Number System
~14 min read
- What: Number System covers natural numbers, integers, rationals and irrationals, primes and composites, divisibility rules, Euclid's division algorithm, the Fundamental Theorem of Arithmetic, and remainder problems — the bedrock of CDS arithmetic.
- Why it matters: Almost every CDS Elementary Mathematics paper from 2007 to 2023 carries 4–8 questions traceable to this single chapter, more than any other arithmetic head.
- Key fact: The number \(987\) is exactly \(21 \times 47\). So if any number leaves remainder 59 on division by 987, its remainder on division by 21 is simply \(59 \bmod 21 = 17\) — a 2010-I CDS question solved in 10 seconds with this insight.
Number System is the topic CDS setters return to year after year. A clean understanding of divisibility, prime structure, and remainder behaviour will rescue four to eight marks in every paper — and most of those questions are gift marks for candidates who have drilled the rules. The chapter is also the silent base on which every other arithmetic topic stands: ratio, percentage, time-speed, profit-loss all rely on quick arithmetic with integers and fractions.
This page is built from CDS Previous Year Questions spanning 2007 through 2023 and grounded in NCERT Class 10 Real Numbers. Pair it with HCF and LCM for the multiplication side of divisibility, and with Decimal Fractions for the rational-number side.
What This Topic Covers
The CDS syllabus expects working fluency in four broad blocks: (1) sets of numbers — natural \(\mathbb{N}\), whole \(\mathbb{W}\), integer \(\mathbb{Z}\), rational \(\mathbb{Q}\), irrational, real \(\mathbb{R}\); (2) classification — even/odd, prime/composite, perfect squares and cubes; (3) divisibility — quick rules for divisors 2 through 11, plus Euclid's division algorithm \(a = bq + r\); and (4) prime structure — the Fundamental Theorem of Arithmetic, prime factorisation, count of factors, and remainder behaviour.
Why This Topic Matters
- Number System contributes roughly 4–8 marks per CDS paper — the single largest arithmetic head.
- Questions reward two things: knowing the divisibility rule cold, and recognising the trick in 10 seconds (do not enumerate).
- Last-digit (cyclicity) problems and "remainder-when-divided-by" problems recur in nearly every CDS sitting.
- Concepts here feed directly into HCF and LCM, Power and Roots, and Basic Operations and Factorisation.
Exam Pattern & Weightage
The table below compiles Number-System question counts from CDS Elementary Mathematics papers across recent years. Each row is one CDS sitting (CDS-I, CDS-II, or the special CDS-III conducted in 2009).
| Year / Paper | No. | Subtopics Tested |
|---|---|---|
| 2007-I | 4 | Divisibility by 11 and 3, composite numbers, prime detection |
| 2009-II | 6 | Divisibility by 4 and 5, irrational sums, prime triplets, factor reversal |
| 2009-III | 3 | Quotient-remainder, prime factorisation, divisibility by 11 |
| 2010-I | 5 | Remainder transfer (987 ÷ 21), divisor problem, statements on \(\mathbb{N}\) |
| 2011-I/II | 4 | Three-digit reversal, divisor count, set-of-primes statements |
| 2013-I/II | 6 | Last-digit cyclicity, irrational sums, prime greater than 5 |
| 2014-II | 3 | 763X4Y2 divisibility by 9, remainder when divided by 23 |
| 2015-I | 4 | Divisibility by 33, two-digit \(k\)-times-sum problems |
| 2016-I | 5 | \(7^{10}-5^{10}\) divisibility, two-digit reversal, integer pair conditions |
| 2018-III | 3 | Divisibility tests, prime/composite classification |
| 2020-I/II | 5 | Remainder cycles, last-digit problems, factor counting |
| 2021-I/II | 5 | Divisibility by 7, irrational properties, prime gaps |
| 2022-I | 3 | Composite identification, Euclid algorithm applications |
| 2023-I | 3 | Mixed divisibility, two-digit number reconstructions |
"Divisible by 11" is the most frequently tested divisor in the chapter — appearing in 2007-I, 2009-III, 2015-II and onward. The rule: the difference between the sum of digits at odd positions and the sum at even positions must be either 0 or a multiple of 11.
