Think of numbers as a family tree. At the top, you have all real numbers. Zoom in and you reach the branch that RRB NTPC actually tests you on — the integers and their properties.
Here is how the classification works, and why it matters for your exam:
p/q where q ≠ 0, including fractions and terminating/repeating decimals.√2, √3, π that cannot be expressed as p/q.The analogy that sticks: imagine natural numbers as the positive floors of a building (Floor 1, Floor 2, …). Whole numbers add the ground floor (Floor 0). Integers add the basement floors (Floor -1, Floor -2, …). Rational numbers add the mezzanines (the fractions between floors). Irrational numbers are the awkward measurements that never land exactly on any floor.
For RRB NTPC, the bulk of number system questions revolve around five practical skills:
n natural numbers)You will rarely get a question asking "which of these is irrational?" — but you will absolutely get questions that require you to know, instinctively, that 4³ = 64 or that LCM(12, 18) = 36. The classification is background knowledge; the computation is the actual exam skill.
Every integer has exactly one of three properties with respect to divisibility by 2: it is even, odd, or zero (zero is even). Every integer greater than 1 is either prime (exactly two factors: 1 and itself) or composite (more than two factors). The number 1 is neither prime nor composite — this trips up candidates in elimination-type questions.
Do not reach for a calculator method in the exam hall. Memorize these directly:
| n | n² | n³ | |----|-----|-------| | 1 | 1 | 1 | | 2 | 4 | 8 | | 3 | 9 | 27 | | 4 | 16 | 64 | | 5 | 25 | 125 | | 6 | 36 | 216 | | 7 | 49 | 343 | | 8 | 64 | 512 | | 9 | 81 | 729 | | 10 | 100 | 1000 | | 11 | 121 | 1331 | | 12 | 144 | 1728 | | 15 | 225 | 3375 | | 20 | 400 | 8000 | | 25 | 625 | 15625 |
For numbers ending in 5, there is a clean identity: (10k+5)² = k(k+1) × 100 + 25. So 15² → k=1, 1×2=2, answer = 225. So 25² → k=2, 2×3=6, answer = 625. No multiplication needed.
Always factorize both numbers completely, then:
For 12 and 18:
12 = 2² × 3¹18 = 2¹ × 3²2¹ × 3¹ = 62² × 3² = 4 × 9 = 36Cross-check with the identity: HCF × LCM = Product of the two numbers. Here: 6 × 36 = 216 = 12 × 18. Always use this to verify quickly.
| Divisor | Rule | |---------|------| | 2 | Last digit is even | | 3 | Sum of digits divisible by 3 | | 4 | Last two digits divisible by 4 | | 5 | Last digit is 0 or 5 | | 6 | Divisible by both 2 and 3 | | 8 | Last three digits divisible by 8 | | 9 | Sum of digits divisible by 9 | | 11 | (Sum of odd-position digits) − (Sum of even-position digits) = 0 or multiple of 11 | | 25 | Last two digits are 00, 25, 50, or 75 |
For any integers a and b (divisor b > 0), there exist unique integers q (quotient) and r (remainder) such that:
a = bq + r, where 0 ≤ r < b
So for 17 ÷ 5: 17 = 5 × 3 + 2, remainder = 2. This is the definition, and it directly produces the answer without any ambiguity.
When the dividend is large, look for cyclicity in remainders. The remainders of powers of a number w.r.t. a fixed divisor follow a repeating cycle. This becomes important in moderate-difficulty remainder questions — for now, Euclid's lemma covers everything at the beginner level.
S = n(n+1)/2
For n = 10: S = 10 × 11 / 2 = 55. This formula is exact, fast, and applies whenever you see "sum of first n natural numbers" or an arithmetic series starting from 1 with common difference 1.
Related formulas worth knowing:
n odd numbers = n²n even numbers = n(n+1)n naturals = n(n+1)(2n+1)/6n naturals = [n(n+1)/2]²Any number ending in 5, when squared, always ends in 25. The digits before 25 come from multiplying the leading digit(s) by (leading digit(s) + 1).
For 15: leading = 1, multiply 1 × 2 = 2, answer = 2|25 = 225. For 35: leading = 3, multiply 3 × 4 = 12, answer = 12|25 = 1225. For 65: leading = 6, multiply 6 × 7 = 42, answer = 42|25 = 4225.
Standard method (long multiplication): ~40 seconds. This pattern: under 8 seconds.
After computing LCM, always verify: HCF × LCM must equal the product of the two numbers. If the answer choices contain a wrong value, this cross-check eliminates it in 3 seconds without re-doing the factorization.
Example: LCM(12, 18). HCF = 6. So LCM must be (12 × 18) / 6 = 216 / 6 = 36. You can get the answer directly this way — no prime factorization needed if HCF is obvious.
Standard method (full prime factorization): ~45 seconds. This direct formula path: ~12 seconds when HCF is visible by inspection.
Never add a sequence of consecutive natural numbers one by one. Use n(n+1)/2 directly.
Sum of 1 to 10: 10 × 11 / 2 = 55. One multiplication, one halving. Sum of 1 to 100: 100 × 101 / 2 = 5050. Same two steps.
