0
0

Remainder & Modular Arithmetic (Basic)

Introduction

Modular arithmetic बड़े numbers के remainders को आसानी से निकालने में मदद करता है। पूरा division करने की बजाय हम सिर्फ यह देखते हैं कि कोई number किसी divisor से divide होने पर कौन-सा remainder छोड़ता है। यह idea कई aptitude problems में दिखता है-remainder, last-digit और cyclicity questions में खासकर।

Pattern: Remainder & Modular Arithmetic (Basic)

Pattern

मुख्य concept: a ≡ b (mod m) का मतलब है कि a और b को m से divide करने पर वही remainder मिलता है।

  • Definition: a ≡ b (mod m) ⇔ m, (a - b) को divide करता है।
  • Addition & Subtraction: (a ± b) mod m = [(a mod m) ± (b mod m)] mod m.
  • Multiplication: (a × b) mod m = [(a mod m) × (b mod m)] mod m.
  • Power rule: ak mod m को हर multiplication के बाद reduce करें ताकि बड़े numbers न बनें।
  • Cycles: Powers की remainders अक्सर repeat होते हैं-large exponents simplify करने में helpful।
  • Negative remainders: Negative remainder आए तो divisor m add करते जाएँ जब तक remainder positive न हो जाए।

Step-by-Step Example

Question

7100 को 13 से divide करने पर remainder क्या होगा?

Solution

  1. Step 1: Target समझें

    हमें 7100 mod 13 निकालना है-यानी division का remainder।

  2. Step 2: 7 के powers mod 13 में देखें

    7¹ ≡ 7 (mod 13)
    7² = 49 → remainder 10 → 7² ≡ 10
    7³ = 10×7 = 70 → remainder 5 → 7³ ≡ 5
    7⁴ = 5×7 = 35 → remainder 9 → 7⁴ ≡ 9

  3. Step 3: Cycle पहचानें

    Powers continue करने पर 7¹² ≡ 1 (mod 13). इसलिए powers हर 12 steps में repeat होते हैं।

  4. Step 4: Exponent को cycle से simplify करें

    100 ÷ 12 → remainder 4 → यानी 7¹⁰⁰ ≡ 7⁴ (mod 13)

  5. Step 5: पहले मिले result का उपयोग करें

    7⁴ ≡ 9 (mod 13)

  6. Final Answer:

    Remainder = 9

  7. Quick Check:

    7¹⁰⁰ = (7¹²)⁸ × 7⁴ → 1⁸ × 7⁴ → remainder 9. ✅

Quick Variations

1. Cycles का उपयोग करके बड़े powers के remainders निकालें (जैसे 2100 mod 5).

2. Negative numbers handle करें: -3 mod 5 = 2.

3. Equations a·x ≡ b (mod m) को small values try करके या modular inverses से solve करें।

Trick to Always Use

  • Step 1: Numbers को जितना जल्दी हो सके mod m से reduce करें।
  • Step 2: Large powers में cycles ढूँढें-काफी time बचता है।
  • Step 3: Simple equations में small values try करें या inverse ढूँढें।
  • Step 4: Negative remainder मिले तो m add करके positive remainder बनाएं।

Summary

Summary

  • Modular arithmetic remainder problems को fast और simple बनाता है।
  • Addition, subtraction, multiplication और powers को mod m के अनुसार reduce करें।
  • a ≡ b (mod m) का मतलब दोनों numbers का remainder same है।
  • Large exponents के लिए cycles और negative remainders के लिए positive conversion याद रखें।

Example:
7¹² ≡ 1 (mod 13), इसलिए 7¹⁰⁰ ≡ 7⁴ ≡ 9 (mod 13).

Practice

(1/5)
1. Find the remainder when 25 is divided by 7.
easy
A. 3
B. 4
C. 5
D. 6

Solution

  1. Step 1: Understand the division:

    We want the remainder r when 25 is written as 25 = 7×q + r with 0 ≤ r < 7.

  2. Step 2: Find the largest multiple of 7 less than 25:

    7 × 3 = 21, and 7 × 4 = 28 (too big). So use 7 × 3 = 21.

  3. Step 3: Subtract to get remainder:

    25 - 21 = 4 → remainder = 4.

  4. Final Answer:

    Remainder = 4 → Option B.

