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A dedicated and disciplined Mathematics educator at Sainik School Nalanda, recognized as a Gemini Certified Educator and Google AI Certified Educator. Dedicated to transforming abstract math into joyful learning through interactive digital tools, NEP‑aligned resources, and creative publications.

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2. Formulate

Crafting digital TLMs and NEP‑aligned worksheets around that gap.

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3. Execute

Implementing interactive, NEP 2020‑aligned methodologies in the classroom.

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I design interactive, NEP-aligned digital worksheets tailored for conceptual clarity, self-paced practice, and immediate feedback.
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Yes! I specialize in creating digital Teaching-Learning Models (TLMs) and virtual visual tools.
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✦ NCERT GRADE 8 GANITA PRAKASH • NCF 2023 ✦

Chapter 5: Number Play

Exploring Parity, Divisibility Rules, Algebraic Generalisation, Digital Roots & Cryptarithms

๐Ÿ”ข 1. Parity Invariance

Even/Odd Behavior

Take any 4 consecutive numbers (such as 3, 4, 5, 6). Placing any combination of + or − across the 8 possibilities always yields an even number!

Algebraic Proof: Switching a sign from +b to −b changes the total by (a + b − c − d) − (a − b − c − d) = 2b. Since 2b is always an even number, the overall parity never changes!

• Odd ± Odd = Even
• Even ± Even = Even
• Odd ± Even = Odd

⚡ 2. Pairs to Make Fours

Remainder Class Logic

Even numbers divided by 4 leave either remainder 0 (form 4p) or remainder 2 (form 4p + 2).

  • Multiple + Multiple: 4p + 4q = 4(p + q) → Multiple of 4.
  • Non-Multiple + Non-Multiple: (4p + 2) + (4q + 2) = 4(p + q + 1) → Multiple of 4 (remainders 2 + 2 = 4)!
  • Multiple + Non-Multiple: 4p + (4q + 2) = 4(p + q) + 2 → Not a multiple of 4.

๐Ÿ“œ 3. Divisibility Theorems

Universal Laws
  • If a divides M and N, then a divides (M + N) and (M − N).
  • If A is divisible by k, then all multiples of A are divisible by k.
  • If A is divisible by k, then A is divisible by all factors of k.
  • If A is divisible by coprime numbers k and m, then A is divisible by LCM(k, m).

➗ 4. Divisibility by 9 & 3

Place Value Decomposition

Powers of 10 are always 1 more than a multiple of 9:
10 = 9 + 1, 100 = 99 + 1, 1000 = 999 + 1.

dcba = d(999 + 1) + c(99 + 1) + b(9 + 1) + a
= 9 × (Integer) + (d + c + b + a)

Rule: A number is divisible by 9 (or 3) if and only if the sum of its digits is divisible by 9 (or 3).

⚖️ 5. Divisibility by 11

Alternating Excess & Deficit

Place values alternate between 1 more (+1) and 1 less (−1) than multiples of 11:
1 = 0 + 1, 10 = 11 − 1, 100 = 99 + 1, 1000 = 1001 − 1.

Rule: (Sum of Odd Place Digits) − (Sum of Even Place Digits). If the difference is 0 or a multiple of 11, the number is divisible by 11!

๐ŸŒฑ 6. Digital Roots & Roots Club

Mahฤsiddhฤnta Heritage

Repeatedly summing digits until a single digit remains yields the Digital Root (e.g., 489710 → 29 → 11 → 2).

  • Codified by Aryabhata II (c. 950 CE) in Mahฤsiddhฤnta for checking calculations.
  • Digital root equals the remainder when divided by 9 (if the root is 9, the remainder is 0).

๐Ÿ” Master Divisibility Shortcut Matrix

Summary of testing rules and algebraic foundations from the Ganita Prakash curriculum:

Divisor Testing Rule Algebraic Basis Example
2, 5, 10 Check only the Units Digit (last digit). 10b, 100c, 1000d are already multiples of 2, 5, 10. 639210 (Divisible by 2, 5, 10)
4 & 8 Last 2 digits divisible by 4; Last 3 digits divisible by 8. 100 is a multiple of 4; 1000 is a multiple of 8. 2856 (56 ÷ 4, 856 ÷ 8)
3 & 9 Sum of all digits is divisible by 3 or 9. 10n leaves remainder 1 for each place value. 405 (4 + 0 + 5 = 9)
6 Divisible by both 2 and 3 simultaneously. LCM(2, 3) = 6 because 2 and 3 are coprime. 429714 (Even and sum = 27)
11 (Sum of Odd Places) − (Sum of Even Places) is 0 or multiple of 11. 10n alternates as (11k + 1) and (11k − 1). 90904 (13 − 2 = 11)
36 & 44 Divisible by (4 and 9) or (4 and 11) respectively. Must use coprime factor pairs. 45036 (Divisible by 36)

๐ŸŽญ 7. Digits in Disguise (Cryptarithms)

Letter-digit puzzles where each distinct letter represents a unique digit (0–9) and the first digit is never 0:

Addition Puzzles

• A1 + 1B = B0 → A = 7, B = 9 (71 + 19 = 90).
• ON + ON + ON = PO → O = 3, N = 1, P = 9 (31 × 3 = 93).

Multiplication Puzzles

• PQ × 8 = RS → 12 × 8 = 96 (Unique non-repeating digits).
• JK × 6 = KKK → 74 × 6 = 444 (J = 7, K = 4).

๐ŸŽฎ 8. Navakankari & Remainder Puzzles

Navakankari (Nine Men's Morris / Sฤlu Mane Aศ›a): Traditional Indian alignment strategy game using 9 pawns to make lines of three and block opponent moves.

The Ancient Pebble Riddle

"Grouped by 2, 3, or 5 leaves remainder 1; divisible perfectly by 7; total < 100."
LCM(2, 3, 5) = 30 → Numbers of the form 30k + 1.
Candidates: 31, 61, 91. Since 91 = 7 × 13, the answer is 91 pebbles!

⚠️ Common Pitfalls & Conceptual Traps

๐Ÿ›‘

Non-Coprime Factors Trap

Checking divisibility by composite non-coprime factors fails! For instance, 12 is divisible by 4 and 6, but not by 24. Always check coprime pairs (e.g., 3 and 8 for 24).

๐Ÿ”„

Digit Reversal Invariance

Reversing digits preserves divisibility only for 3 and 9 (sum of digits remains identical). It does not hold for 2, 4, 5, 8, 10, or 11 where place position determines the value!

0️⃣

Leading Zero & Root 9

In multi-digit letter representations like abc, the first letter a ≠ 0. Also, a digital root of 9 corresponds to a remainder of 0 when divided by 9.

Designed by Akash Srivastava | Mathematics Hub • Visual Learning Series
๐Ÿš€ Explore Interactive TLMs, Worksheets & Visual Notes at: www.akashmaths.online