The number system is the backbone of all arithmetic in CDS. It is not about memorising classifications — you already know those. What CDS actually tests is your ability to exploit the structure of numbers: when does aⁿ + bⁿ become divisible by something predictable? Why does the HCF of two numbers not change when you replace one of them with their sum or difference? How do you find the greatest divisor leaving the same remainder across three numbers? These are not arbitrary tricks — they follow from a small set of properties that, once internalised, let you solve what looks like a hard problem in under 30 seconds.
Here is the mental map you need:
Classification (for reference, not rote learning):
ℕ: {1, 2, 3, …}W: {0, 1, 2, …}ℤ: {…, -2, -1, 0, 1, 2, …}ℚ: numbers expressible as p/q where p, q ∈ ℤ and q ≠ 0√2, π)ℝ: all rational + irrational numbersThe analogy that helps: think of the number system as a set of concentric circles. Every natural number is a whole number, every whole number is an integer, every integer is a rational, and every rational is a real. The containment is strict in one direction only.
What CDS is actually interested in is the arithmetic of integers — specifically divisibility, remainders, and the HCF-LCM framework. The classification question comes up once in ten years; divisibility and HCF-LCM questions appear every single paper.
CDS loves questions of the form "which of these is aⁿ ± bⁿ divisible by?" The key identities are:
Identity 1: For any positive integer n, (aⁿ - bⁿ) is always divisible by (a - b).
Identity 2: (aⁿ + bⁿ) is divisible by (a + b) if and only if n is odd.
Identity 3: (aⁿ - bⁿ) is also divisible by (a + b) when n is even.
These three identities eliminate 90% of the work in CDS divisibility questions. You do not need to compute the actual numbers — you just check the parity of the exponent and identify a and b.
Look — 41^\{43\} + 43^\{43\} is immediately aⁿ + bⁿ with n = 43 (odd), so it is divisible by 41 + 43 = 84. Done. No calculation.
Property 1 — Linear combination: If HCF(p, q) = G, then G divides every linear combination ap + bq where a, b are integers. Consequently:
HCF(p, p+q) = HCF(p, q) = GHCF(p, p-q) = HCF(p, q) = GThis is the Euclidean algorithm at its core. Because G | p and G | q, it must divide p + q and p - q. And since any common divisor of p and p + q must divide their difference q, the set of common divisors doesn't change.
Property 2 — Same remainder divisor: If a number N divides each of a, b, c leaving the same remainder, then N divides the differences (b - a), (c - b), (c - a). So N must be a divisor of HCF(b-a, c-b, c-a). For the greatest such N, you want that HCF itself.
Property 3 — Product relationship: For any two positive integers p and q:
HCF(p, q) × LCM(p, q) = p × q
This is the most-used formula in HCF-LCM problems. Given any three of the four quantities, the fourth is determined.
The trick that most students miss: convert all decimals to fractions with the same denominator, compute LCM(numerators)/common denominator.
For 3.6, 0.009, 0.27 — bring all to thousandths: 3600/1000, 9/1000, 270/1000. Now just find LCM(3600, 9, 270) and divide by 1000.
To find LCM(3600, 9, 270): prime-factorise each.
3600 = 2^4 × 3^2 × 5^29 = 3^2270 = 2 × 3^3 × 5LCM takes the highest power of each prime: 2^4 × 3^3 × 5^2 = 16 × 27 × 25 = 10800.
So LCM of the decimals = 10800/1000 = 10.8.
When you see a sum of like powers (same base), factor out the lowest power first.
4^\{61\} + 4^\{62\} + 4^\{63\} + 4^\{64\} = 4^\{61\}(1 + 4 + 16 + 64) = 4^\{61\} × 85 = 4^\{61\} × 5 × 17
Now you can read off divisibility directly — this is divisible by 5 and 17. If the option choices include 17, that is your answer without touching the giant exponents.
The same factoring technique applies to expressions like 3^\{81\} - 3^\{80\} - 3^\{79\}. Factor out 3^\{79\}: you get 3^\{79\}(3^2 - 3 - 1) = 3^\{79\} × 5. The full expression is divisible by 3^\{79\} and by 5. Natural-number divisors less than 10 are therefore 1, 3, 5, 9 — that is four values, which is "more than 3".
Questions like "how many minutes in x weeks and x days?" are pure number manipulation:
x weeks = 7x days, so total days = 8x8x days × 24 hours × 60 minutes = 8 × 24 × 60 × x = 11520xThe trap is forgetting to add x days to 7x days — students who rush often compute just 7x × 24 × 60 and pick 10080x, which is not even an option here, but the error of ignoring the x days is real.
When you see aⁿ + bⁿ: check if n is odd. If yes, the expression is divisible by (a + b). When you see aⁿ - bⁿ: it is always divisible by (a - b), and also by (a + b) when n is even.
Micro-example: 41^\{43\} + 43^\{43\}. n = 43 is odd, so divisible by 41 + 43 = 84. Identify a, b, check parity, write answer.
Standard method (modular arithmetic from scratch): ~90 seconds. This scan: 8 seconds. 11× faster.
For any sum/difference of terms sharing a base (e.g., 4^\{61\} + 4^\{62\} + 4^\{63\} + 4^\{64\}), always factor out the smallest power first. What remains is a small bracket you can compute mentally.
Micro-example: Factor out 4^\{61\} to get 4^\{61\}(1 + 4 + 16 + 64) = 4^\{61\} × 85. Now 85 = 5 × 17. Divisibility by 17 is immediately visible.
Without this: you would need to check each option individually by trying to find a pattern. With this: 3 steps, ~15 seconds vs ~60 seconds.
