Divisibility rules — 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 (with proofs where possible)

2.5.1 · Maths › Number Theory (Intermediate)

Core Idea

Ek number ko ek recipe ki tarah socho: $n = d_{k} \cdot 1 0^{k} + d_{k - 1} \cdot 1 0^{k - 1} + \dots + d_{1} \cdot 10 + d_{0}$ . Divisibility rules "Kya ye recipe evenly divide hoti hai?" ko "Kya ye ingredients (digits) ek simpler check satisfy karte hain?" mein translate kar dete hain.

Foundation: Place Value Representation

Koi bhi positive integer is tarah likha ja sakta hai: $n = d_{k} \cdot 1 0^{k} + d_{k - 1} \cdot 1 0^{k - 1} + \dots + d_{1} \cdot 10 + d_{0}$

jahan $d_{i}$ digits hain (0–9). For example, $n = 1234 = 1 \cdot 1 0^{3} + 2 \cdot 1 0^{2} + 3 \cdot 10 + 4$ .

Key insight: Agar hum $m$ se divisibility test karna chahte hain, to hum poochte hain ki kya $n \equiv 0 (mod m)$ hai. Hum 10 ki har power ko modulo $m$ reduce kar sakte hain aur sirf digits ke saath kaam kar sakte hain.

The Rules & Proofs

Rule 1: Divisibility by 2

Kyun? $n = all multiples of 10, hence even d_{k} \cdot 1 0^{k} + \dots + d_{1} \cdot 10 + d_{0}$

Kyunki $10 \equiv 0 (mod 2)$ hai, 10 ki har power 2 se divisible hai. To: $n \equiv d_{0} (mod 2)$

Example: $n = 1238$ . Last digit 8 hai (even) → 2 se divisible. Check: $1238 = 2 \times 619$ . ✓

Rule 2: Divisibility by 3

Proof from first principles: $n = d_{k} \cdot 1 0^{k} + d_{k - 1} \cdot 1 0^{k - 1} + \dots + d_{1} \cdot 10 + d_{0}$

Observe karo ki $10 \equiv 1 (mod 3)$ hai (kyunki $10 = 3 \times 3 + 1$ ). Isliye: $1 0^{2} \equiv 1^{2} = 1 (mod 3), 1 0^{3} \equiv 1 (mod 3), etc.$

To: $n \equiv d_{k} \cdot 1 + d_{k - 1} \cdot 1 + \dots + d_{1} \cdot 1 + d_{0} (mod 3)$ $n \equiv d_{k} + d_{k - 1} + \dots + d_{1} + d_{0} (mod 3)$

Ye step kyun? Humne har $1 0^{i}$ ko 1 se replace kiya kyunki sab mod 3 mein 1 ke congruent hain.

Rule 3: Divisibility by 4

Proof: $n = multiples of 100 d_{k} \cdot 1 0^{k} + \dots + d_{2} \cdot 1 0^{2} + d_{1} \cdot 10 + d_{0}$

Kyunki $100 = 4 \times 25$ hai, to $100 \equiv 0 (mod 4)$ , aur isliye $i \geq 2$ ke liye $1 0^{i}$ wale sab terms vanish ho jaate hain: $n \equiv d_{1} \cdot 10 + d_{0} (mod 4)$

Ye step kyun? Humne last two digits ko isolate kiya kyunki 10 ki higher powers sab 100 ke multiples hain, jo 4 se divisible hai.

Rule 4: Divisibility by 5

Proof: $n \equiv d_{0} (mod 5)$ kyunki $10 \equiv 0 (mod 5)$ hai. Saari higher powers vanish ho jaati hain.

Example: $n = 1235$ . Last digit 5 hai → 5 se divisible. Check: $1235 = 5 \times 247$ . ✓

Rule 5: Divisibility by 6

Kyun? $6 = 2 \times 3$ aur $g cd (2, 3) = 1$ hai. Agar $n \equiv 0 (mod 2)$ aur $n \equiv 0 (mod 3)$ , to Chinese Remainder Theorem se $n \equiv 0 (mod 6)$ .

Rule 6: Divisibility by 7

Proof sketch: Maano $n = 10 a + b$ jahan $b$ last digit hai aur $a$ bacha hua part hai.

Hum ek aisa linear combination dhundhna chahte hain jo $b$ wale factor ko modulo 7 eliminate kare. Note karo ki $10 \equiv 3 (mod 7)$ hai, to: $n \equiv 3 a + b (mod 7)$

Hum ek naya number $a - k b$ construct karna chahte hain jo divisibility preserve kare. Agar hum $k = 2$ choose karein: $a - 2 b \equiv a - 2 b (mod 7)$

Humhe verify karna hai: $n \equiv 0 (mod 7)$ iff $a - 2 b \equiv 0 (mod 7)$ .

