Number System and Arithmetic Foundations

Before a single percentage, ratio or interest problem makes sense, an aspirant must be fluent in how numbers are classified, how they divide, and how their remainders behave. The number system is the bedrock of every quantitative section in judiciary preliminary papers, the CLAT-PG aptitude component and allied competitive screenings. This chapter rebuilds that foundation from first principles — natural numbers to irrationals, the Fundamental Theorem of Arithmetic, divisibility shortcuts, HCF and LCM, the remainder theorem, unit-digit cyclicity and factorials — with the exam-grade shortcuts that turn a two-minute calculation into a ten-second observation.

Why the Number System Anchors Every Aptitude Section

Quantitative aptitude in judicial-service and CLAT-PG screening is rarely about heavy calculation; it rewards the candidate who recognises structure. A question that looks like long division collapses to a one-line answer once you spot a divisibility pattern, and a problem on the last digit of a colossal power needs no power at all once you know its cycle. The number system supplies this structural literacy. It is the grammar beneath every other topic on the hub — you cannot reason about a ratio or proportion without understanding factors, you cannot simplify a profit or loss fraction without HCF, and you cannot track a recurring schedule in time and work without LCM.

The examiner exploits this dependency deliberately. A great many "hard" questions in these papers are number-system questions wearing a costume — a word problem about distributing sweets is an HCF question, a problem about bells tolling together is an LCM question, and a problem about the number of trailing zeroes in a factorial is a prime-counting question. This chapter therefore aims not merely to list definitions but to build the recognition reflex that lets you classify a problem on sight and reach for the right tool. For a full map of the topics that build on this base, see the aptitude and reasoning hub.

Classifying Numbers: From Naturals to Reals

The real number system is a nested hierarchy, and precision about which set a number belongs to prevents most conceptual errors. Natural numbers are the counting numbers 1, 2, 3, … . Adding zero gives the whole numbers 0, 1, 2, 3, … . Including the negatives produces the integers …, −2, −1, 0, 1, 2, … . A rational number is any number expressible as a quotient p/q of two integers with q ≠ 0; its decimal expansion either terminates or eventually repeats. An irrational number — such as √2, √3, π or e — cannot be so expressed and has a non-terminating, non-repeating decimal expansion. Rationals and irrationals together form the real numbers.

Two distinctions trip up candidates most often. First, every integer is rational (an integer n equals n/1), but not every rational is an integer. Second, a terminating decimal like 0.75 is rational because it equals 3/4, and a purely recurring decimal like 0.333… is rational because it equals 1/3. The test of rationality is not whether the decimal "ends" but whether it terminates or repeats; only a pattern that never settles — like the digits of π — signals irrationality. A reliable exam heuristic: the square root of a non-perfect-square positive integer is always irrational, so √4 is rational (it is 2) while √5 is not.

A further refinement that examiners probe is the test for when a rational number has a terminating rather than a recurring decimal. A fraction in its lowest terms terminates if and only if the denominator's prime factorisation contains no primes other than 2 and 5. Thus 7/8 = 7/2³ terminates (giving 0.875) while 7/6 = 7/(2×3) recurs, because the factor of 3 in the denominator cannot be cleared into a power of ten. This single criterion lets you predict the decimal behaviour of any fraction without performing the division, and it is itself a direct consequence of the prime factorisation principle developed in the next sections. Aspirants should also note the closure properties tested in objective questions: the sum or product of two rationals is always rational, but the sum of a rational and an irrational is always irrational, while the sum of two irrationals may be either — √2 + (−√2) = 0 is rational, yet √2 + √3 is not.

Even, Odd, Prime and Composite Numbers

An even number is divisible by 2; an odd number is not. The parity rules are exam staples: even ± even = even, odd ± odd = even, even ± odd = odd; even × anything = even, while odd × odd = odd. A prime number is a natural number greater than 1 with exactly two distinct positive divisors, 1 and itself. A composite number is a natural number greater than 1 with more than two divisors. The number 1 is neither prime nor composite — it has exactly one divisor — and this is not an arbitrary convention but a requirement for the uniqueness of prime factorisation discussed below.

