Number System (RRB) — Core

The number system is the bedrock of every quantitative section — and in RRB exams it shows up as fast, formula-driven questions on LCM/HCF, divisibility, unit digits and trailing zeros. This lesson builds the intuition behind each tool so you can solve these in seconds, not minutes.

Definition: The number system classifies numbers (natural, whole, integer, rational, real) and studies their properties — divisibility, factors, primes, and the patterns hidden in powers and factorials.

Classifying numbers

Numbers nest inside one another like boxes:

• Natural numbers (N): 1, 2, 3, … (counting numbers).
• Whole numbers (W): 0, 1, 2, 3, … (naturals plus 0).
• Integers (Z): …, −2, −1, 0, 1, 2, … (whole numbers plus negatives).
• Rational numbers (Q): any number expressible as p/q with q ≠ 0 (includes fractions and terminating/recurring decimals).
• Real numbers (R): all rationals plus irrationals (like √2, π) — every point on the number line.

Why it matters: Many "is this prime/composite/rational?" questions test only whether you know the right box a number belongs to.

Divisibility and the LCM–HCF link

Divisibility rules (for 2, 3, 4, 5, 6, 7, 8, 9, 10, 11) let you test factors without dividing — for example, a number is divisible by 3 if its digit-sum is divisible by 3, and by 11 if the alternating digit-sum is divisible by 11.

The single most-tested identity is:

LCM × HCF = product of the two numbers (i.e. LCM(a,b) × HCF(a,b) = a × b).

Two numbers are co-prime when their HCF = 1 (they share no common factor except 1, e.g. 8 and 15).

Worked example:
Question: The HCF of two numbers is 6 and their LCM is 36. If one number is 12, find the other.
Solution:
Step 1: Use LCM × HCF = product of the numbers → 36 × 6 = 12 × (other).
Step 2: 216 = 12 × (other), so other = 216 ÷ 12 = 18.
Conclusion: The other number is 18 (check: HCF(12,18)=6, LCM(12,18)=36 ✓).

Unit-digit (cyclicity) of large powers

When you raise a number to a high power, only the last digit follows a repeating cycle. Memorise the cycle lengths:

• Ending in 2: cycle of 4 → 2, 4, 8, 6.
• Ending in 3: cycle of 4 → 3, 9, 7, 1.
• Ending in 7: cycle of 4 → 7, 9, 3, 1.
• Ending in 8: cycle of 4 → 8, 4, 2, 6.
• Ending in 9: cycle of 2 → 9, 1.
• Ending in 0, 1, 5, 6: the unit digit stays the same for every power.
• Ending in 4: cycle of 2 → 4, 6.

The trick: for a 4-cycle, divide the exponent by 4 and use the remainder to pick the digit (remainder 0 means the last digit of the cycle).

Worked example:
Question: Find the unit digit of 7^123.
Solution:
Step 1: 7 has a 4-cycle: 7, 9, 3, 1.
Step 2: 123 ÷ 4 leaves remainder 3.
Step 3: The 3rd term in the cycle is 3.
Conclusion: The unit digit of 7^123 is 3.

Trailing zeros in a factorial

A trailing zero comes from a factor of 10 = 2 × 5. In any factorial there are far more 2s than 5s, so the number of trailing zeros equals the count of factors of 5 in n!:

Number of zeros = ⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + …

Worked example:
Question: How many trailing zeros does 100! have?
Solution:
Step 1: ⌊100/5⌋ = 20.
Step 2: ⌊100/25⌋ = 4.
Step 3: ⌊100/125⌋ = 0; stop.
Step 4: Add: 20 + 4 = 24.
Conclusion: 100! ends in 24 zeros.

Common misconception: Students count factors of 10 directly, or count 2s. Always count 5s — they are the scarcer factor and therefore the limiting one.

The five RRB-favourite question types

1. Find the LCM or HCF of two (or more) numbers.
2. Find the smallest number to ADD to make a number divisible by X (remainder method).
3. Find the largest n-digit number divisible by Y (divide, subtract the remainder).
4. Find the unit digit of a large power (cyclicity).
5. Identify a number as prime or composite, or as co-prime.

Number System is the quiet backbone of RRB NTPC Maths: 2–3 direct questions in CBT-1, and its ideas (LCM, remainders, unit digits) leak into Time & Work, TSD and Simplification too. Master it and you also speed up half the paper.

Why it matters

Questions here are formula-light and trick-heavy — a candidate who knows divisibility, unit-digit cycles and HCF/LCM shortcuts solves them in 20–30 seconds while others do long division.