Ye step kyun? $n = 10 a + b$ se: $n \equiv 0 (mod 7) ⟺ 10 a + b \equiv 0 (mod 7) ⟺ 3 a + b \equiv 0 (mod 7)$

Dono sides ko 5 se multiply karo (kyunki $5 \times 3 = 15 \equiv 1 (mod 7)$ hai, to 5, 3 ka modular inverse hai): $5 (3 a + b) \equiv 0 (mod 7) ⟺ 15 a + 5 b \equiv 0 (mod 7) ⟺ a + 5 b \equiv 0 (mod 7)$

Lekin $5 b \equiv - 2 b (mod 7)$ hai (kyunki $5 \equiv - 2 (mod 7)$ ), to: $a - 2 b \equiv 0 (mod 7)$

Ek aur: $n = 1057$

$a = 105$ , $b = 7$
$a - 2 b = 105 - 14 = 91$
$a = 9$ , $b = 1$ : $9 - 2 = 7$ ✓
To $1057$ divisible hai 7 se. Check: $1057 = 7 \times 151$ . ✓

Rule 7: Divisibility by 8

Proof: $n \equiv d_{2} \cdot 1 0^{2} + d_{1} \cdot 10 + d_{0} (mod 8)$ kyunki $1000 = 8 \times 125 \equiv 0 (mod 8)$ hai.

Ye step kyun? $1 0^{3}$ aur uski higher powers sab 8 se divisible hain.

Rule 8: Divisibility by 9

Proof: 3 wale divisibility proof se bilkul identical. $10 \equiv 1 (mod 9) ⟹ n \equiv \sum d_{i} (mod 9)$

Rule 9: Divisibility by 10

Proof: $n \equiv d_{0} (mod 10)$ , to $n \equiv 0 (mod 10) ⟺ d_{0} = 0$ .

Example: $n = 1230$ . Last digit 0 hai → 10 se divisible. ✓

Rule 10: Divisibility by 11

Proof: $n = d_{k} \cdot 1 0^{k} + d_{k - 1} \cdot 1 0^{k - 1} + \dots + d_{1} \cdot 10 + d_{0}$

Observe karo ki $10 \equiv - 1 (mod 11)$ hai. Isliye: $1 0^{0} \equiv 1, 1 0^{1} \equiv - 1, 1 0^{2} \equiv 1, 1 0^{3} \equiv - 1, \dots$

To: $n \equiv d_{k} \cdot (- 1)^{k} + d_{k - 1} \cdot (- 1)^{k - 1} + \dots - d_{1} + d_{0} (mod 11)$

Ye step kyun? Humne 10 ki har power ko $(- 1)^{i}$ se replace kiya, jo ek alternating pattern create karta hai.

Practical use ke liye compute karo: $(d_{0} - d_{1} + d_{2} - d_{3} + \dots)$ ya equivalently (odd positions ke digits ka sum) − (even positions ke digits ka sum).

Visual Summary

Common Mistakes

Kyun galat hai: $6 = 2 \times 3$ . Number DONO se divisible hona chahiye. Example: 15, 3 se divisible hai lekin 2 se nahi, isliye 6 se bhi nahi.

Fix: DONO conditions check karo. Even last digit AUR sum 3 se divisible.

Kyun galat hai: Key hai $10 \equiv - 1 (mod 11)$ , na ki $10 \equiv 1$ . Isse ek alternating pattern create hota hai, plain sum nahi.

Fix: Signs alternate karo: $(d_{0} - d_{1} + d_{2} - d_{3} + \dots)$ . Pattern matter karta hai!

Kyun galat hai: Algebraic proof specifically subtraction require karta hai taaki modular relationship preserve ho.

Fix: Yaad rakho: "7 hai heaven, to wahan pahunchne ke liye SUBTRACT karo." Hamesha $a - 2 b$ .

Active Recall

Recall Divisibility rules ek 12-saal ke bacche ko explain karo

Socho tumhare paas ek bada number hai, jaise tumhare phone ka passcode ya tumhara favourite cricket score. Tum jaanna chahte ho ki kya tum ise evenly groups mein split kar sakte ho (jaise dosto mein chocolates baantna) kuch numbers se — 2, 3, 5, etc.

Instead of doing long division every time, kuch shortcuts hain!

Divisibility by 2: Bas last digit check karo. Agar wo even hai (0, 2, 4, 6, 8), to tum 2 se divide kar sakte ho. Kyun? Kyunki number mein baaki sab kuch already 10 ka multiple hai, jo 2 se divisible hai.
Divisibility by 3 ya 9: Saare digits add kar lo. Agar wo sum 3 (ya 9) se divide ho jaaye, to pura number bhi ho sakta hai! Ye ek cool pattern ki wajah se kaam karta hai: 10, 100, 1000… sab 3 ya 9 se divide hone par remainder 1 dete hain.
Divisibility by 5: Last digit 0 ya 5 hai? Done! Bilkul paison ki tarah (₹5, ₹10).
Divisibility by 11: Ye wala fun hai! Digits ko alternately add aur subtract karo (right se start karo). Agar 0 ya 11 ka multiple mile, to number 11 se divisible hai.

Ye rules time bachate hain aur tumhe math wizard jaisa dikhate hain!

