📚 Mathematics • Class 8 • NCERT Ganita Prakash
🔢 Chapter 5: Number Play
Explore the magical world of numbers — patterns, divisibility, digital roots, and more!
📌 Parity
➗ Divisibility Rules
🔍 Remainders
🌱 Digital Roots
🔐 Cryptarithms
🧮 Algebra
➗ Divisibility Rules
🔍 Remainders
🌱 Digital Roots
🔐 Cryptarithms
🧮 Algebra
📋 Contents
- Introduction to Number Play
- Sums of Consecutive Numbers
- Parity — Even & Odd
- Breaking Even: Even Number Expressions
- Pairs to Make Fours
- Always, Sometimes, or Never True
- What Remains? — Remainders & Algebra
- Section 5.2 — Checking Divisibility Quickly
- Divisibility by 9
- Divisibility by 3
- Divisibility by 11
- Digital Roots
- Section 5.3 — Digits in Disguise (Cryptarithms)
- Chapter Summary
- Important Exam Questions
Introduction to Number Play
This chapter explores fascinating patterns and properties of numbers using algebra, reasoning, and visualisation. We will learn how numbers behave under different operations and discover clever shortcuts.
What will you learn?
In this chapter, you will explore sums of consecutive numbers, understand parity (even/odd), learn divisibility shortcuts for 3, 9, and 11, discover digital roots, and solve fun cryptarithms (digit puzzles).
In this chapter, you will explore sums of consecutive numbers, understand parity (even/odd), learn divisibility shortcuts for 3, 9, and 11, discover digital roots, and solve fun cryptarithms (digit puzzles).
Key Idea
Mathematics is not just about getting answers — it’s about understanding why something works. We use algebra and visualisation to explain patterns that seem magical!
Mathematics is not just about getting answers — it’s about understanding why something works. We use algebra and visualisation to explain patterns that seem magical!
5.1 Sums of Consecutive Numbers
Consecutive numbers are numbers that follow each other one after another (e.g., 3, 4, 5, 6).
📌 Examples from the Textbook
7 = 3 + 4 (sum of 2 consecutive)
10 = 1 + 2 + 3 + 4 (sum of 4 consecutive)
12 = 3 + 4 + 5 (sum of 3 consecutive)
15 = 7 + 8 = 4+5+6 = 1+2+3+4+5 (multiple ways!)
10 = 1 + 2 + 3 + 4 (sum of 4 consecutive)
12 = 3 + 4 + 5 (sum of 3 consecutive)
15 = 7 + 8 = 4+5+6 = 1+2+3+4+5 (multiple ways!)
🤔 Interesting Questions to Explore
- Can every natural number be written as a sum of consecutive numbers?
- Which numbers can be written as a sum of consecutive numbers in more than one way?
- All odd numbers can be written as a sum of two consecutive numbers. Can all even numbers?
- Powers of 2 (like 2, 4, 8, 16…) cannot be written as sum of consecutive natural numbers!
➕➖ Four Consecutive Numbers with + and – Signs
Take any 4 consecutive numbers (e.g., 3, 4, 5, 6). Place + or – between them in all possible ways. There are 8 different expressions.
With 4 consecutive numbers a, b, c, d → There are exactly 8 expressions of the form a ± b ± c ± d
Key Discovery!
No matter which 4 consecutive numbers you choose and no matter how you place + and – between them, the result is always an even number! This is called even parity.
No matter which 4 consecutive numbers you choose and no matter how you place + and – between them, the result is always an even number! This is called even parity.
3 + 4 – 5 + 6 = 8 (even ✓)
3 – 4 – 5 – 6 = –12 (even ✓)
5 + 6 – 7 + 8 = 12 (even ✓)
5 – 6 – 7 – 8 = –16 (even ✓)
3 – 4 – 5 – 6 = –12 (even ✓)
5 + 6 – 7 + 8 = 12 (even ✓)
5 – 6 – 7 – 8 = –16 (even ✓)
💡 Why Is This Always Even? (3 Explanations)
Explanation 1 — Sign Switch
When you switch the sign of any term (e.g., +b → –b), the value changes by 2b, which is always even. So all 8 expressions have the same parity.
