Classification of Numbers & Divisibility Rules

Natural numbers (N): 1,2,3,... Whole numbers (W): 0,1,2,3,... (add zero to N). Integers (Z): ...-2,-1,0,1,2... Rational numbers: can be written as p/q (q not 0), e.g. 3/4, 0.25, 0.333. Irrational numbers: non-terminating non-repeating, e.g. root2, pi. Prime number: exactly two factors (1 and itself); 2 is the smallest and ONLY even prime. Composite: more than two factors. 1 is NEITHER prime nor composite. Even: divisible by 2; Odd: not. Memory aid: every Natural is Whole, every Whole is Integer, every Integer is Rational. There are 25 prime numbers below 100. Co-prime numbers share only 1 as common factor, e.g. (8,15).

By 2: last digit even (0,2,4,6,8). By 3: digit-sum divisible by 3. By 4: last two digits divisible by 4. By 5: last digit 0 or 5. By 6: divisible by 2 AND 3. By 8: last three digits divisible by 8. By 9: digit-sum divisible by 9. By 10: ends in 0. By 11: (sum of odd-position digits) minus (sum of even-position digits) is 0 or a multiple of 11. By 7 shortcut: double the last digit, subtract from rest; if result divisible by 7 the number is. Trick for 6,12,15: check the two co-prime factors (e.g. 12 = 3 and 4). Always test smaller co-prime factors rather than the big number itself.

Eleven is the trickiest of the school-age divisibility rules — and that is exactly why RPF Constable papers love to test it. Once you understand why the alternating-sum rule works, you can decide divisibility of any six- or seven-digit number in under fifteen seconds. Let us walk through the two worked examples you must internalise.

Definition: A number n is divisible by 11 if and only if the difference between the sum of digits in odd positions and the sum of digits in even positions is either 0 or a multiple of 11 (i.e., ±11, ±22, ±33, …). Position counting may start from the right or the left — only consistency matters.

Why the Rule Works (the Quick Why)

Powers of 10, taken mod 11, alternate between +1 and −1:

10⁰ = 1, 10¹ = 10 ≡ −1, 10² = 100 ≡ +1, 10³ = 1000 ≡ −1, …

So any number written in decimal as d₀ + d₁·10 + d₂·10² + … reduces, mod 11, to d₀ − d₁ + d₂ − d₃ + …, which is exactly the alternating sum of its digits. That alternating sum is what the textbook calls "odd-position sum minus even-position sum". This is why the rule is bulletproof — it is just modular arithmetic of 10 mod 11.

Worked Example 1: Is 4,83,771 Divisible by 11?

The number is 4 8 3 7 7 1 (six digits). We label positions starting from the rightmost digit.

Question: Check whether 483771 is divisible by 11.

Solution:
Step 1: Number the positions from the right.

Step 2: Sum the digits in odd positions: 1 + 7 + 8 = 16.
Step 3: Sum the digits in even positions: 7 + 3 + 4 = 14.
Step 4: Difference = 16 − 14 = 2.
Step 5: Is 2 equal to 0 or a multiple of 11? No.
Conclusion: 483771 is not divisible by 11. (For confirmation, 483771 ÷ 11 = 43,979.18…)

Worked Example 2: Is 4,83,769 Divisible by 11?

Same six-digit shape, last two digits changed. Watch how a tiny tweak flips the answer.

Question: Check whether 483769 is divisible by 11.

Solution:
Step 1: List positions from the right for 4 8 3 7 6 9.

Step 2: Sum the digits in odd positions: 9 + 7 + 8 = 24.
Step 3: Sum the digits in even positions: 6 + 3 + 4 = 13.
Step 4: Difference = 24 − 13 = 11.
Step 5: Is 11 a multiple of 11? Yes.
Conclusion: 483769 is divisible by 11. (Confirmation: 483769 ÷ 11 = 43,979.)

Why It Matters — and the One Trap to Avoid

The rule itself is simple, but every year candidates lose marks by changing the direction of counting in the middle of the calculation. The rule does not care whether you start from the right or the left — but you must stay consistent within a single problem. If you start position-1 from the left for one number and from the right for another, you may flip the sign of the difference. The magnitude stays the same, so the divisibility conclusion is unchanged (because we only care whether the answer is a multiple of 11), but you can confuse yourself by reading negative differences as wrong.

Why it matters: in RPF Constable / SSC arithmetic sections, divisibility by 7, 11 and 13 routinely appears as a "find the smallest number to be added/subtracted to make N divisible by 11" problem. Without a fast rule, you would do long division — and lose 30 seconds per question. With the alternating-sum rule, every such item should take you under 20 seconds.

Real-world example: bank account numbers, ISBN-10 book codes and many product barcodes use modulo-11 check digits to detect typing mistakes. The very same alternating-sum logic is what your phone uses to validate that you typed an ISBN correctly.

Common misconception: "Difference of 11 means the number is not divisible by 11." Wrong. Eleven is itself a multiple of 11, so a difference of 11, 22, 33, … all confirm divisibility by 11.

Common misconception: "Difference must equal exactly 0 to mean divisibility." Wrong. Zero or any multiple of 11 works.

Common misconception: "Position counting must start from the left." It can start from either end, but stay consistent. The magnitude of the difference does not change with direction.