Candidates should commit the primes up to 100 to memory; there are exactly 25 of them. The only even prime is 2, which makes every other prime odd. A frequent shortcut for testing whether a number n is prime is to check divisibility only by primes up to √n: if no prime at or below the square root divides n, then n is prime. Thus to test 113 one checks 2, 3, 5, 7 (since √113 ≈ 10.6) and, finding none divides it, concludes 113 is prime. Two numbers are co-prime (relatively prime) when their only common factor is 1 — for instance 8 and 15 — even though neither need be prime itself.

The Fundamental Theorem of Arithmetic

The Fundamental Theorem of Arithmetic, also called the unique factorisation theorem, states that every integer greater than 1 is either prime or can be expressed as a product of primes, and that this representation is unique up to the order of the factors. Carl Friedrich Gauss gave the first complete proof of both existence and uniqueness in his 1801 Disquisitiones Arithmeticae, where he named it the fundamental theorem. The insistence that the factors be prime is essential: factorisations allowing composite factors are not unique, which is precisely why 1 is excluded from the primes — admitting it would let any factorisation be padded with extra 1s and destroy uniqueness.

For exam purposes the theorem is the engine behind almost every divisor problem. Writing a number in its canonical (standard) form N = p₁^a × p₂^b × p₃^c … lets you read off three quantities instantly. The number of factors of N is (a+1)(b+1)(c+1)… . The sum of factors is the product of the geometric series for each prime. And the count of trailing zeroes, perfect-square divisors, or co-prime pairs all flow from the same exponents. For example, 360 = 2³ × 3² × 5¹ has (3+1)(2+1)(1+1) = 24 factors. This single decomposition replaces brute-force listing and is the reason the theorem deserves close mastery.

Divisibility Rules That Save Minutes

Divisibility tests let you reject options or simplify fractions without dividing. A number is divisible by 2 if its last digit is even; by 4 if its last two digits form a number divisible by 4; by 8 if its last three digits do. It is divisible by 5 if it ends in 0 or 5, and by 10 if it ends in 0. The digit-sum tests cover 3 (digit sum divisible by 3) and 9 (digit sum divisible by 9). For 6, the number must pass both the 2 and 3 tests.

The harder primes have elegant tests. For 11, take the difference between the sum of the digits in odd positions and the sum in even positions; the number is divisible by 11 if that difference is 0 or a multiple of 11. For 7, double the last digit and subtract it from the rest; repeat until you reach a number whose divisibility by 7 is obvious. For 13, multiply the last digit by 4 and add it to the rest; if the result is a multiple of 13, so is the original. A celebrated consequence ties these three together: 7 × 11 × 13 = 1001, so writing any three-digit number twice in succession (forming a six-digit number such as 482482) yields a value divisible by 7, 11 and 13, and dividing it by all three returns the original block. Examiners love this identity because it looks mysterious and resolves in one line.

Two further patterns round out the toolkit. A number is divisible by a composite such as 12 only if it is divisible by co-prime factors whose product is 12 — here 3 and 4, not 2 and 6, because 2 and 6 share a factor and would under-test. The general rule is that to test divisibility by a composite, split it into co-prime components and apply each test; divisibility by 36 means passing both the 4 and 9 tests, since 4 and 9 are co-prime and multiply to 36. The second pattern concerns successive subtraction: the divisibility rule for 7 can be iterated indefinitely, so to test 1,073 you compute 107 − (2×3) = 101, then 10 − (2×1) = 8, which is not a multiple of 7, so 1,073 is not divisible by 7. These rules are not mere curiosities; in a timed paper they let you eliminate three of four options before the slower candidate has finished setting up a long division.

HCF and LCM: Two Sides of Factorisation

The Highest Common Factor (HCF, also GCD) of two or more numbers is the largest number that divides each of them; the Lowest Common Multiple (LCM) is the smallest number that each of them divides. Both are read directly off prime factorisations: the HCF takes the lowest power of each common prime, while the LCM takes the highest power of every prime appearing in any number. For 12 = 2²×3 and 18 = 2×3², the HCF is 2×3 = 6 and the LCM is 2²×3² = 36.

The single most useful identity for two numbers is that HCF × LCM = product of the two numbers. With 12 and 18, that is 6 × 36 = 216 = 12 × 18. This lets you find the LCM the instant you have the HCF, or recover a missing number. A critical caveat the examiner exploits: this product relationship holds only for two numbers, not for three or more. A second standard result is that the HCF of co-prime numbers is 1, so their LCM equals their product — the basis of many "least number" puzzles. HCF answers questions of the form "largest size that divides evenly" (cutting cloth, grouping students), while LCM answers "smallest quantity reachable by all" (bells tolling together, gears meshing) — the same reasoning that drives synchronised-cycle problems in time and work.