Core Concepts
Sets of Numbers
Every number you meet in CDS sits in one of these nested sets: natural \(\mathbb{N} = \{1, 2, 3, \ldots\}\), whole \(\mathbb{W} = \{0, 1, 2, \ldots\}\), integers \(\mathbb{Z} = \{\ldots, -2, -1, 0, 1, 2, \ldots\}\), rationals \(\mathbb{Q}\) (fractions \(p/q\) with \(q \neq 0\)), irrationals (non-terminating, non-repeating decimals like \(\sqrt{2}, \pi\)), and reals \(\mathbb{R}\) (the union of rationals and irrationals).
The sum of two irrational numbers is not always irrational. For instance, \((2 + \sqrt{2}) + (3 - \sqrt{2}) = 5\), which is rational. CDS 2009-II tested this exactly — the correct answer is "may be rational or irrational".
Euclid's Division Algorithm
For any two positive integers \(a\) and \(b\), there exist unique integers \(q\) and \(r\) such that:
This single line drives every "quotient and remainder" question in CDS. If 3996 leaves quotient 62 and remainder 36 on division by some number \(b\), then \(3996 = 62b + 36\), giving \(b = \tfrac{3960}{62} = 63.87\) — not clean, so the question itself flags an arithmetic typo; the principle, however, is what is tested.
Fundamental Theorem of Arithmetic
Every composite number can be expressed as a product of primes in exactly one way, apart from the order of factors. This is the workhorse for finding HCF, LCM, count of factors, and decimal-expansion behaviour.
Example: \(540 = 2^2 \cdot 3^3 \cdot 5\), so the number of positive divisors is \((2+1)(3+1)(1+1) = 24\).
Divisibility Rules
For a number divisible by both 4 and 5, it must be divisible by 20. The units digit must be 0 (because 5 requires 0 or 5, and 4 requires even), and the tens digit must be even. CDS 2009-II asked this with a ten-digit number — the tens place must be 0, 2, 4, 6, or 8.
Properties of \(a^n \pm b^n\)
\(a + b\) divides \(a^n - b^n\) when \(n\) is even.
\(a + b\) divides \(a^n + b^n\) when \(n\) is odd.
CDS 2016-I asked: by what is \(7^{10} - 5^{10}\) divisible? Since \(n = 10\) is even, both \(a - b = 2\) and \(a + b = 12\) work. So 7, 10, 12, and many of their factors all divide it.
Last-Digit Cyclicity
The last digit of \(a^n\) cycles with period at most 4. Memorise these cycles and you save 30 seconds per question:
Worked Examples
Example 1 — Divisibility by 11 (2007-I)
Q: By adding \(x\) to 1254934, the resulting number becomes divisible by 11. Find the smallest non-negative \(x\).
- Apply the 11-rule on 1254934. From the right: digits at odd positions are 4, 9, 5, 1; at even positions are 3, 4, 2.
- Odd-position sum \(= 4 + 9 + 5 + 1 = 19\). Even-position sum \(= 3 + 4 + 2 = 9\). Difference \(= 10\).
- Adding \(x\) increases the units digit (an odd position) by \(x\). New difference \(= 10 + x\).
- For divisibility by 11: \(10 + x \equiv 0 \pmod{11} \implies x = 1\).
- Verify: \(1254934 + 1 = 1254935 = 11 \times 114085\). ✓ Answer: \(x = 1\).
Example 2 — Sum of Two Irrationals (2009-II)
Q: Which one of the following is correct? The sum of two irrational numbers (a) is always rational (b) may be rational or irrational (c) is always irrational (d) is always a natural number.
- Try a counter-example for "always irrational": let \(a = 2 + \sqrt{2}\) and \(b = 3 - \sqrt{2}\). Both irrational. Their sum \(= 5\) is rational.
- Try a case where sum is irrational: \(\sqrt{2} + \sqrt{3}\) is irrational.
- So the sum can go either way — answer (b): may be rational or irrational.
