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JAC • Class 8 • Mathematics • Ch 5
Estimated Time: 45 Mins
Study Progress: In Progress

Playing with Numbers

JCERT · Class 8 · Chapter 5 · Playing with Numbers. Syllabus-aligned topic notes and teaching examples, not a complete textbook answer key.

Why This Chapter Matters

JCERT · Class 8 · Chapter 5 · Playing with Numbers. Syllabus-aligned topic notes and teaching examples, not a complete textbook answer key.

Before You Begin (Prerequisites)

  • Revise the relevant ideas from the previous grade.

What You Will Learn (Core Objectives)

  • Write a two-digit number as10a+b and a three-digit number as100a+10b+c. Use place value and carries in reversals, puzzles and cryptarithms. Explain divisibility tests using place value; repeated letters represent the same digit.

Chapter Roadmap & Progression

1 Syllabus, textbook and notes alignm...
2 Concept and method
3 Check and common pitfall

Complete Concept Guide (100% Curriculum Coverage)

Syllabus, textbook and notes alignment

JCERT · Regular schools (Non-SOE) · Class 8 · Chapter 5 — Playing with Numbers
Official syllabus pages 87–97. Official syllabus · Prescribed Hindi book (17 chapters).

These topic notes and teaching examples are not a complete answer key to every textbook exercise. English notes use this same JCERT chapter order; a matching English textbook PDF has not been verified.

Official Hindi PDF for this exact chapter

Concept and method

Write a two-digit number as10a+b and a three-digit number as100a+10b+c. Use place value and carries in reversals, puzzles and cryptarithms. Explain divisibility tests using place value; repeated letters represent the same digit.

Check and common pitfall

For example72−27=45=9×5. A leading digit cannot be zero when the number of digits is specified.

Key Formulas, Identities & Theorems

Key relationships
(10a+b)−(10b+a)=9(a−b); divisibility by9 uses digit sum
Observe formula conditions and units.

Conceptual Solved Examples & Case Studies

Example 1
Why is the difference between a two-digit number and its reversal divisible by9?
Step-by-Step Solution:

(10a+b)−(10b+a)=9(a−b)

For example72−27=45=9×5. A leading digit cannot be zero when the number of digits is specified.

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