Core facts & formulas

• Divisibility: by 3 (digit-sum ÷3), by 9 (digit-sum ÷9), by 4 (last 2 digits ÷4), by 8 (last 3 digits ÷8), by 11 (odd−even place sum ÷11).
• HCF × LCM = product of the two numbers. So LCM = (a×b)/HCF.
• Number of factors: if N = pᵃ·qᵇ then factors = (a+1)(b+1).
• Unit-digit cycles (period 4): 2→2,4,8,6 · 3→3,9,7,1 · 7→7,9,3,1 · 8→8,4,2,6.
• Remainder: (a×b) mod n = [(a mod n)(b mod n)] mod n.

Digit	Cycle of unit digit	Digit	Cycle
2	2,4,8,6	7	7,9,3,1
3	3,9,7,1	8	8,4,2,6
4	4,6	9	9,1
5	5	6	6

Exam Tricks & Tips

• 🎯 Unit digit of aⁿ: divide n by 4; use remainder as position in the cycle (remainder 0 → last term). Unit digit of 7¹⁰² : 102÷4 leaves 2 → 9.
• 🎯 Factor count fast: 72 = 2³·3² → (3+1)(2+1) = 12 factors — no listing needed.
• 🎯 LCM shortcut: for two numbers just compute (a×b)/HCF; e.g. LCM(12,18) = 216/6 = 36.
• 🎯 Sum of first n naturals = n(n+1)/2; of squares = n(n+1)(2n+1)/6 — memorise both.
• 🎯 Divisibility by 11: difference of alternate digit sums; 918082 → (9+8+8)−(1+0+2)=25−3=22, divisible.
• ❌ Common mistake: treating a remainder-0 case as "position 0" and reading the first cycle term instead of the last.

Expected exam pattern

Typically "find the unit digit", "smallest number that when divided leaves remainder r", or "largest number dividing a, b, c" (HCF-based). Usually 2–3 questions, all quick if the shortcuts are automatic.

Quick recap

Learn divisibility rules, HCF×LCM=product, factor-count formula and the four unit-digit cycles. These convert slow long-division questions into 20-second wins.

Cover the answer, recall, then check. 12 cards on the highest-yield Number System facts for RRB NTPC.

Q1. Unit digit of 7¹⁰²?
A1. Cycle of 7 is 7,9,3,1 (period 4). 102 ÷ 4 leaves remainder 2 → 2nd term = 9.

Q2. State HCF–LCM relation for two numbers.
A2. HCF × LCM = product of the numbers. So LCM = (a×b)/HCF.

Q3. LCM of 12 and 18 using HCF?
A3. HCF(12,18)=6 → LCM = (12×18)/6 = 216/6 = 36.

Q4. How many factors does 72 have?
A4. 72 = 2³·3² → (3+1)(2+1) = 12 factors.

Q5. Divisibility rule for 8?
A5. Number formed by the last three digits is divisible by 8 (e.g. 17,912 → 912÷8=114 ✓).

Q6. Is 918082 divisible by 11?
A6. Alternate sums: (9+8+8)=25, (1+0+2)=3, difference 22 → divisible by 11 ✓.

Q7. Sum of first 20 natural numbers?
A7. n(n+1)/2 = 20·21/2 = 210.

Q8. Unit digit of 2⁵⁰?
A8. Cycle 2,4,8,6. 50÷4 leaves 2 → 2nd term = 4.

Q9. Smallest number leaving remainder 3 when divided by 4,6,8?
A9. LCM(4,6,8)=24 → 24+3 = 27.

Q10. Sum of squares of first 10 naturals?
A10. n(n+1)(2n+1)/6 = 10·11·21/6 = 385.

Q11. Divisibility rule for 3 and 9?
A11. Digit-sum divisible by 3 → number divisible by 3; digit-sum divisible by 9 → divisible by 9.

Q12. Remainder when 17×19 is divided by 5?
A12. (17 mod5=2)(19 mod5=4)=8; 8 mod5 = 3.

Key formulas

• Natural, whole, integer, rational (p/q), irrational, real, prime, composite numbers.
• Sum of first n naturals = n(n+1)/2; of squares = n(n+1)(2n+1)/6; of cubes = [n(n+1)/2]².
• Even + even = even; odd + odd = even; product of two odds is odd.
• A prime > 3 is of the form 6k ± 1.
• Number of primes is infinite; 2 is the only even prime.
• (aⁿ − bⁿ) is divisible by (a − b) for all n; by (a + b) for even n.

Usage: use the standard series sums and divisibility identities to shortcut number problems.

Worked Example

Problem: Find the unit digit of 7^103.

Solution:
Unit digits of powers of 7 repeat in a cycle of 4:
7¹ = 7 → 7
7² = 49 → 9
7³ = 343 → 3
7⁴ = 2401 → 1
Then the cycle 7, 9, 3, 1 repeats. To find where 7^103 lands, divide the exponent by 4 and take the remainder:
103 ÷ 4 = 25 remainder 3.
A remainder of 3 corresponds to the 3rd term of the cycle, which is 3.

Answer: The unit digit of 7^103 is 3.