Chapter 5 • Ganita Prakash | Grade 8
Number Play
🔢 ➕ ➖ 🎯
📋 Table of Contents
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➕ Sum of Consecutive Numbers
🔍 What are Consecutive Numbers?
Numbers that come one after another — like 3, 4, 5 or 7, 8, 9. Their sum shows interesting patterns!
7 = 3 + 4
10 = 1 + 2 + 3 + 4
12 = 3 + 4 + 5
15 = 7 + 8 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5
10 = 1 + 2 + 3 + 4
12 = 3 + 4 + 5
15 = 7 + 8 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5
💡
Key Facts:
• All odd numbers can be written as sum of 2 consecutive numbers (e.g. 7 = 3+4)
• Powers of 2 (2, 4, 8, 16…) CANNOT be written as sum of consecutive natural numbers
• Numbers that ARE NOT powers of 2 can always be written as such a sum
• All odd numbers can be written as sum of 2 consecutive numbers (e.g. 7 = 3+4)
• Powers of 2 (2, 4, 8, 16…) CANNOT be written as sum of consecutive natural numbers
• Numbers that ARE NOT powers of 2 can always be written as such a sum
🌿 4 Consecutive Numbers — Sign Play
Take 4 consecutive numbers (e.g. 3, 4, 5, 6). Place + and − between them in all possible ways → 8 expressions are possible.
3 + 4 + 5 + 6 = 18 | 3 + 4 + 5 − 6 = 6
3 + 4 − 5 + 6 = 8 | 3 + 4 − 5 − 6 = −4
3 − 4 + 5 + 6 = 10 | 3 − 4 + 5 − 6 = −2
3 − 4 − 5 + 6 = 0 | 3 − 4 − 5 − 6 = −12
3 + 4 − 5 + 6 = 8 | 3 + 4 − 5 − 6 = −4
3 − 4 + 5 + 6 = 10 | 3 − 4 + 5 − 6 = −2
3 − 4 − 5 + 6 = 0 | 3 − 4 − 5 − 6 = −12
🌟
Big Discovery! No matter which 4 consecutive numbers you pick, and no matter how you place + and − between them, ALL 8 results are always EVEN numbers! This is called having the same parity.
⚖️ Parity – Even & Odd
📌 What is Parity?
Parity means whether a number is Even or Odd. Two numbers have the same parity if they're both even or both odd.
Parity rules for addition and subtraction
🧠 Why All 8 Expressions Have Same Parity
Explanation 1 (Algebra): When you switch any sign in an expression like a + b − c − d, the value changes by ±2b (or ±2c etc.) — always an even number. So parity never changes!
Explanation 2 (Direct): Since a ± b always has same parity as a + b, extending: all expressions a ± b ± c ± d must have the same parity!
Even Parity of 4 Consecutive Numbers: Let 4 consecutive numbers be n, n+1, n+2, n+3. Their sum = 4n + 6 = 2(2n+3) which is always even. So the base sum is even, and all 8 variations stay even!
4️⃣ Pairs to Make Fours
When does even + even = multiple of 4?
Every even number is either a multiple of 4 (remainder 0) or not a multiple of 4 (remainder 2). This decides what happens when you add two even numbers!
| Case | Form | Sum | Result | Example |
|---|---|---|---|---|
| Multiple of 4 + Multiple of 4 | 4p + 4q | 4(p+q) | ✅ Multiple of 4 | 12 + 16 = 28 |
| Non-multiple + Non-multiple | (4p+2)+(4q+2) | 4(p+q+1) | ✅ Multiple of 4 | 2 + 6 = 8 |
| Multiple of 4 + Non-multiple | 4p+(4q+2) | 4(p+q)+2 | ❌ NOT Multiple of 4 | 4 + 6 = 10 |
🎯
Pattern: Two even numbers add to give a multiple of 4 if and only if they are both multiples of 4 OR both non-multiples of 4 (i.e., they have the same "4-remainder"). This is similar to parity!
🔎 Always, Sometimes, Never
Key Divisibility Properties
We examine statements about factors and multiples to decide if they are Always True, Sometimes True, or Never True.