When you switch the sign of any term (e.g., +b → –b), the value changes by 2b, which is always even. So all 8 expressions have the same parity.
Explanation 2 — Parity Rules
odd ± odd = even, even ± even = even, odd ± even = odd. Since a ± b always has the same parity, all expressions a ± b ± c ± d share the same parity.
odd ± odd = even, even ± even = even, odd ± even = odd. Since a ± b always has the same parity, all expressions a ± b ± c ± d share the same parity.
General Rule
This is NOT limited to 4 numbers! For any set of numbers, all expressions formed by placing + and – between them have the same parity.
This is NOT limited to 4 numbers! For any set of numbers, all expressions formed by placing + and – between them have the same parity.
Parity — Understanding Even & Odd
Parity refers to whether a number is even or odd.
Even number = any integer of the form 2n | Odd number = any integer of the form 2n + 1
📊 Parity Rules for Operations
| Operation | Result | Example |
|---|---|---|
| odd ± odd | even | 3 + 5 = 8 ✓ |
| even ± even | even | 4 + 6 = 10 ✓ |
| odd ± even | odd | 3 + 4 = 7 ✓ |
| odd × odd | odd | 3 × 5 = 15 ✓ |
| even × any | even | 4 × 7 = 28 ✓ |
Quick Tip
a + b and a − b always have the SAME parity, regardless of what a and b are. This is a very useful fact!
a + b and a − b always have the SAME parity, regardless of what a and b are. This is a very useful fact!
Breaking Even — Even Number Expressions
We can check if arithmetic or algebraic expressions always give even numbers without actually calculating.
🔢 Arithmetic Expressions — Which Are Even?
| Expression | Even or Odd? | Reason |
|---|---|---|
| 43 + 37 | ✅ Even | odd + odd = even |
| 672 − 348 | ✅ Even | even − even = even |
| 4 × 347 × 3 | ✅ Even | has factor 4 (even) |
| 708 − 477 | ❌ Odd | even − odd = odd |
| 809 + 214 | ❌ Odd | odd + even = odd |
| 119 × 303 | ❌ Odd | odd × odd = odd |
| 543 − 479 | ✅ Even | odd − odd = even |
| 513³ | ❌ Odd | odd³ = odd |
🔡 Algebraic Expressions — Always Even?
| Expression | Always Even? | Reason |
|---|---|---|
| 2a + 2b | ✅ Always | = 2(a+b), has factor 2 |
| 3g + 5h | ❌ Sometimes | depends on g, h values |
| 4m + 2n | ✅ Always | = 2(2m+n), factor 2 |
| 2u − 4v | ✅ Always | = 2(u−2v), factor 2 |
| 13k − 5k | ✅ Always | = 8k, factor 2 |
| 6m − 3n | ❌ Sometimes | = 3(2m−n), even only when n is even |
| x² + 2 | ❌ Sometimes | odd when x is odd (e.g. 3² + 2 = 11) |
| b² + 1 | ❌ Sometimes | even when b is odd only |
| 4k × 3j | ✅ Always | = 12kj, has factor 2 (and 4 and 12!) |
Rule to Remember
An algebraic expression is always even if it can be written as 2 × (some integer expression). Check if 2 is a factor!
An algebraic expression is always even if it can be written as 2 × (some integer expression). Check if 2 is a factor!
Pairs to Make Fours — Divisibility by 4
Even numbers can be divided into two types based on their remainder when divided by 4:
Type A: Multiples of 4
Numbers like 4, 8, 12, 16, 24, 36…
Leave remainder 0 when divided by 4.
Written as 4p
Numbers like 4, 8, 12, 16, 24, 36…
Leave remainder 0 when divided by 4.
Written as 4p
Type B: Non-multiples of 4
Numbers like 2, 6, 10, 14, 18, 22…
Leave remainder 2 when divided by 4.
Written as 4p + 2
Numbers like 2, 6, 10, 14, 18, 22…
Leave remainder 2 when divided by 4.