If N leaves the same remainder when dividing a, b, c, then N divides (b-a), (c-b), (c-a). The greatest such N is HCF of those differences.
Micro-example: Numbers 600, 631, 724. Differences: 31, 93, 124. HCF(31, 93, 124): 31 divides all three (31×1, 31×3, 31×4). Answer: 31.
Standard method (trial division of all differences): ~60 seconds. Pattern recognition + single HCF check: ~20 seconds.
Convert all decimals to fractions sharing the same power-of-10 denominator. Apply LCM(numerators) ÷ common denominator.
Micro-example: LCM(3.6, 0.009, 0.27) → LCM(3600, 9, 270)/1000. Prime factorise: max powers give 10800. Answer: 10800/1000 = 10.8.
Working with decimals directly leads to errors in ~40% of attempts. This substitution reduces it to a clean integer-LCM problem: 4 steps instead of 7+.
HCF(p, q) = HCF(p, p+q) = HCF(p, p-q) = HCF(p+q, q). This is because the HCF divides any linear combination.
Micro-example: HCF(220, 924). Instead of prime-factorising both, use: HCF(220, 924-4×220) = HCF(220, 44) = 44. This is Euclid's algorithm in one step.
Standard prime factorisation of 924: 6 division steps. Euclid reduction: 2 steps.
When you encounter a number-system question in the exam hall, run this decision tree:
Step 1 — Identify the question type:
aⁿ ± bⁿ? → Use algebraic identity (odd/even exponent check).Step 2 — Execute:
For aⁿ ± bⁿ problems, you need only know: (a) the parity of n, and (b) a ± b. No exponent arithmetic required.
For same-remainder problems, compute the three pairwise differences and find their HCF. Often one of the differences is already the HCF.
For HCF-LCM product problems, always extract the known number from the word problem first (e.g., "quotient 4 remainder 0 when divided by 55" → number = 220), then apply HCF × LCM = p × q.
Time target: Any number-system question in this chapter should be answered in under 60 seconds. If you are past 45 seconds, you are not using the right property — backtrack and re-identify the type.
Why this question: Tests the core algebraic identity aⁿ + bⁿ divisible by (a + b) when n is odd. Appears in multiple CDS years.
Solving path: Recognise the form aⁿ + bⁿ with a = 41, b = 43, n = 43. Since n is odd, divisible by a + b = 84. Match with option (c). Time: 8 seconds.
Why this question: Tests the "factor out the lowest power" technique on a sum of consecutive powers. The bracket (1 + 4 + 16 + 64) must be computed correctly and then factored.
Solving path: 4^\{61\}(1 + 4 + 16 + 64) = 4^\{61\} × 85 = 4^\{61\} × 5 × 17. Check options — 17 is present. Done. Note: 85 is also divisible by 5, so 5 would work too, but 5 is not among the options here.
Why this question: Tests the same-remainder divisor concept using pairwise differences and HCF, with options close enough together to punish casual guessing.
Solving path: Differences: 631 - 600 = 31, 724 - 631 = 93, 724 - 600 = 124. Check: 93 = 3 × 31, 124 = 4 × 31. So HCF = 31. Answer: 31.
Why this question: Tests the HCF × LCM = p × q relationship with a disguised way of giving you p (via quotient-remainder statement).
Solving path: "Divided by 55, quotient 4, remainder 0" → p = 55 × 4 = 220. Then q = (44 × 4620)/220 = 203280/220 = 924. Verify: HCF(220, 924) — use Euclid: 924 = 4 × 220 + 44, 220 = 5 × 44 + 0. HCF = 44. Confirmed.
Why this question: Tests whether you understand HCF invariance under sum and difference — a property that looks abstract but is directly provable and is a recurring CDS assertion question type.
Solving path: Let p = Ga, q = Gb with gcd(a, b) = 1. Then p + q = G(a + b). HCF(p, p+q) = G × gcd(a, a+b). Since gcd(a, b) = 1, any common factor of a and a+b must divide b — but gcd(a, b) = 1, so gcd(a, a+b) = 1. Therefore HCF(p, p+q) = G. Same logic for p - q. Both statements I and II are correct.
Applying aⁿ + bⁿ identity without checking parity. aⁿ + bⁿ is divisible by (a + b) only when n is odd. For even n, this identity does not hold. Students who do not check the exponent parity will pick a wrong answer and not understand why.
Forgetting to add x days when x weeks and x days are both given. x weeks already gives 7x days; you must add the separate x days to get 8x total. This error is silent — no alarm goes off — and the wrong value (7x × 24 × 60) may still look plausible.
Using only two of the three pairwise differences in a same-remainder problem. You must compute all three differences (b-a, c-b, c-a) and take their HCF. Sometimes one difference is a multiple of another, and only checking two differences gives you a divisor larger than the correct answer.
Treating HCF × LCM = p × q as valid for three numbers. This product formula holds only for two numbers. For three numbers p, q, r, there is no such clean formula. Students who extend it blindly to three numbers will get nonsense.
Factoring out the wrong power in sum-of-like-base problems. Always factor out the lowest power present. If you factor out a higher power, you will get a non-integer bracket and the factorisation breaks.
Converting decimals to fractions with different denominators before computing LCM. LCM(3600/1000, 9/1000, 270/1000) works cleanly. But if you write 36/10, 9/1000, 27/100 with different denominators, the formula LCM(numerators)/HCF(denominators) applies — and most students forget this form and make errors. The safest approach: equalise all denominators to the highest power of 10 required, then take LCM(numerators)/common denominator.