The Number System is the foundation of CDS & OTA Maths. Almost every other chapter — HCF-LCM, simplification, algebra, even data questions — quietly rests on how confidently you handle integers, factors, primes and remainders. This Cavalier guide rebuilds the topic from the ground up with exam-ready rules, shortcuts and solved questions so you stop losing easy marks.
Why Number System decides your CDS score
In the CDS Elementary Mathematics paper you face 100 questions in 120 minutes, with negative marking. You cannot afford to grind through arithmetic the long way. The Number System chapter is where speed is built — divisibility, unit digits and factor counting turn 90-second problems into 15-second ones.
It is also a direct scorer: every CDS paper carries several questions purely on number properties (primes, remainders, last digit, HCF-LCM word problems). Master this and you also unlock simplification, ratio and algebra.
Examiners like this chapter because the questions are short to print but reward a candidate who knows the underlying property. A student who has internalised divisibility, prime factorisation and cyclicity simply looks at the number and answers, while an unprepared candidate reaches for long division and burns precious minutes. Over a 100-question paper, that difference in pace can decide whether you finish the section or leave easy marks on the table.
Number System is not a single chapter you revise once. It is the toolkit you reuse in every other arithmetic and algebra question. Time spent here pays compound interest.
The family of numbers
Knowing exactly which set a number belongs to prevents silly errors. Build the hierarchy outward from the simplest set.
- Natural numbers (N): 1, 2, 3, … — counting numbers.
- Whole numbers (W): 0, 1, 2, 3, … — naturals plus zero.
- Integers (Z): … −2, −1, 0, 1, 2 … — positives, negatives and zero.
- Rational numbers (Q): any number of the form p÷q where q ≠ 0. Decimals that terminate or recur are rational.
- Irrational numbers: non-terminating, non-recurring decimals such as √2, √3 and π. They cannot be written as p÷q.
- Real numbers (R): all rationals and irrationals together.
0 is a whole number and an integer but not a natural number. Every integer is rational (e.g. 5 = 5÷1), but not every rational is an integer.
A useful sub-classification of rationals is by their decimal behaviour. A fraction in lowest terms gives a terminating decimal only when its denominator has no prime factors other than 2 and 5; otherwise the decimal recurs. For instance 3÷8 = 0.375 terminates because 8 = 23, while 1÷3 = 0.333… recurs. Both are rational. Recognising this saves you from wrongly tagging a recurring decimal as irrational.
Irrational numbers, by contrast, never settle into a repeating block. The square root of any non-perfect-square (like √2, √5, √7) is irrational, and so are π and the number e. When an exam option offers √9, remember it equals 3 — a perfect square root is rational, not irrational.
Even, odd, prime and composite
These four labels appear constantly in CDS objective questions.
- Even: divisible by 2 (… −4, −2, 0, 2, 4 …).
- Odd: not divisible by 2 (1, 3, 5, 7 …).
- Prime: a natural number greater than 1 with exactly two factors — 1 and itself.
- Composite: a natural number greater than 1 with more than two factors.
1 is neither prime nor composite. Also, 2 is the only even prime — every other prime is odd. Many candidates wrongly call 1 a prime or forget 2 when listing even numbers.
There are 25 primes below 100. Memorise the first few: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37… A quick test: to check whether n is prime, try dividing it only by primes up to √n. For example, to test 97 you only need to try 2, 3, 5 and 7 (since 72 = 49 ≤ 97 but 112 = 121 > 97); none divide it, so 97 is prime.
Two numbers are co-prime (relatively prime) if their HCF is 1, even if neither is prime — for example 8 and 15. Co-primality is the property that makes split divisibility tests valid, so it is worth keeping distinct in your mind from primality itself.
Place value, face value and the decimal system
The Indian and international numeral systems are base-10 (decimal) positional systems. Each digit's contribution depends on its position.
- Face value of a digit = the digit itself, regardless of position.
- Place value = digit × (value of its place).
In 4752, the face value of 7 is 7, but its place value is 7 × 100 = 700.
Expanded form makes this clear: 4752 = 4×1000 + 7×100 + 5×10 + 2×1. This idea drives divisibility rules and many "find the digit" questions.
Divisibility rules you must know cold
These let you avoid long division. A number is divisible by:
- 2 — if its last digit is even (0, 2, 4, 6, 8).
- 3 — if the sum of its digits is divisible by 3.
- 4 — if the last two digits form a number divisible by 4.
- 5 — if the last digit is 0 or 5.
- 6 — if divisible by both 2 and 3.
- 8 — if the last three digits form a number divisible by 8.
- 9 — if the digit sum is divisible by 9.
- 11 — if the difference between the sum of digits in odd places and even places is 0 or a multiple of 11.
For divisibility by a composite like 12, split it into co-prime factors (12 = 3 × 4) and check each. Do not test 2 and 6 for 12 — they are not co-prime, so the rule fails.
Factors, multiples and the HCF-LCM link
A factor divides a number exactly; a multiple is the result of multiplying a number by an integer. Prime factorisation is the master tool.
Write each number as a product of primes, e.g. 360 = 23 × 32 × 5. From this you can extract HCF, LCM and the number of factors.
For any two numbers a and b: HCF × LCM = a × b. Also, the number of factors of N = (p+1)(q+1)(r+1)… where N = ap × bq × cr in prime form.
