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Prelims GS-II (CSAT) · Numeracy · Arithmetic

Number system

The number system is a foundational CSAT topic covering integers, divisibility, prime factorisation, factors and multiples, remainders, decimal representation and digit patterns. Success depends less on lengthy calculation than on identifying numerical structure and applying the correct rule. These ideas also support questions on fractions, ratios, percentages, time and work, and data interpretation.

The International Standard Book Number (ISBN) is a unique number in the world that is assigned to each format of book published.
ISBN For Book

The International Standard Book Number (ISBN) is a unique number in the world that is assigned to each format of book published. ISBN For Book

Credit: RatulAsian · CC BY-SA 4.0 · source

Animation that visualizes the "Sieve of Eratosthenes" algorithm.
The Sieve of Eratosthenes is an method for efficiently finding all prime numbers up to a number, 120 in this case, by eliminating (colo

Animation that visualizes the "Sieve of Eratosthenes" algorithm. The Sieve of Eratosthenes is an method for efficiently finding all prime numbers up to a number, 120 in this case, by eliminating (colo

Credit: SKopp at German Wikipedia · CC BY-SA 3.0 · source

1. Types of numbers and their properties

Natural numbers are the counting numbers 1, 2, 3 and so on; whole numbers additionally include 0. Some mathematical conventions include 0 among natural numbers, so follow any definition supplied in the question. Integers include negative whole numbers, zero and positive whole numbers. Rational numbers can be expressed as p/q, where p and q are integers and q is non-zero. Irrational numbers, such as √2 and π, cannot be expressed in this form. Rational and irrational numbers together constitute the real numbers.

A rational number has a decimal expansion that either terminates or eventually repeats. An irrational number has a non-terminating, non-repeating decimal expansion. After a rational fraction is reduced to its lowest terms, its decimal terminates exactly when the denominator contains no prime factors other than 2 and 5. Thus, 7/40 terminates, whereas 7/30 repeats. The reduction condition is essential: 3/6 terminates because it simplifies to 1/2.

An integer is even if divisible by 2 and odd otherwise; zero is even. Even plus even and odd plus odd are even, while even plus odd is odd. A product is odd only when every integer factor is odd. These parity rules can eliminate options without finding exact values. However, do not transfer integer properties to arbitrary fractions or decimals.

  • Zero is neither positive nor negative. Division by zero is undefined.
  • The sum or product of two rational numbers is rational.
  • Two irrational numbers may have a rational sum or product: √2 + (−√2) = 0 and √2 × √2 = 2.
  • The square of every integer is non-negative; the principal square root symbol denotes the non-negative root.

2. Divisibility, primes and prime factorisation

A prime number is a positive integer greater than 1 with exactly two positive factors: 1 and itself. A composite number has more than two positive factors. The number 1 is neither prime nor composite; 2 is the only even prime. To test whether an integer n greater than 1 is prime, check divisibility only by primes not exceeding √n. Any composite number must have at least one prime factor within that limit.

Divisibility tests reduce calculation. A number is divisible by 2 if its units digit is even, by 5 if it ends in 0 or 5, and by 10 if it ends in 0. Divisibility by 3 or 9 depends on the digit sum. For 4 and 8, examine the last two and last three digits respectively. For 11, the difference between the sums of digits in alternating positions must be zero or a multiple of 11.

Prime factorisation converts many questions into comparisons of exponents. For example, 360 = 2³ × 3² × 5. A perfect square has even exponents for all prime factors; a perfect cube has exponents divisible by 3. Therefore, the smallest positive integer by which 360 must be multiplied to obtain a perfect square is 10: this makes the exponents of both 2 and 5 even.

  • To test divisibility by 6, check both 2 and 3; for 12, check both 3 and 4.
  • Divisibility by a and b guarantees divisibility by ab only when a and b are coprime.
  • Coprime numbers have HCF 1 but need not be prime: 8 and 15 are coprime.

Number-system problem-solving sequence

  1. 1. Identify the number domain and all restrictions.
  2. 2. Classify the task: divisibility, factors, remainder, digits or powers.
  3. 3. Apply a structural rule before attempting long calculation.
  4. 4. Compute using factorisation, cycles or option elimination.
  5. 5. Verify all conditions and select the answer.