Standard method (sequential addition for 1 to 10): ~20 seconds, error-prone. Formula: ~5 seconds, zero error risk.
To find remainder of N ÷ d, mentally find the largest multiple of d that does not exceed N, then subtract.
For 17 ÷ 5: nearest multiple of 5 below 17 is 15, so remainder = 17 − 15 = 2. For 83 ÷ 9: nearest multiple of 9 below 83 is 81, so remainder = 83 − 81 = 2. For 127 ÷ 11: nearest multiple of 11 below 127 is 121, so remainder = 127 − 121 = 6.
Standard method (long division setup): ~25 seconds. Mental nearest-multiple subtraction: ~6 seconds.
To check divisibility by 9 (or find remainder when divided by 9), add all digits repeatedly until you get a single digit. That single digit is the remainder when the number is divided by 9. If it reduces to 9, the remainder is 0.
Example: Is 738 divisible by 9? 7+3+8 = 18, 1+8 = 9. Yes, remainder = 0. Example: Remainder when 4567 is divided by 9? 4+5+6+7 = 22, 2+2 = 4. Remainder = 4.
This replaces division entirely for the specific divisor 9. Standard long division: ~30 seconds. Digit-sum shortcut: ~8 seconds.
When you see a number system question in the exam hall, run through this decision flow:
Is it a square or cube question? Recall from memory directly. If the number ends in 5, use the pattern trick. Do not compute — retrieve.
Is it an LCM/HCF question? Check if HCF is obvious by inspection. If yes, use LCM = (Product) / HCF directly. If not, go to prime factorization — it takes 30 seconds max for two-digit numbers.
Is it a remainder question? Write a = bq + r mentally. Find the nearest multiple of the divisor below the dividend. Subtract. Done.
Is it a sum of consecutive numbers? Identify n, apply n(n+1)/2. Never add term by term.
Is it a divisibility question? Apply the relevant rule from the table — no long division needed.
Are the answer options far apart? Use estimation to eliminate 2-3 options before computing. This cuts the probability of a careless error affecting your final answer.
If none of these apply, re-read the question — there is almost always a shortcut hiding in the structure of RRB NTPC number questions.
Why this question: This tests whether you know the fastest method for squaring numbers ending in 5, rather than grinding through multiplication.
Solving path: Identify that 15 ends in 5. Apply the pattern: leading digit = 1, multiply 1 × 2 = 2, attach 25 to get 225. Alternatively, use (10+5)² = 100 + 2(10)(5) + 25 = 100 + 100 + 25 = 225. Either way, you reach option (C) in under 10 seconds.
Why this question: LCM questions appear regularly in NTPC. This one tests prime factorization and the HCF×LCM identity as a cross-check.
Solving path: Factorize: 12 = 2² × 3, 18 = 2 × 3². LCM takes the highest power of each prime: 2² × 3² = 4 × 9 = 36. Verify: HCF(12,18) = 6, and 6 × 36 = 216 = 12 × 18. Answer is option (A), 36.
Why this question: Remainder questions test Euclid's division lemma directly. The trap here is confusing quotient with remainder.
Solving path: Write 17 = 5 × q + r. The largest multiple of 5 not exceeding 17 is 15 (5 × 3). So q = 3, r = 17 − 15 = 2. The remainder is 2, not the quotient 3. Answer is option (B).
Why this question: Cube recall is a direct memory test. Knowing 4³ = 64 cold prevents wasted computation time.
Solving path: 4³ = 4 × 4 × 4. First: 4 × 4 = 16. Then: 16 × 4 = 64. Answer is option (D), 64. If you have this memorized, this is a 3-second question. Do not skip the cubes table.
Why this question: Sum of natural numbers questions are formula-dependent. The trap is computing term by term and making an addition error under time pressure.
Solving path: n = 10. Apply S = n(n+1)/2 = 10 × 11 / 2 = 110 / 2 = 55. Answer is option (C). The wrong answer 50 (option B) catches candidates who use n²/2 instead of n(n+1)/2 — note the +1 in the formula.
Confusing quotient and remainder. In 17 ÷ 5, the quotient is 3 and the remainder is 2. Options are often designed to make you select 3 instead of 2. Always re-read: the question asks for the remainder, not how many times the divisor goes in.
Treating 1 as a prime number. One has exactly one factor (itself), which disqualifies it from being prime. Questions about the "smallest prime" or "prime factorization" rely on this distinction. The smallest prime is 2.
Using wrong LCM formula. HCF × LCM = Product holds only for exactly two numbers, not three or more. Applying it to three numbers and expecting HCF × LCM = a × b × c is incorrect.
Off-by-one in natural number sums. n(n+1)/2 gives the sum of natural numbers from 1 to n. If the question asks for the sum from 1 to 10, n = 10 and the answer is 55, not 50 (which is 10²/2, a different and incorrect formula).
Forgetting that 0 is a whole number but not a natural number. Any question asking whether 0 belongs to a specific set of numbers hinges on this. Natural numbers start at 1.
Not verifying LCM with the cross-product identity. If you compute LCM and do not cross-check via HCF × LCM = Product, you have no safety net against factorization errors. The cross-check takes 5 seconds and catches most errors.