Example: 360 = 23 × 32 × 51, so the number of factors = (3+1)(2+1)(1+1) = 4 × 3 × 2 = 24 factors.
Remainders, last digits and cyclicity
CDS loves questions like "find the remainder" or "find the unit digit of a large power." The trick is cyclicity — the unit digit of powers repeats in a short cycle.
- Powers of 2: 2, 4, 8, 6, 2, 4… (cycle of 4).
- Powers of 3: 3, 9, 7, 1… (cycle of 4).
- Powers of 7: 7, 9, 3, 1… (cycle of 4).
- Digits 0, 1, 5, 6 always end in themselves.
To find the unit digit of a power, divide the exponent by the cycle length (usually 4) and use the remainder to pick the position. A remainder of 0 means you take the last digit of the cycle.
For division remainders, the remainder theorem helps: the remainder when a sum or product is divided by d equals the same operation done on the individual remainders, taken mod d. So to find the remainder of 17 × 23 when divided by 5, replace 17 by 2 and 23 by 3, multiply to get 6, then take 6 mod 5 = 1. This breaks frightening-looking products into one-line mental arithmetic.
A related favourite is the negative remainder shortcut. When a number is just below a multiple of the divisor, treat its remainder as a small negative. For 99 divided by 10, instead of remainder 9 you may use −1, which makes powers easy: 992 leaves remainder (−1)2 = 1 when divided by 10. Used carefully, this converts heavy power-remainder questions into trivial sign tracking.
Factorials and trailing zeros
The factorial n! = 1 × 2 × 3 × … × n. A favourite CDS question asks for the number of trailing zeros in a factorial.
Trailing zeros come from factors of 10 = 2 × 5. Since 2s are plentiful, count the number of 5s: zeros in n! = [n÷5] + [n÷25] + [n÷125] + … (each term is the integer part).
So 100! has [100÷5] + [100÷25] = 20 + 4 = 24 trailing zeros. This single formula answers a whole family of questions.
Worked example: putting it together
Find the number of factors of 720, the sum of its prime factors, and the unit digit of 7203.
One factorisation gave us three separate answers — that is the power of the prime-factor method.
Common mistakes that cost marks
Most Number System errors are avoidable. Watch for these traps.
- Treating 0 as a natural number or 1 as prime.
- Forgetting negative integers when a question says "integers," not "natural numbers."
- Applying divisibility of 12 by checking 2 and 6 (not co-prime) instead of 3 and 4.
- In remainder questions, forgetting to take the final answer mod d again after combining.
- Miscounting cyclicity: a remainder of 0 points to the last term of the cycle, not the first.
Read whether the question wants natural numbers, whole numbers, or integers. The answer often hinges on whether 0 and negatives are allowed.
Previous-year style question
Q. The number 357xy is divisible by both 5 and 9. What is the least possible value of (x + y)?
Answer: For divisibility by 5, the last digit y must be 0 or 5. The digit sum is 3+5+7+x+y = 15+x+y, which must be divisible by 9. Trying y = 0: 15 + x must be a multiple of 9, so x = 3 (sum 18), giving x+y = 3. Trying y = 5: 20 + x divisible by 9 needs x = 7 (sum 27), giving x+y = 12. The least value is 3.
When two divisibility conditions apply, lock down the most restrictive one first (here, divisibility by 5 fixes y to two choices), then test each against the second rule.
Quick revision
- N ⊂ W ⊂ Z ⊂ Q ⊂ R; irrationals fill the gaps in R.
- 0 is whole/integer but not natural; 1 is neither prime nor composite; 2 is the only even prime.
- Divisibility: 3 and 9 by digit sum; 4 by last two digits; 8 by last three; 11 by alternating-sum difference.
- HCF × LCM = product of the two numbers; factor count = (p+1)(q+1)… from prime form.
- Unit digit via cyclicity (cycle 4 for 2, 3, 7, 8); trailing zeros in n! = count of 5s.
Drill divisibility rules and unit-digit cyclicity until they are automatic. They convert long calculations into instant answers and protect you from negative marking under time pressure.
Frequently asked questions
Is 0 a natural number for CDS purposes?
No. In the CDS syllabus, natural numbers start from 1. Zero is a whole number and an integer, but it is not counted as a natural number.
How many prime numbers are there below 100?
There are 25 prime numbers below 100. The first few are 2, 3, 5, 7, 11, 13, 17, 19 and 23. Remember that 2 is the only even prime.
What is the fastest way to find the unit digit of a large power?
Use cyclicity. Most digits repeat their unit digit in a cycle of length 4. Divide the exponent by 4, and the remainder tells you which term of the cycle gives the unit digit; a remainder of 0 means the last term.
How do I count the number of factors of a number?
Write the number in prime-factor form, say N = a^p x b^q x c^r. The total number of factors is (p+1)(q+1)(r+1). For example, 360 = 2^3 x 3^2 x 5 has (3+1)(2+1)(1+1) = 24 factors.
Why is divisibility by 12 checked using 3 and 4, not 2 and 6?
Because 3 and 4 are co-prime and multiply to 12. The factor pair you test must be co-prime; 2 and 6 share the factor 2, so passing both does not guarantee divisibility by 12.
How are HCF and LCM related?
For any two numbers, HCF x LCM equals the product of the two numbers. This lets you find one quickly if you know the other and the two numbers.
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