Written as 4p + 2
📋 Three Cases: When Does Sum of Two Even Numbers = Multiple of 4?
| Case | Algebra | Result | Examples |
|---|---|---|---|
| Both are multiples of 4 | 4p + 4q = 4(p+q) | ✅ Multiple of 4 | 12 + 16 = 28 ✓ |
| Both are non-multiples of 4 | (4p+2)+(4q+2) = 4(p+q+1) | ✅ Multiple of 4 | 6 + 10 = 16 ✓ |
| One of each type | 4p+(4q+2) = 4(p+q)+2 | ❌ NOT a multiple of 4 | 8 + 6 = 14 ✗ |
Key Rule
The sum of two even numbers is a multiple of 4 if and only if both are multiples of 4 OR neither is a multiple of 4.
The sum of two even numbers is a multiple of 4 if and only if both are multiples of 4 OR neither is a multiple of 4.
Always, Sometimes, or Never True?
We test statements about divisibility to find if they are Always True (A), Sometimes True (S), or Never True (N).
📜 Key General Rules of Divisibility
If a divides M and a divides N, then a divides (M + N) and (M − N) too.
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 both k and m, then A is divisible by LCM(k, m).
📝 Statements and Their Truth Values
| Statement | Verdict | Reason |
|---|---|---|
| If 8 divides two numbers separately, it must divide their sum. | ✅ Always True | 8a + 8b = 8(a+b) |
| If a number is divisible by 8, then 8 also divides any two numbers that add up to it. | ⚠️ Sometimes True | 72 = 48+24 ✓ but 72 = 50+22 ✗ |
| If a number is divisible by 7, then all its multiples are divisible by 7. | ✅ Always True | (7j)×m = 7(jm) |
| If divisible by 12, it is also divisible by all factors of 12. | ✅ Always True | 12m = 2×6×m = 3×4×m |
| If divisible by 7, it is also divisible by any multiple of 7. | ⚠️ Sometimes True | 42 ÷ 14 = 3 ✓ but 42 ÷ 28 ✗ |
| If divisible by both 9 and 4, it must be divisible by 36. | ✅ Always True | LCM(9,4) = 36; 9 and 4 are coprime |
| If divisible by both 6 and 4, it must be divisible by 24. | ⚠️ Sometimes True | LCM(6,4) = 12, not 24. e.g. 12 ÷ 24 ✗ |
| Adding an odd number to an even number gives a multiple of 6. | ❌ Never True | odd + even = odd; multiples of 6 are even |
Common Mistake!
To check divisibility by 24, checking divisibility by 4 AND 6 is NOT enough (since LCM(4,6) = 12, not 24). Instead, check divisibility by 3 AND 8 (since LCM(3,8) = 24 and they are coprime).
To check divisibility by 24, checking divisibility by 4 AND 6 is NOT enough (since LCM(4,6) = 12, not 24). Instead, check divisibility by 3 AND 8 (since LCM(3,8) = 24 and they are coprime).
What Remains? — Remainders & Algebra
Numbers that leave the same remainder when divided by a given number can be described by an algebraic expression.
📌 Numbers Leaving Remainder 3 When Divided by 5
Such numbers are: 3, 8, 13, 18, 23, 28… (they are 3 more than multiples of 5)
Numbers leaving remainder 3 when divided by 5 = 5k + 3, where k = 0, 1, 2, 3, …
k = | 0 | 1 | 2 | 3 | 4
5k+3 = | 3 | 8 | 13 | 18 | 23
5k+3 = | 3 | 8 | 13 | 18 | 23
Another way to see it!
The same numbers can also be written as 5k − 2 (where k ≥ 1), because they are 2 less than multiples of 5.
So 5k + 3 and 5k − 2 describe the same set of numbers!
The same numbers can also be written as 5k − 2 (where k ≥ 1), because they are 2 less than multiples of 5.
So 5k + 3 and 5k − 2 describe the same set of numbers!
🧩 General Pattern
Numbers leaving remainder r when divided by d = dk + r, where k = 0, 1, 2, 3, …
Exam Tip
If a number leaves remainder r when divided by d, it can also be seen as (d − r) short of the next multiple of d. So it can be written as dk + r or equivalently dk − (d − r) for k ≥ 1.
If a number leaves remainder r when divided by d, it can also be seen as (d − r) short of the next multiple of d. So it can be written as dk + r or equivalently dk − (d − r) for k ≥ 1.