A standard family of HCF questions asks for the largest number that divides several given numbers leaving the same remainder in each case, or different specified remainders. When the remainders are equal but unknown, the trick is that the required divisor divides the differences of the numbers, so you take the HCF of the pairwise differences. When each number leaves a stated remainder, you first subtract that remainder from each number and then take the HCF of the results. The mirror-image LCM question asks for the smallest number that, when divided by several divisors, leaves the same remainder in each case — the answer is the LCM of the divisors plus the common remainder. Recognising which of these templates a word problem fits is far more valuable than any single calculation, because the arithmetic itself is trivial once the structure is named. The Euclidean algorithm — repeatedly replacing the larger number by its remainder on division by the smaller until one becomes zero — gives the HCF of any pair efficiently and underlies the relationship above.

HCF and LCM of Fractions and Decimals

Many candidates freeze when an HCF or LCM problem involves fractions, yet the rules are clean. The HCF of fractions equals the HCF of the numerators divided by the LCM of the denominators. The LCM of fractions equals the LCM of the numerators divided by the HCF of the denominators. So the LCM of 2/3, 4/9 and 6/5 is LCM(2,4,6) / HCF(3,9,5) = 12 / 1 = 12, while their HCF is HCF(2,4,6) / LCM(3,9,5) = 2 / 45.

Before applying either rule, the fractions should be in lowest terms, which itself depends on dividing numerator and denominator by their HCF. For decimals, convert to like decimal places, treat the digits as integers to take the HCF or LCM, then reinsert the decimal point. These conversions sit at the boundary of number theory and the fraction-handling skills you will reuse throughout percentage, ratio and proportion, where reducing a ratio to its simplest form is nothing more than dividing through by an HCF.

The Remainder Theorem and Modular Thinking

The division algorithm states that for any integers a (dividend) and d (divisor, d > 0) there exist unique integers q (quotient) and r (remainder) with a = dq + r and 0 ≤ r < d. Almost every remainder shortcut is an application of this single identity together with the fact that remainders combine: the remainder of a product equals the product of the remainders, and the remainder of a sum equals the sum of the remainders, each taken modulo the divisor. This is the heart of modular arithmetic, and it lets you replace huge numbers by their small remainders before computing.

For example, to find the remainder when 17 × 23 is divided by 5, replace 17 by 2 and 23 by 3, multiply to get 6, and reduce to a remainder of 1 — no multiplication of the full numbers needed. Fermat-style and cyclicity arguments extend this to powers: remainders of successive powers of a fixed base under a fixed modulus repeat in a cycle, so finding the position within the cycle answers the question. The negative-remainder trick is especially powerful: since 100 leaves remainder −1 modulo 101, 100^n leaves remainder (−1)^n, which is 1 for even n and 100 for odd n. Mastering these reductions turns intimidating power-and-remainder questions into arithmetic on single digits.

Three named results recur often enough to memorise. The Chinese Remainder Theorem guarantees that if you know a number's remainders with respect to several pairwise co-prime divisors, the number is uniquely determined modulo their product — the principle behind "find the smallest number that leaves remainder 2 on division by 3 and 3 on division by 5" puzzles. Fermat's Little Theorem states that if p is prime and does not divide a, then a^(p−1) leaves remainder 1 on division by p; this instantly collapses large exponents, so 3^100 modulo 7 reduces because 3⁶ ≡ 1, leaving 3^(100 mod 6) = 3⁴ ≡ 4. Finally, the remainder when a number is divided by 9 equals the remainder of its digit sum divided by 9, which is why the digit-sum divisibility test for 9 works and why "casting out nines" can verify multiplication. Each of these is a labour-saving device, not an abstraction to be feared.

Unit-Digit Determination and Cyclicity

Finding the unit (last) digit of a large power is a classic question that needs zero heavy computation once cyclicity is understood. The unit digit of a^n depends only on the unit digit of a and on n, and the unit digits of successive powers repeat in a short cycle. Four digits are fixed points — 0, 1, 5 and 6 raised to any positive power keep the same unit digit (so 6^100 ends in 6). The digits 4 and 9 have a cycle of length 2: powers of 4 end in 4, 6, 4, 6… and powers of 9 end in 9, 1, 9, 1… . The remaining digits 2, 3, 7 and 8 have a cycle of length 4.