Mnemonic

4, 8: Last 2 ya 3 digits dekho

4: last 2 digits
8: last 3 digits

3, 9: Saare digits ka Sum karo

Pattern: powers of 10 ≡ 1 hain

11: Alternating sum (kyunki $10 \equiv - 1 (mod 11)$ )

6: 2 aur 3 dono pass karne chahiye

7: Mushkil wala — last digit ka do guna subtract karo

"2-5-10 end dekho, 4-8 ko do ya teen doston ki zaroorat, 3-9 sum se aage badho, 11 alternate karta rehta hai, 6 ko bachane ko do rules chahiye, 7 woh hai jahan subtract karke neeche aao."

Connections Modular Arithmetic — divisibility rules, congruence ke applications hain

Chinese Remainder Theorem — isliye divisibility by 6 = divisibility by 2 AND 3
Place Value System — digit manipulation samajhne ki foundation
GCD and LCM — composite divisors jaise 6, 10, 12 prime factorization par rely karte hain
Casting Out Nines — digit sum techniques jo error checking ke liye use hoti hain
Fermat's Little Theorem — modular arithmetic wale deeper number-theoretic patterns

Summary Table

| Divisor | Rule | Key Modular Fact | |---------|---------------| | 2 | Last digit even | $10 \equiv 0 (mod 2)$ | | 3 | Digits ka sum 3 se divisible | $10 \equiv 1 (mod 3)$ | | 4 | Last 2 digits 4 se divisible | $100 \equiv 0 (mod 4)$ | | 5 | Last digit 0 ya 5 | $10 \equiv 0 (mod 5)$ | | 6 | 2 aur 3 dono se divisible | $g cd (2, 3) = 1$ | | 7 | $a - 2 b$ divisible by 7 | $10 \equiv 3 (mod 7)$ | | 8 | Last 3 digits 8 se divisible | $1000 \equiv 0 (mod 8)$ | | 9 | Digits ka sum 9 se divisible | $10 \equiv 1 (mod 9)$ | | 10 | Last digit 0 hai | $n \equiv d_{0} (mod 10)$ | | 11 | Alternating digit sum 11 se divisible | $10 \equiv - 1 (mod 11)$ |

#flashcards/maths

2 ke liye divisibility rule kya hai? :: Ek number 2 se divisible hai agar uska last digit even ho (0, 2, 4, 6, 8).

3 ke liye divisibility rule kyun kaam karta hai?

Kyunki

10 \equiv 1 (mod 3)

hai, isliye 10 ki har power bhi ≡ 1 mod 3 hai. Iska matlab hai

n \equiv \sum digits (mod 3)

4 ke liye divisibility rule kya hai?

Ek number 4 se divisible hai agar uske last two digits se bana number 4 se divisible ho, kyunki

100 \equiv 0 (mod 4)

6 ke liye divisibility rule kya hai?

Ek number 6 se divisible hai agar wo 2 AUR 3 DONO se divisible ho (kyunki

6 = 2 \times 3

aur

g cd (2, 3) = 1

7 se divisibility kaise test karte hain?

Number ka last digit hata do (

a

), aur last digit ka do guna subtract karo (

2 b

). Agar

a - 2 b

divisible hai 7 se, to original number bhi hai.

7 ke rule mein subtraction kyun hai, addition kyun nahi?

Algebraic proof require karta hai

a - 2 b

taaki equivalence

n \equiv 0 (mod 7)

preserve ho, jo

10 \equiv 3 (mod 7)

aur modular inverses se derive hoti hai.

8 ke liye divisibility rule kya hai?

Ek number 8 se divisible hai agar uske last three digits se bana number 8 se divisible ho, kyunki

1000 \equiv 0 (mod 8)

9 ke liye divisibility rule kya hai?

Ek number 9 se divisible hai agar uske digits ka sum 9 se divisible ho, kyunki

10 \equiv 1 (mod 9)

11 ke liye divisibility rule kya hai?

Ek number 11 se divisible hai agar uske digits ka alternating sum (right se start karke:

d_{0} - d_{1} + d_{2} - d_{3} + \dots

) 11 se divisible ho.

11 ke rule mein alternating signs kyun hote hain?

Kyunki

10 \equiv - 1 (mod 11)

hai, isliye 10 ki powers alternate karti hain:

1 0^{0} \equiv 1

1 0^{1} \equiv - 1

1 0^{2} \equiv 1

, etc.

Kya 1236 divisible hai 6 se? Kaise check karoge?

Haan. 2 se divisibility check karo: last digit 6 hai (even) ✓. 3 se divisibility check karo: digit sum =

1 + 2 + 3 + 6 = 12

, 3 se divisible ✓. To 6 se divisible hai.

Kya 2728 divisible hai 11 se? :: Haan. Alternating sum: $8 - 2 + 7 - 2 = 11$ , jo 11 se divisible hai.

Common mistake: "6 se divisible matlab 2 YA 3 se divisible." Ye galat kyun hai?

DONO se divisible hona chahiye — 2 AUR 3. Example: 15, 3 se divisible hai lekin 2 se nahi, isliye 6 se bhi nahi.