✅ Always True
If a divides M and a divides N → a divides M+N and M−N
If A is divisible by k → all multiples of A are divisible by k
If A is divisible by k → A is divisible by all factors of k
If A is divisible by k → all multiples of A are divisible by k
If A is divisible by k → A is divisible by all factors of k
🔶 Sometimes True
If a number is divisible by 8, then 8 also divides any two numbers that add up to it (only if both are multiples of 8)
If a number is divisible by 7, then it is also divisible by any multiple of 7 (only if that multiple is a factor)
If a number is divisible by 7, then it is also divisible by any multiple of 7 (only if that multiple is a factor)
❌ Never True
Adding an odd number to an even number gives a multiple of 6 → Never! (because odd + even = odd, and multiples of 6 are always even)
Algebraic proof: 2n + (2m+1) = 6j leads to even = odd — impossible!
Algebraic proof: 2n + (2m+1) = 6j leads to even = odd — impossible!
📐
General Rule for LCM: If A is divisible by k and also divisible by m, then A is divisible by the LCM of k and m. This is because A's prime factorisation must contain the prime factorisation of LCM(k, m).
🧩 What Remains?
Numbers with a Fixed Remainder
Numbers that leave a remainder of 3 when divided by 5 are 3 more than multiples of 5.
Multiples of 5 → 0, 5, 10, 15, 20…
3 more than multiples of 5 → 3, 8, 13, 18, 23…
General form → 5k + 3 (where k = 0, 1, 2, 3…)
3 more than multiples of 5 → 3, 8, 13, 18, 23…
General form → 5k + 3 (where k = 0, 1, 2, 3…)
Same numbers can also be written as 5k − 2 (where k ≥ 1). Both give the same set!
How to write "remainder r when divided by n": The form is nk + r where k = 0, 1, 2, 3... For example: remainder 2 when divided by 7 → numbers are 7k + 2 = 2, 9, 16, 23, 30…
✂️ Divisibility Shortcuts
Why shortcuts work — Algebra!
Any number can be written as: …+ 1000d + 100c + 10b + a where a = units digit, b = tens digit, etc.
Since 10b, 100c, 1000d… are all multiples of 10, the divisibility by 10 depends only on the units digit a!
÷ 9 Shortcut
- Add all digits of the number
- If sum divisible by 9 → number divisible by 9
- Keep adding digits till single digit → that's the remainder!
- Example: 427 → 4+2+7=13 → 1+3=4, so remainder = 4
- Why? Because 10 = 9+1, 100 = 99+1, 1000 = 999+1…
÷ 3 Shortcut
- Add all digits of the number
- If sum divisible by 3 → number divisible by 3
- Works same way as ÷9 (since powers of 10 leave remainder 1 when divided by 3)
- Example: 123 → 1+2+3=6 → divisible by 3 ✓
÷ 11 Shortcut
- Alternating pattern: 1 is 1 more, 10 is 1 less, 100 is 1 more…
- Add digits at ODD positions (units, hundreds…): call it S1
- Add digits at EVEN positions (tens, thousands…): call it S2
- If S1 − S2 = 0 or multiple of 11 → divisible by 11
- Example: 462 → (2+4)−6 = 0 → divisible by 11 ✓
÷ 6, 24 etc.
- Div by 6 → check div by 2 AND div by 3
- Div by 24 → check div by 3 AND div by 8 (NOT 4 and 6!)
- Key: use factors that are coprime (share no common factor)
- Div by 2 → units digit even
- Div by 5 → units digit 0 or 5
🎯 Shortcut Method for ÷ 11
Step 1: Write alternating signs starting from units digit: −, +, −, +…
Step 2: Evaluate the expression
Step 3: Result = remainder when number is divided by 11
Example: 328105
−5 + 0 − 1 + 8 − 2 + 3 = 3
So 328105 is 3 short of (or 8 more than) a multiple of 11
−5 + 0 − 1 + 8 − 2 + 3 = 3
So 328105 is 3 short of (or 8 more than) a multiple of 11
🌱 Digital Roots
What is a Digital Root?
