11. Remainders & Modulo Arithmetic

Remainder concepts + modulo arithmetic (mod): division algorithm, remainder rules, cyclicity base, negative mod basics, SSC CGL level examples & practice questions with answers.

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1) Division Algorithm (base concept)

Kisi bhi integers a (dividend) aur b (>0) (divisor) ke liye: a = bq + r jahan 0 ≤ r < b. Yahi r remainder hota hai.

• Example: 29 = 7×4 + 1 ⇒ remainder 1
• Important: remainder always divisor se chhota hota hai.

2) Modulo (mod) kya hota hai?

a mod n ka matlab: a ko n se divide karne par jo remainder aaye.

• 17 mod 5 = 2
• 100 mod 8 = 4

3) Congruence (≡) — SSC friendly meaning

Agar a aur b ko n se divide karne par same remainder aaye, to: a ≡ b (mod n)

• Example: 17 ≡ 2 (mod 5) because both leave remainder 2 when divided by 5.
• Shortcut: a ≡ b (mod n) ⇔ (a − b) divisible by n.

4) Most-used modulo rules (must learn)

• (a + b) mod n = ((a mod n) + (b mod n)) mod n
• (a − b) mod n = ((a mod n) − (b mod n)) mod n (negative aaya to n add kar do)
• (a × b) mod n = ((a mod n) × (b mod n)) mod n
• (a^k) mod n: pehle a mod n reduce, then power/cyclicity use

5) Quick examples (reduce first)

Example 1

(123 + 789) mod 10

• 123 mod 10 = 3, 789 mod 10 = 9
• (3 + 9) mod 10 = 12 mod 10 = 2

Example 2

(98 × 77) mod 7

• 98 mod 7 = 0 ⇒ (0 × 77) mod 7 = 0

6) Negative modulo (simple handling)

Exam me kabhi kabhi subtraction me negative aa jata hai. Rule: answer ko 0 to n−1 ke beech laane ke liye n add kar do.

• Example: (2 − 5) mod 7 = (−3) mod 7 = 4 (because −3 + 7 = 4)

7) Standard remainder patterns (SSC)

A) (a + b) divisible by n

Agar a mod n = r, to (n − r) mod n = n − r (unless r=0). So a + (n − r) is divisible by n.

• Example: 17 mod 5 = 2 ⇒ add 3 to make divisible by 5: 17+3=20

B) If a ≡ b (mod n) then a^k ≡ b^k (mod n)

• Example: 12 ≡ 2 (mod 5) ⇒ 12⁴ mod 5 = 2⁴ mod 5 = 16 mod 5 = 1

8) Classic SSC-type remainder questions

Example 1: Big power remainder

Find remainder when 3¹⁰⁰ is divided by 7.

• 3 mod 7 = 3
• 3²=9 ≡ 2 (mod 7)
• 3³≡ 6, 3⁴≡ 4, 3⁵≡ 5, 3⁶≡ 1 (cycle length 6)
• 100 mod 6 = 4 ⇒ remainder = 3⁴ mod 7 = 4

Example 2: Expression remainder

Find remainder when (10⁵⁰ + 3) is divided by 7.

• 10 mod 7 = 3 ⇒ 10⁵⁰ mod 7 = 3⁵⁰ mod 7
• 3 cycle mod 7 has length 6 (as above): 50 mod 6 = 2 ⇒ 3⁵⁰ mod 7 = 3² mod 7 = 2
• (2 + 3) mod 7 = 5

9) Common traps

• Remainder always 0 to (divisor−1) ke beech hota hai.
• Before big calculation, numbers ko mod n me reduce karo (life easy).
• Subtraction me negative aaye to n add karke positive remainder lo.

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10) Practice (SSC CGL) + Answers

1. Find remainder when 257 is divided by 9.
2. Find (3456 mod 8).
3. Find remainder when 2²⁵ is divided by 7.
4. Find remainder when (99⁵ + 1) is divided by 10.
5. (15×17 + 19) mod 6 ?
6. (5 − 19) mod 7 ?

Show Answers

1. 257 = 9×28 + 5 ⇒ remainder 5
2. 3456 divisible by 8? last3=456, 456÷8=57 ⇒ remainder 0
3. 2 mod 7 cycle: 2,4,1,2... length 3. 25 mod 3 = 1 ⇒ remainder 2
4. 99 mod 10 = 9 ⇒ 9^odd mod 10 = 9 ⇒ (9 + 1) mod 10 = 0
5. 15 mod 6=3, 17 mod 6=5 ⇒ 3×5=15 mod 6=3; 19 mod 6=1 ⇒ (3+1)=4
6. (5−19)=-14; -14 mod 7 = 0