The working method is to divide the exponent by the cycle length and use the remainder to locate the position. Take 7^123: powers of 7 cycle through 7, 9, 3, 1 with period 4, and 123 leaves remainder 3 on division by 4, so the unit digit is the third in the cycle, namely 3. A subtlety worth remembering: when the remainder is 0, you take the last element of the cycle, not the first. This dovetails with the remainder cyclicity discussed above and is one of the highest-yield, lowest-effort techniques in the entire syllabus.

Factorials and Counting Trailing Zeroes

The factorial n! is the product of all positive integers up to n, with the convention that 0! = 1. A recurring exam question asks for the number of trailing zeroes in n!, which is simply the number of times 10 divides it. Since 10 = 2 × 5 and factorials always contain more factors of 2 than of 5, the count of trailing zeroes equals the number of times 5 appears in the prime factorisation of n!.

That count is given by Legendre's formula: add the integer parts of n/5, n/25, n/125, and so on, until the divisor exceeds n. For 100!, this is ⌊100/5⌋ + ⌊100/25⌋ = 20 + 4 = 24 trailing zeroes. The same logic finds the highest power of any prime dividing a factorial — just sum the floor terms for that prime. This single formula answers a family of questions that look formidable ("how many zeroes does 250! end in?") in two lines, and it is a direct corollary of the Fundamental Theorem of Arithmetic applied to a product of consecutive integers.

A common variation asks for the highest power of a composite number dividing a factorial — say the highest power of 12 in 50!. Here you decompose 12 = 2² × 3, count the available 2s and 3s separately with Legendre's formula, then take whichever supply runs out first after accounting for the required multiplicity. The factor of 2 is abundant and the factor of 3 (or in this case the pair-up of 2² with each 3) becomes the binding constraint, so the answer is governed by the scarcer resource. The governing idea throughout is that a factorial is a structured product whose prime content is entirely predictable, so any question about what divides it, how many zeroes it ends in, or how many factors it has reduces to counting exponents rather than multiplying out an astronomically large number.

Number of Factors, Their Sum and Product

Once a number is in canonical form N = p₁^a × p₂^b × …, three formulae follow directly. The number of factors is (a+1)(b+1)…, counting each independent choice of exponent. The sum of all factors is the product over each prime of (p^(exp+1) − 1)/(p − 1), the closed form of the geometric series of its powers. The product of all factors is N^(d/2), where d is the number of factors.

These let you answer compound questions in one stroke. The number of even factors of N is found by reserving at least one factor of 2; the number of odd factors is the factor count of N with all 2s removed. The number of ways to write N as a product of two factors is d/2 when d is even, and (d+1)/2 when N is a perfect square (because the square root pairs with itself). For 36 = 2² × 3² there are (2+1)(2+1) = 9 factors, an odd count confirming 36 is a perfect square, and it can be written as a product of two factors in (9+1)/2 = 5 ways. Recognising which formula a question is really asking for is the whole battle.

Common Pitfalls and Exam Strategy

Several recurring errors cost candidates easy marks. Treating 1 as prime corrupts factorisation problems. Forgetting that 2 is prime — and the only even prime — derails parity reasoning. Applying the HCF × LCM = product identity to three numbers gives wrong answers, because it holds only for two. In unit-digit cyclicity, taking the first element instead of the last when the remainder is 0 is a frequent slip. And in trailing-zero problems, counting factors of 2 instead of 5 over-counts the zeroes.

Strategically, the number system rewards pattern recognition over calculation. Before reaching for long division, ask whether a divisibility rule, a remainder reduction or a cyclicity pattern collapses the problem. Memorise the primes to 100, the cyclicity table for unit digits, and the divisibility tests for 7, 11 and 13. Keep canonical factorisation as your default first move on any factor, HCF, LCM or zero-counting question. These same instincts — reduce before you compute, factorise before you multiply — carry directly into the interest growth patterns of simple and compound interest and the structured reading of data interpretation. The number system is not a topic to be revised once; it is the operating system on which the rest of the quantitative paper runs.