Add the digits of a number repeatedly until you get a single digit. That single digit is the Digital Root.
Example: 489710
4+8+9+7+1+0 = 29
2+9 = 11
1+1 = 2 ← Digital Root!
4+8+9+7+1+0 = 29
2+9 = 11
1+1 = 2 ← Digital Root!
Digital root patterns for different number families
📜
Historical Fact: Aryabhata II (c. 950 CE) in his work Mahāsiddhānta mentioned computing digital roots by repeatedly adding digits. This was used to check arithmetic calculations — an ancient calculator trick!
🔐 Digits in Disguise (Cryptarithms)
What are Cryptarithms?
Puzzles where each letter stands for a digit (0–9). Rules: each digit is represented by at most one letter, and the first digit of a number is never 0.
Example Cryptarithms to remember:
A1 + 1B = B0 → A=7, B=9ON + ON + ON = PO → N=1, O=3, P=9
PQ × 8 = RS → PQ=12, RS=96
BYE × 6 = RAY → B=1, Y is even, <7
🔑
How to solve cryptarithms:
1. Use carry logic — check how digits carry over
2. Use parity — multiplication/addition rules for even/odd
3. Use divisibility shortcuts to narrow options
4. Use place value — a 2-digit × single digit can only give 2 digits in a limited range
1. Use carry logic — check how digits carry over
2. Use parity — multiplication/addition rules for even/odd
3. Use divisibility shortcuts to narrow options
4. Use place value — a 2-digit × single digit can only give 2 digits in a limited range
📋 Quick Summary
1
Consecutive Number Sums
All odd numbers = sum of 2 consecutive numbers. Powers of 2 cannot be expressed as consecutive sums. 4 consecutive numbers with ± always give even results.
2
Parity Rules
odd±odd=even, even±even=even, odd±even=odd. a±b always has same parity as a+b, so all a±b±c±d have same parity.
3
Divisibility Properties
If a|M and a|N → a|(M+N) and a|(M−N). If a|k → a divides all multiples of k. If a|k → a divisible by all factors of k. If a|k and a|m → a divisible by LCM(k,m).
4
Divisibility Shortcuts
÷9: digit sum divisible by 9. ÷3: digit sum divisible by 3. ÷11: alternating digit sum = 0 or multiple of 11. ÷6: divisible by both 2 and 3.
5
Digital Roots & Remainders
Digital root = repeatedly add digits till single digit. It equals the remainder when the number is divided by 9. Numbers of form nk+r all leave remainder r when divided by n.
🗝️ Key Properties at a Glance
| Property | Rule | Example |
|---|---|---|
| Divisibility sum | If a|M and a|N → a|(M+N) | 8|16, 8|24 → 8|40 |
| Divisibility multiples | If a|k → a divides all multiples of k | 7|14 → 7|28, 7|42… |
| Divisibility factors | If a|k → a divisible by all factors of k | 24|96 → 2,3,4,6,8,12|96 |
| Divisibility LCM | If a|k and a|m → a|LCM(k,m) | 36|9 and 36|4 → 36|LCM(9,4)=36 |
| Digit sum rule (9) | Number div by 9 ↔ digit sum div by 9 | 7+3+0+9=19→1+9=10→not div by 9 |
| Digit sum rule (3) | Number div by 3 ↔ digit sum div by 3 | 1+2+3=6 → 123 div by 3 ✓ |
| Alternating sum (11) | Alternating digit sum = 0 or mult of 11 | 462: (2+4)−6=0 → div by 11 ✓ |
| Digital root | = remainder when divided by 9 | 489710 → root=2, 489710÷9 rem=2 |
"Several problems in mathematics can be thought about and solved in different ways. While the method you came up with may be dear to you, it can be amusing and enriching to know how others thought about it. Two tidbits: share and listen."
— Ganita Prakash, Chapter 5
— Ganita Prakash, Chapter 5
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📚 Notes by @edugrown
Chapter 5 • Ganita Prakash Grade 8 • Reprint 2026–27