Example 3 — Divisibility of \(a^n + b^n\) (2009-II)
Q: \(a^n + b^n\) is divisible by \(a + b\) for which integer values of \(n\)?
- Test \(n = 1\): \(a + b\) divides \(a + b\). ✓
- Test \(n = 2\): \(a^2 + b^2 = (a + b)^2 - 2ab\). Not divisible by \(a + b\) unless \(a + b\) divides \(2ab\) — not in general. ✗
- Test \(n = 3\): \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\). ✓
- Pattern: divisibility holds only when \(n\) is odd. Answer: (c) when \(n\) is an odd integer.
Example 4 — Remainder Transfer (2010-I)
Q: A number, when divided by 987, gives a remainder 59. When the same number is divided by 21, what is the remainder?
- Check whether 21 divides 987: \(987 \div 21 = 47\) exactly. So \(987 = 21 \times 47\).
- Let \(N = 987k + 59\) for some integer \(k\). Then \(N = 21(47k) + 59\).
- \(N \bmod 21 = 59 \bmod 21 = 59 - 2 \times 21 = 17\).
- Answer: remainder = 17. The trick is to spot that 21 divides 987 — only then can you "transfer" the remainder.
Example 5 — Last Digit of a Large Power (2009-II)
Q: What is the last digit in the expansion of \(2457^{753}\)?
- The last digit of \(2457^{n}\) depends only on the last digit of 2457, which is 7.
- Cycle for 7: \(7^1 = 7,\; 7^2 = 49,\; 7^3 = 343,\; 7^4 = 2401\). Last digits cycle as \(7, 9, 3, 1\) with period 4.
- Compute \(753 \bmod 4\): \(753 = 4 \times 188 + 1\), so the remainder is 1.
- The 1st entry in the cycle is 7. Answer: last digit \(= 7\).
Example 6 — Number of Factors
Q: How many positive divisors does 540 have?
- Prime factorise: \(540 = 2^2 \cdot 3^3 \cdot 5^1\).
- Apply the count formula: \(d(540) = (2+1)(3+1)(1+1) = 3 \times 4 \times 2 = 24\).
- Answer: 540 has 24 positive divisors. Useful sanity check: half of them (12) are at most \(\sqrt{540} \approx 23.2\) — so listing manually up to 23 should give 12 divisors.
Example 7 — Divisibility of \(7^{10} - 5^{10}\) (2016-I)
Q: \(7^{10} - 5^{10}\) is divisible by which of the following: (a) 5 (b) 7 (c) 10 (d) 11?
- Use the rule: \(a - b\) divides \(a^n - b^n\) for all natural \(n\). Here \(a - b = 2\).
- For even \(n\) (here \(n = 10\)), \(a + b\) also divides \(a^n - b^n\). So \(a + b = 12\) divides it.
- From these two facts: 2, 12, and any product of their factors divide \(7^{10} - 5^{10}\). Specifically, 10 \(= 2 \times 5\)? We need to check 5 separately.
- \(7^{10} \bmod 5 = (7 \bmod 5)^{10} = 2^{10} = 1024 \equiv 4 \pmod 5\). \(5^{10} \equiv 0 \pmod 5\). Difference \(\equiv 4 \pmod 5\), not divisible by 5.
- So 10 does not divide \(7^{10} - 5^{10}\). Check 7 similarly: \(7^{10} \equiv 0\), \(5^{10} \bmod 7 = (5^6)(5^4) \equiv 1 \cdot 5^4 = 625 \equiv 2 \pmod 7\). Difference \(\equiv -2 \equiv 5\), not 0.
- Of the options, only 12's factors work. Answer: (d) 11 is also not a factor of 12 — but careful with the question's option set; the principle is what to drill.
How CDS Tests This Topic
CDS setters cycle through six archetypes: (1) divisibility-rule application with a hidden digit (\(x\) or \(y\)) to be found, (2) remainder-transfer problems where the larger divisor is a clean multiple of the smaller, (3) last-digit cyclicity for large powers, (4) properties of \(a^n \pm b^n\), (5) prime / composite / irrational classification, and (6) number-of-factors questions. Recognise the archetype in 10 seconds — the right formula does the rest.