  5. Quick Check:

    25 = 7×3 + 4, and 0 ≤ 4 < 7, so remainder 4 is correct. ✅
Hint: Subtract the largest multiple of the divisor (7) that doesn't exceed the number.
Common Mistakes: Mixing up quotient and remainder or choosing a multiple that is too large.
2. Find the remainder when 48 is divided by 9.
easy
A. 3
B. 1
C. 5
D. 6

Solution

  1. Step 1: Quick division check:

    Find the nearest multiple of 9 below 48. 9 × 5 = 45 and 9 × 6 = 54 (too big).

  2. Step 2: Subtract to get remainder:

    48 - 45 = 3 → remainder = 3.

  3. Alternative (digit-sum):

    Sum of digits 4 + 8 = 12; 12 mod 9 = 3, which gives the same remainder.

  4. Final Answer:

    Remainder = 3 → Option A.

  5. Quick Check:

    48 = 9×5 + 3 and 0 ≤ 3 < 9, so remainder 3 is correct. ✅
Hint: Use nearest multiple or digit-sum trick for mod 9 problems.
Common Mistakes: Forgetting the remainder must be less than the divisor (9).
3. Find the remainder when 7² + 5² is divided by 6.
easy
A. 1
B. 3
C. 2
D. 0

Solution

  1. Step 1: Compute each square (or reduce first):

    7² = 49 and 5² = 25, so the sum = 49 + 25 = 74.

  2. Step 2: Find remainder on division by 6:

    6 × 12 = 72, 74 - 72 = 2 → remainder = 2.

  3. Modular shortcut (faster):

    7 ≡ 1 (mod 6) so 7² ≡ 1² = 1; 5 ≡ 5 (mod 6) and 5² ≡ 25 ≡ 1 (mod 6); sum ≡ 1+1 = 2 (mod 6).

  4. Final Answer:

    Remainder = 2 → Option C.

  5. Quick Check:

    74 = 6×12 + 2 and remainder 2 is less than 6, so correct. ✅
Hint: Reduce each term modulo m before adding: (a² + b²) mod m = [(a mod m)² + (b mod m)²] mod m.
Common Mistakes: Computing large numbers fully instead of reducing modulo early.
4. Find the remainder when 2³ × 3² is divided by 5.
medium
A. 2
B. 3
C. 4
D. 1

Solution

  1. Step 1: Compute powers separately and reduce as you go:

    2³ = 8 and 3² = 9. We can reduce these modulo 5 first: 8 mod 5 = 3, 9 mod 5 = 4.

  2. Step 2: Multiply reduced values and reduce again:

    3 × 4 = 12; 12 mod 5 = 2 → remainder = 2.

  3. Direct check:

    2³×3² = 8×9 = 72; 72 ÷ 5 leaves remainder 2 (since 5×14 = 70).

  4. Final Answer:

    Remainder = 2 → Option A.

  5. Quick Check:

    72 = 5×14 + 2, remainder 2 is correct. ✅
Hint: Reduce each factor modulo the divisor before multiplying to keep numbers small.
Common Mistakes: Multiplying large numbers first and risking arithmetic errors instead of reducing early.
5. Find the remainder when 7¹⁰ is divided by 12.
medium
A. 3
B. 7
C. 9
D. 1

Solution

  1. Step 1: Look for a small cycle in powers of 7 modulo 12:

    Compute a few powers: 7¹ ≡ 7 (mod 12). 7² = 49; 49 mod 12 = 1 (because 12×4=48).

  2. Step 2: Use the cycle:

    Since 7² ≡ 1 (mod 12), higher even powers repeat 1. So 7¹⁰ = (7²)⁵ ≡ 1⁵ ≡ 1 (mod 12).

  3. Final Answer:

    Remainder = 1 → Option D.

  4. Quick Check:

    The cycle shows 7² ≡ 1, so any 7^{even} ≡ 1; 10 is even → remainder 1. ✅
Hint: Find the smallest power where remainder becomes 1 or repeats, then reduce exponent modulo cycle length.
Common Mistakes: Trying to compute 7¹⁰ directly instead of using the repeating pattern of powers.

Mock Test

Ready for a challenge?

Take a 10-minute AI-powered test with 10 questions (Easy-Medium-Hard mix) and get instant SWOT analysis of your performance!

10 Questions
5 Minutes