3. Factors, HCF and LCM

The highest common factor, also called the greatest common divisor, is the largest positive integer dividing each of the given integers. The least common multiple is their smallest common positive multiple. In prime factorisation, the HCF takes the minimum exponent of each prime across the numbers, treating an absent prime as having exponent zero. The LCM takes the maximum exponent. For 72 = 2³ × 3² and 120 = 2³ × 3 × 5, the HCF is 24 and the LCM is 360.

Use HCF when dividing quantities into the largest equal-sized units without any remainder. Use LCM when finding the earliest recurrence of events with fixed integer periods, assuming they begin together. For two positive integers, their product equals the product of their HCF and LCM. This identity does not generally extend to three or more integers, a frequent trap in objective questions.

If n = pᵃqᵇrᶜ for distinct primes, its number of positive factors is (a + 1)(b + 1)(c + 1), because each prime exponent can be independently selected from zero to its maximum. Thus, 72 has 4 × 3 = 12 positive factors. A positive integer has an odd number of positive factors exactly when it is a perfect square, since all other factors pair with distinct complementary factors.

  • Euclid’s algorithm uses repeated division: HCF(252, 105) = HCF(105, 42) = HCF(42, 21) = 21.
  • The smallest positive integer divisible by each number in a given set is their LCM.
  • Read whether the question asks for positive factors, prime factors or distinct prime factors; these counts differ.
Recognising the appropriate number-system tool
Question patternMain toolImportant condition
Largest equal-sized groupingHCFNo quantity should leave a remainder
First simultaneous recurrenceLCMEvents start together and have fixed periods
Number of positive divisorsProduct of exponent-plus-one termsBegin with prime factorisation
Terminating decimalInspect the denominatorReduce the fraction to lowest terms first
Units digit of a large powerRepeating units-digit cycleA zero exponent remainder selects the cycle’s last entry
Trailing zeros of n!Count factors of 5Include contributions from 25, 125 and higher powers

4. Remainders, cyclicity and factorials

The division algorithm writes an integer N as N = dq + r, where d is a positive divisor, q is an integer and 0 ≤ r < d. Numbers with the same remainder on division by d are congruent modulo d. In sums and products, large numbers can be replaced by their remainders before calculation. For example, 47 × 53 leaves the same remainder modulo 5 as 2 × 3, namely 1.

Powers often follow repeating remainder cycles. The units digits of powers of 7 repeat as 7, 9, 3, 1. Since 103 leaves remainder 3 when divided by 4, the units digit of 7¹⁰³ is 3. If the exponent is divisible by the cycle length, use the last member of the cycle, not a nonexistent zeroth member.

A factorial n! is the product of positive integers from 1 to n, with 0! defined as 1. Trailing zeros come from factors of 10, each formed by pairing 2 with 5. In n!, factors of 5 are scarcer, so count them using ⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ and further terms until zero. Consequently, 100! has 20 + 4 = 24 trailing zeros.

  • If several numbers leave the same remainder on division by d, d divides their pairwise differences.
  • For negative integers, retain a non-negative standard remainder: −7 = 5 × (−2) + 3.
  • Do not cancel a common factor in a congruence while retaining the modulus unless that factor is coprime to the modulus.

5. Digit problems and an efficient CSAT approach

Place value translates digit puzzles into algebra. A two-digit number with tens digit a and units digit b equals 10a + b, where a ranges from 1 to 9 and b from 0 to 9. Its reversed value is 10b + a. Their difference is 9(a − b), necessarily divisible by 9. Similarly, a three-digit number abc equals 100a + 10b + c; digit symbols are not multiplied together.

Before calculating, identify the domain, constraints and required quantity. Check whether zero, negative integers, repeated digits or leading zeros are allowed. Apply parity, divisibility and size bounds to eliminate impossible options. Use substitution from options when it is faster than solving an equation, but verify every condition. A candidate answer satisfying only one of several remainder conditions is insufficient.

  • Write short working notes: divisor, permitted remainder, digit restrictions and whether the least positive value is required.
  • Estimate magnitude before exact calculation to detect misplaced zeros or incorrect powers.
  • Under negative marking, prioritise questions with a clear method; avoid spending disproportionate time on an unfamiliar pattern.

Real-world case studies

ISBN-13: remainders used in error detection

ISBN-13 book identifiers use alternating weights of 1 and 3 on the first twelve digits. The final check digit makes the weighted total divisible by 10. For ISBN 9780306406157, the weighted sum of the first twelve digits is 93; adding the check digit 7 produces 100. This illustrates modular arithmetic in practice. Such a checksum detects many transcription errors but does not establish a book’s authenticity.