5.2 — Checking Divisibility Quickly
We can write any number in the Indian place value system as:
Any number = … + 1000d + 100c + 10b + a, where a, b, c, d are digits (units, tens, hundreds, thousands)
📋 Basic Divisibility Rules (Revision)
| Divisible By | Rule | Why It Works |
|---|---|---|
| 2 | Units digit is 0, 2, 4, 6, or 8 | All place values except units are multiples of 2 |
| 5 | Units digit is 0 or 5 | All place values except units are multiples of 5 |
| 10 | Units digit is 0 | All place values except units are multiples of 10 |
| 4 | Last 2 digits form a number divisible by 4 | All place values ≥ 100 are multiples of 4 |
| 8 | Last 3 digits form a number divisible by 8 | All place values ≥ 1000 are multiples of 8 |
Divisibility by 9 — The Digit Sum Trick
🔍 The Key Observation
Every power of 10 is exactly 1 more than a multiple of 9:
1 = 0×9 + 1
10 = 1×9 + 1
100 = 11×9 + 1
1000 = 111×9 + 1
10000 = 1111×9 + 1
10 = 1×9 + 1
100 = 11×9 + 1
1000 = 111×9 + 1
10000 = 1111×9 + 1
This means each digit contributes its face value as remainder when divided by 9!
📐 Example: Find Remainder When 7309 ÷ 9
7 × 1000 + 3 × 100 + 0 × 10 + 9
= 7×(999+1) + 3×(99+1) + 0×(9+1) + 9×1
= [7×999 + 3×99 + 0×9] + (7 + 3 + 0 + 9)
= [multiple of 9] + 19
→ 1+9 = 10 → 1+0 = 1
∴ Remainder = 1
= 7×(999+1) + 3×(99+1) + 0×(9+1) + 9×1
= [7×999 + 3×99 + 0×9] + (7 + 3 + 0 + 9)
= [multiple of 9] + 19
→ 1+9 = 10 → 1+0 = 1
∴ Remainder = 1
Divisibility Rule for 9: A number is divisible by 9 if and only if the sum of its digits is divisible by 9.
🧪 Test Yourself
| Number | Digit Sum | Divisible by 9? |
|---|---|---|
| 123 | 1+2+3 = 6 | ❌ No |
| 405 | 4+0+5 = 9 | ✅ Yes |
| 8888 | 8+8+8+8 = 32 → 3+2 = 5 | ❌ No |
| 93547 | 9+3+5+4+7 = 28 → 2+8 = 10 → 1 | ❌ No |
| 358095 | 3+5+8+0+9+5 = 30 → 3+0 = 3 | ❌ No |
Four Key Facts about Divisibility by 9
(i) If divisible by 9 → digit sum divisible by 9. ✅ Always True
(ii) If digit sum divisible by 9 → number divisible by 9. ✅ Always True
(iii) If NOT divisible by 9 → digit sum NOT divisible by 9. ✅ Always True
(iv) If digit sum NOT divisible by 9 → number NOT divisible by 9. ✅ Always True
(i) If divisible by 9 → digit sum divisible by 9. ✅ Always True
(ii) If digit sum divisible by 9 → number divisible by 9. ✅ Always True
(iii) If NOT divisible by 9 → digit sum NOT divisible by 9. ✅ Always True
(iv) If digit sum NOT divisible by 9 → number NOT divisible by 9. ✅ Always True
Divisibility by 3 — Similar to 9
Every power of 10 leaves a remainder of 1 when divided by 3 (just like for 9!). So the digit sum trick works for 3 too.
Divisibility Rule for 3: A number is divisible by 3 if and only if the sum of its digits is divisible by 3.
Example: 123
Digit sum = 1+2+3 = 6
6 ÷ 3 = 2 ✓
So 123 is divisible by 3 ✅
Digit sum = 1+2+3 = 6
6 ÷ 3 = 2 ✓
So 123 is divisible by 3 ✅
Example: 115
Digit sum = 1+1+5 = 7
7 is not divisible by 3
So 115 is NOT divisible by 3 ❌
Digit sum = 1+1+5 = 7
7 is not divisible by 3
So 115 is NOT divisible by 3 ❌
Relationship Between 3 and 9
All multiples of 9 are also multiples of 3, but NOT all multiples of 3 are multiples of 9. For example, 15, 33, 87 are divisible by 3 but NOT by 9.
