... + (1/99×100)?
- 99/100
- 100/101
- 1/100
- 1/99
Q4 · medium · AI-verified
What is the largest 4-digit number that is exactly divisible by 88?
- 9900
- 9976
- 9856
- 9944
Q5 · medium · AI-verified
The product of two co-prime numbers is 117. What is their LCM?
- 1
- 117
- 9
- 13