Gregorian leap years: divisibility with exceptions

In the Gregorian calendar, a year divisible by 4 is generally a leap year, but a century year must also be divisible by 400. Thus, 1900 was not a leap year, whereas 2000 was. A 400-year cycle contains 100 − 4 + 1 = 97 leap years. This is a practical example of combining divisibility conditions rather than applying a single rule mechanically.

Previous year questions

No source-linked PYQ is available here yet. Use the practice questions below.

Practice questions

Practice MCQ 1 · Easy

Which one of the following fractions has a terminating decimal expansion?

  • A. 13/60
  • B. 21/84
  • C. 17/45
  • D. 19/70

Practice MCQ 2 · Easy

The four-digit number 52x4, where x is a digit, is divisible by 12. What is the value of x?

  • A. 2
  • B. 3
  • C. 4
  • D. 7

Practice MCQ 3 · Easy

How many positive factors does 540 have?

  • A. 12
  • B. 18
  • C. 20
  • D. 24

Practice MCQ 4 · Easy

What is the remainder when 7^43 is divided by 10?

  • A. 3
  • B. 7
  • C. 9
  • D. 1

Practice MCQ 5 · Easy

What is the smallest positive integer by which 432 must be multiplied to obtain a perfect square?

  • A. 2
  • B. 3
  • C. 6
  • D. 12

Practice MCQ 6 · Moderate

The HCF and LCM of two positive integers are 12 and 1260, respectively. What is the smallest possible sum of the two integers?

  • A. 252
  • B. 276
  • C. 288
  • D. 264

Practice MCQ 7 · Moderate

What is the greatest positive integer that divides 187, 331 and 475 leaving the same remainder in each case?

  • A. 48
  • B. 72
  • C. 144
  • D. 288

Practice MCQ 8 · Moderate

For how many integers n satisfying 1 ≤ n ≤ 180 does n/180 have a terminating decimal expansion?

  • A. 20
  • B. 24
  • C. 30
  • D. 36

Practice MCQ 9 · Moderate

How many integers from 1 to 300, inclusive, are divisible by at least one of 6 and 10, but are not divisible by 15?

  • A. 50
  • B. 60
  • C. 70
  • D. 80

Practice MCQ 10 · Moderate

How many trailing zeros are there in the decimal representation of the integer 100!/(40! × 60!)? Here, n! denotes the product of the positive integers from 1 to n.

  • A. 0
  • B. 2
  • C. 3
  • D. 1

Practice MCQ 11 · Difficult

A perfect square N greater than 10000 leaves remainders 1, 4 and 9 when divided by 8, 9 and 25, respectively. What is the least possible value of N?

  • A. 10609
  • B. 38809
  • C. 124609
  • D. 162409

Practice MCQ 12 · Difficult

Let N be the smallest positive integer such that N/12 is the square of a positive integer and N/18 is the cube of a positive integer. How many positive factors of N are divisible by 6 but not by 36?

  • A. 6
  • B. 10
  • C. 12
  • D. 8

Practice MCQ 13 · Difficult

How many four-digit positive integers can be formed using four distinct digits from the set {0, 2, 4, 6, 8} such that the resulting number is divisible by 72?

  • A. 6
  • B. 8
  • C. 10
  • D. 12

Practice MCQ 14 · Difficult

How many positive integers not exceeding 1000 leave remainder 2 when divided by 7 and remainder 3 when divided by 5, and are divisible by neither 3 nor 4?

  • A. 12
  • B. 13
  • C. 14
  • D. 16

Practice MCQ 15 · Difficult

How many unordered pairs of positive integers have HCF 18 and LCM 22680, with both integers lying between 100 and 2000, inclusive? An unordered pair is counted only once, irrespective of the order of its members.

  • A. 3
  • B. 4
  • C. 5
  • D. 6

Further reading

  • UPSC: Civil Services Examination notification, examination scheme and Preliminary Examination Paper II syllabus, upsc.gov.in.
  • UPSC: Previous Question Papers, Civil Services Preliminary Examination, General Studies Paper II, upsc.gov.in.
  • NCERT Mathematics, Class VI, Playing with Numbers.
  • NCERT Mathematics, Class IX, Number Systems.
  • NCERT Mathematics, Class X, Real Numbers.
  • International ISBN Agency: ISBN Users’ Manual, isbn-international.org.

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