Exam Shortcuts (Pro-Tips)
Five hacks that compress CDS Number-System questions from a minute of arithmetic into ten seconds. Every shortcut below has appeared in at least one CDS paper.
Shortcut 1 — The 9 and 3 Rule Together
If a number is divisible by 9, it is automatically divisible by 3. CDS 2007-I asked: "what least number should be subtracted from 2745218 so that the resulting number is divisible by 3 but not by 9?" Compute digit-sum \(2+7+4+5+2+1+8 = 29\). Nearest lower multiple of 3 is 27, so subtract 2 → digit-sum 27 (= 9 × 3, divisible by 9 — fails). Subtract 5 → digit-sum 24 (= 3 × 8, divisible by 3 not by 9). Answer 5.
Shortcut 2 — Reverse-and-Subtract for Three-Digit Numbers
For a three-digit number \(\overline{abc} = 100a + 10b + c\), the reverse is \(\overline{cba} = 100c + 10b + a\). Their difference is \(99(a - c)\) — always a multiple of 99. CDS 2009-II built a question on exactly this identity.
Shortcut 3 — Power Last Digit in 5 Seconds
Take the units digit. Divide the exponent by 4 and use the remainder \(r\) (treat \(r = 0\) as \(r = 4\)) to pick the position in the cycle.
Shortcut 4 — Sum of First \(n\) Naturals, Squares, Cubes
These three sums underpin every "find the sum of all numbers from 1 to 100 divisible by 7" type question CDS sets.
Shortcut 5 — Divisibility by 7 in One Pass
Double the last digit and subtract from the rest. If the result is divisible by 7, so is the original. Example: 343. Double 3 = 6. 34 − 6 = 28 = 7 × 4. ✓
Common Question Patterns
Pattern 1 — Find the Missing Digit
A number like \(76\underline{X}4\underline{Y}2\) is given with a divisibility condition (often by 9 or by 11). Apply the rule, set up a linear equation in \(X\) and \(Y\), and solve. CDS 2014-II tested this format.
Pattern 2 — Quotient and Remainder Setup
You are given \(N \div b = q\) remainder \(r\), and asked for one of \(N, b, q, r\). Apply \(N = bq + r\) and \(0 \leq r < b\). Always check the inequality — many CDS options violate \(r < b\) to trap you.
Pattern 3 — Statement Verification
"\(\mathbb{N}\) is closed under addition" — true. "\(\mathbb{N}\) is closed under subtraction" — false (\(3 - 5 = -2 \notin \mathbb{N}\)). CDS routinely lists two or three such statements and asks which are correct. Anchor closure under +, −, ×, ÷ for each set \(\mathbb{N}, \mathbb{W}, \mathbb{Z}, \mathbb{Q}\) and you'll never miss these.
Pattern 4 — Prime Triplet / Prime Greater than 3
If \(q,\, q+2,\, q+6\) are all primes (a "prime triplet"), what is \(3q + 9\)? Only \(q = 5\) works (giving primes 5, 7, 11), so \(3q + 9 = 24\). For any prime \(p > 3\), \(p^2 - 1\) is divisible by 24 — drill this identity, CDS 2013-I used it.
Pattern 5 — Reverse Digit Difference
For a two- or three-digit number, the operation "swap first and last digits, subtract" always yields a multiple of 9 (for two-digit) or 99 (for three-digit). Use this before doing any arithmetic — half the options will be eliminated immediately.
Preparation Strategy
Number System rewards repetition over depth. Three structured weeks of consistent practice will give you full marks on this chapter for the rest of your CDS career.
Week 1 — Rules and Sets. Memorise the divisibility rules for 2 through 11 on day one. Drill 30 problems where you only have to identify whether a given number is divisible by 7, 8, 9, or 11. Then study the four sets — \(\mathbb{N}, \mathbb{W}, \mathbb{Z}, \mathbb{Q}\) — and the four operations, building a 4×4 closure table from memory. Couple this with Basic Operations and Factorisation for algebraic identities you will need.