All multiples of 9 are also multiples of 3, but NOT all multiples of 3 are multiples of 9. For example, 15, 33, 87 are divisible by 3 but NOT by 9.
Divisibility by 11 — The Alternating Pattern
🔍 Key Observation for Place Values
| Place Value | Relation to 11 |
|---|---|
| Units (1) | 1 more than a multiple of 11 (11×0 + 1) |
| Tens (10) | 1 less than a multiple of 11 (11×1 − 1) |
| Hundreds (100) | 1 more than a multiple of 11 (11×9 + 1) |
| Thousands (1000) | 1 less than a multiple of 11 (11×91 − 1) |
| Ten-thousands (10000) | 1 more than a multiple of 11 |
The pattern alternates: +1, −1, +1, −1… starting from the units place.
Divisibility Rule for 11: Place alternating + and − signs before each digit, starting from the units digit. If the result is 0 or a multiple of 11, the number is divisible by 11.
📐 Step-by-Step Method (for 320185)
- Write digits with alternating signs from units place: −3 + 2 − 8 + 1 − 0 + 5
- Calculate the sum: −3 + 2 − 8 + 1 − 0 + 5 = −3
- If result is 0 or multiple of 11 → divisible by 11. Otherwise, the result (or its positive value) is related to the remainder.
🧪 Examples
| Number | Alternating Sum | Divisible by 11? |
|---|---|---|
| 158 | 8 − 5 + 1 = 4 | ❌ No (remainder 4) |
| 841 | 1 − 4 + 8 = 5 | ❌ No |
| 121 | 1 − 2 + 1 = 0 | ✅ Yes |
| 5529 | 9 − 2 + 5 − 5 = 7 | ❌ No |
| 90904 | 4 − 0 + 9 − 0 + 9 = 22 | ✅ Yes (22 = 2×11) |
| 857076 | 6 − 7 + 0 − 7 + 5 − 8 = −11 | ✅ Yes (−11 = −1×11) |
Easy Way to Remember
For divisibility by 11: (Sum of digits at odd positions) − (Sum of digits at even positions) = 0 or multiple of 11.
Count positions from right: position 1 (units), 2 (tens), 3 (hundreds)…
For divisibility by 11: (Sum of digits at odd positions) − (Sum of digits at even positions) = 0 or multiple of 11.
Count positions from right: position 1 (units), 2 (tens), 3 (hundreds)…
Digital Roots
The digital root of a number is the single-digit number obtained by repeatedly adding its digits until you get a one-digit result.
Digital Root of 489710:
4+8+9+7+1+0 = 29 → 2+9 = 11 → 1+1 = 2
4+8+9+7+1+0 = 29 → 2+9 = 11 → 1+1 = 2
🔢 Examples
| Number | Steps | Digital Root |
|---|---|---|
| 45 | 4+5 = 9 | 9 |
| 387 | 3+8+7 = 18 → 1+8 = 9 | 9 |
| 124 | 1+2+4 = 7 | 7 |
| 999 | 9+9+9 = 27 → 2+7 = 9 | 9 |
| 100 | 1+0+0 = 1 | 1 |
🌟 Key Properties of Digital Roots
- Digital root of a multiple of 9 is always 9
- Digital root is the same as the remainder when divided by 9 (except when remainder = 0, digital root = 9)
- Digital roots of 12 consecutive numbers cycle through 1 to 9 repeatedly
- Digital root of a multiple of 3 is always 3, 6, or 9
🏛️
Historical Connection!
The Indian mathematician Aryabhata II (c. 950 CE) described the digital root method in his work Mahāsiddhānta. It was used as a check on arithmetic calculations — like a built-in calculator verification!
The Indian mathematician Aryabhata II (c. 950 CE) described the digital root method in his work Mahāsiddhānta. It was used as a check on arithmetic calculations — like a built-in calculator verification!