A Faster Mental Shortcut

For exam speed, alternate +/− signs as you sweep left-to-right and sum:

For 483771: +4 − 8 + 3 − 7 + 7 − 1 = −2. |−2| = 2, not a multiple of 11. Not divisible.

For 483769: +4 − 8 + 3 − 7 + 6 − 9 = −11. |−11| = 11. Divisible by 11.

This is identical to the textbook rule, but you write fewer numbers and you finish under a heartbeat. Many toppers prefer it.

Cover the answer, recall, then check. 12 must-know cards on number types and divisibility.

Q1. Divisibility rule for 3?
A1. The digit sum is divisible by 3. e.g. 5,742 → 5+7+4+2 = 18 (÷3) → 5,742 is divisible by 3.

Q2. Divisibility rule for 4?
A2. The last two digits form a number divisible by 4 (or are 00). e.g. 7,316 → 16 ÷ 4 = 4 → divisible.

Q3. Divisibility rule for 8?
A3. The last three digits are divisible by 8. e.g. 25,096 → 096 = 96 = 8×12 → divisible.

Q4. Divisibility rule for 9?
A4. The digit sum is divisible by 9. e.g. 6,831 → 6+8+3+1 = 18 → divisible by 9.

Q5. Divisibility rule for 11?
A5. (Sum of digits in odd places) − (sum in even places) is 0 or a multiple of 11. e.g. 61,809 → (9+8+6) − (0+1) = 23 − 1 = 22 = 11×2 → divisible (61,809 = 11×5,619).

Q6. Divisibility rule for 6?
A6. Divisible by BOTH 2 and 3. e.g. 4,512 → even, and digit sum 12 (÷3) → divisible by 6.

Q7. Divisibility rule for 7?
A7. Double the last digit and subtract it from the rest; if the result is divisible by 7, so is the number. e.g. 672 → 67 − (2×2) = 63 = 7×9 → divisible.

Q8. Divisibility rule for 12?
A8. Divisible by 3 AND 4 together. e.g. 3,516 → digit sum 15 (÷3) and last two digits 16 (÷4) → divisible.

Q9. Divisibility rule for 25?
A9. The last two digits are 00, 25, 50 or 75. e.g. 13,975 → ends in 75 → divisible.

Q10. How many primes are below 100, and which are below 20?
A10. 25 primes below 100. Below 20: 2, 3, 5, 7, 11, 13, 17, 19 (8 primes). Note: 1 is NOT prime; 2 is the only even prime.

Q11. What are co-prime numbers? Give an example.
A11. Two numbers whose HCF is 1, e.g. 8 and 15. They need not be prime individually.

Q12. Classify natural, whole and integer numbers.
A12. Natural = 1,2,3,…; Whole = 0,1,2,3,… (adds 0); Integers = …,−2,−1,0,1,2,… (adds negatives). Every natural number is also whole and an integer.

In the RPF Constable exam the Maths section is short and speed decides everything (~35 questions in ~35 minutes). Number System is the silent hero: divisibility rules don't just answer 1–2 direct questions, they speed up almost every simplification, HCF/LCM and remainder problem. Spotting that a number is divisible by 8 in two seconds saves 20 seconds of long division — and that compounding is why toppers drill these cold.

Why it matters / RPF weightage

Expect 1–3 direct questions on divisibility, primes or number classification, plus indirect use in another 4–5 questions across the paper.

Must-know classification

• Natural: 1, 2, 3, …
• Whole: 0, 1, 2, 3, … (natural + zero)
• Integers: …, −2, −1, 0, 1, 2, …
• Prime: exactly two factors (1 and itself); 2 is the only even prime; 25 primes below 100.
• Composite: more than two factors. Co-prime: HCF = 1.

Divisibility rules

Divisor	Rule
2	Last digit even (0,2,4,6,8)
3	Digit sum divisible by 3
4	Last two digits divisible by 4
5	Ends in 0 or 5
6	Divisible by 2 and 3
8	Last three digits divisible by 8
9	Digit sum divisible by 9
11	(Odd-place sum − even-place sum) = 0 or ×11
25	Last two digits 00/25/50/75

Exam Tricks & Tips

• 🎯 Combine rules for composite divisors: for 12 test 3 & 4, for 15 test 3 & 5, for 18 test 2 & 9 — never learn a separate rule.
• 🎯 For 11, sweep the digits alternately +,−,+,… ; a running total of 0 or ±11 means divisible.
• 🎯 Count factors by prime-factorising and multiplying (power+1)s: 72 = 2³×3² → (3+1)(2+1) = 12 factors.
• 🎯 A perfect square never ends in 2, 3, 7 or 8 — eliminate wrong options at a glance.
• 🎯 Sum of the first n natural numbers = n(n+1)/2; memorise it for series stems.
• ❌ Don't call 1 a prime, and don't forget 2 is prime — both are classic trap options.

Expected exam pattern

Direct stems: "Which of these is divisible by 11?", "How many primes between 1 and 50?", "34x5 is divisible by 9, find x." The missing-digit type is the RPF favourite — crack it with the digit-sum rule in seconds.

Quick recap

Master the nine divisibility rules, remember 25 primes below 100, that 2 is the only even prime and 1 is neither prime nor composite, and use prime factorisation to count factors. These fundamentals pay off across the whole paper.