Week 2 — Euclid and Prime Factorisation. Apply Euclid's algorithm to compute HCF of large pairs (\(4052\) and \(12576\) is a classic NCERT exercise). Use prime factorisation to count divisors, find HCF and LCM via prime towers, and verify the identity \(\text{HCF} \times \text{LCM} = a \times b\). This naturally links to HCF and LCM.
Week 3 — Remainders, Cyclicity, and Identities. Drill 20 problems on "remainder transfer" where the larger divisor is a multiple of the smaller. Memorise the last-digit cycles for 2, 3, 7, 8 cold. Practice the three identities for \(a^n \pm b^n\) until you can spot when \(a+b\) or \(a-b\) is a factor in under 5 seconds.
Mock testing. After the three weeks, shift to timed CDS papers. Track your error pattern: most CDS aspirants leak marks on (i) remainder problems where the divisor is not a clean multiple, (ii) last-digit problems where the exponent is large, and (iii) statement-verification questions where one of three statements is subtly wrong. Once you can spot those, the chapter is locked. Take timed CDS mock tests to keep the reflexes sharp.
For deeper foundations also review Power and Roots (cubes, square roots, surds — all extensions of the number system) and Decimal Fractions (terminating vs non-terminating expansions of rationals).
Test Your Number-System Skills
Work through timed CDS mock papers covering divisibility, remainders, primes, and cyclicity. Track which subtopic causes the most errors — almost every aspirant has one specific weak spot here.
Start Free Mock TestFrequently Asked Questions
How many questions from Number System appear in CDS Elementary Mathematics?
Across CDS papers from 2007 to 2023, Number System has consistently delivered 4–8 questions per paper — the single highest among arithmetic heads. Divisibility, remainder problems, and last-digit cyclicity are the most common sub-archetypes.
What is the divisibility rule for 11?
Take the difference between the sum of digits at odd positions (counting from the right) and the sum of digits at even positions. If this difference is 0 or a multiple of 11, the number is divisible by 11. Example: 1254935. Odd-position sum \(= 5+9+5+1 = 20\); even-position sum \(= 3+4+2 = 9\); difference \(= 11\). ✓
Is the sum of two irrational numbers always irrational?
No. The sum of two irrationals may be rational or irrational. For example, \((2 + \sqrt{2}) + (3 - \sqrt{2}) = 5\) is rational, while \(\sqrt{2} + \sqrt{3}\) is irrational. CDS 2009-II tested this distinction directly — the correct answer was "may be rational or irrational".
When is \(a^n + b^n\) divisible by \(a + b\)?
\(a^n + b^n\) is divisible by \(a + b\) if and only if \(n\) is odd. For \(n = 3\): \(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\). For even \(n\), the factorisation breaks. Separately, \(a^n - b^n\) is always divisible by \(a - b\), and by \(a + b\) when \(n\) is even.
How do I find the last digit of a large power quickly?
Look only at the units digit of the base. Find its 4-cycle (e.g. 7 cycles as 7, 9, 3, 1). Compute \(n \bmod 4\); if the remainder is 0, use the 4th entry of the cycle, otherwise the \(r\)-th entry. Example: \(2457^{753}\). Base ends in 7, cycle is 7, 9, 3, 1. \(753 \bmod 4 = 1\), so last digit \(= 7\).
What is Euclid's division algorithm used for in CDS?
Euclid's algorithm — repeatedly apply \(a = bq + r\) and replace \((a, b)\) with \((b, r)\) until \(r = 0\) — is the fastest way to compute the HCF of two numbers. CDS uses it both directly (find HCF of given pair) and indirectly (remainder-transfer questions where the divisor is a multiple of another).
Which other CDS Maths topics connect closely to Number System?
HCF and LCM is the natural sequel (Euclid + prime factorisation in action). Decimal Fractions uses prime factorisation of denominators to predict terminating vs repeating decimals. Power and Roots extends the system to surds and irrationals. Build your divisibility intuition first; the rest will follow.