Digital Root Formula
For any expression like 9a + 36b + 13: Since 9a and 36b are multiples of 9 (digital root = 9), only the remaining constant 13 matters. Digital root of 13 = 1+3 = 4.
For any expression like 9a + 36b + 13: Since 9a and 36b are multiples of 9 (digital root = 9), only the remaining constant 13 matters. Digital root of 13 = 1+3 = 4.
5.3 — Digits in Disguise (Cryptarithms)
A cryptarithm is a mathematical puzzle where letters represent digits.
Rules of Cryptarithms
(1) Each letter stands for a unique digit (0–9).
(2) Each digit is represented by at most one letter.
(3) The first digit of any number is never 0.
(1) Each letter stands for a unique digit (0–9).
(2) Each digit is represented by at most one letter.
(3) The first digit of any number is never 0.
🔑 Solved Cryptarithms
Solve: A1 + 1B = B0
Ans: A=2, B=3 → 21 + 13 = 34 ✓ (B0 = 30, A1 = 21, 1B = 13 → wait, 21+13=34=B0 → B=3, A=2)
Solve: PQ × 8 = RS (both 2-digit numbers)
Ans: PQ = 12, RS = 96 → 12 × 8 = 96. (P≠Q and P≠R≠S≠Q, all different digits ✓)
Solve: GH × H = 9K
Ans: 16 × 6 = 96 → G=1, H=6, K=6. (Units digit of GH and H must match — H×H ends in K, and GH×H is in 90s)
Solve: BYE × 6 = RAY
Ans: B=1 (must be, to keep product 3-digit). Y is even and ≤ 6. Solution: 126 × 6 = 756 → B=1, Y=2, E=6, R=7, A=5 ✓
🧮 More Cryptarithms
| Cryptarithm | Solution |
|---|---|
| UT × 3 = PUT | U=1, T=5 → 15 × 3 = 045? No → U=1, T=8 → 18×3=54 → No. Proper: 37×3=111? No. UT × 3 = PUT means 3-digit. So UT ≥ 34. 34×3=102 → P=1,U=0,T=? → re-check carefully; answer: UT=37, 37×3=111 doesn’t work; standard answer: U=1, T=5, P=0 is invalid. Correct: 37→111 ✗; answer is UT = 37 doesn’t hold. Accept exploring approach. |
| JK × 6 = KKK | KKK = K×111 = K×3×37. So JK×6 = K×111 → JK = K×18.5 (not integer unless K=2: JK=37, 37×6=222 ✓ → J=3, K=7? 37×6=222, K=7 but KKK=222≠777. Try K=2: JK=37, 37×6=222=KKK with K=2? No. K=7: 37×6=222≠777. Standard: 18×6=108 ✗; correct: J=1,K=8, 18×6=108 ✗. Answer: 21×6=126 ✗. Textbook answer: JK=37 ✗. Systematic: KKK must be 111,222,…,999; divisible by 6 → 222,444,666; 222÷6=37 → JK=37 ✓ K=7 but K in KKK=2 → contradiction. 444÷6=74 → J=7,K=4, KKK=444 ✓ → 74×6=444 ✅ |
Strategy for Solving Cryptarithms
Start with constraints on the number of digits. Use units digit patterns. Work from what you know (e.g., carry values, largest/smallest possible digits). Use trial and error systematically.
Start with constraints on the number of digits. Use units digit patterns. Work from what you know (e.g., carry values, largest/smallest possible digits). Use trial and error systematically.
Chapter Summary — Quick Revision
📌 Parity
4 consecutive numbers with any +/- always give even results. odd±odd=even, even±even=even, odd±even=odd.
4 consecutive numbers with any +/- always give even results. odd±odd=even, even±even=even, odd±even=odd.
➗ Divisibility Properties
If a|M and a|N → a|(M±N). If a|A → a divides all multiples of A and all factors of A.
If a|M and a|N → a|(M±N). If a|A → a divides all multiples of A and all factors of A.
9️⃣ Divisibility by 9
A number is divisible by 9 iff the sum of its digits is divisible by 9. Works by expressing each place value as 9×□ + 1.
A number is divisible by 9 iff the sum of its digits is divisible by 9. Works by expressing each place value as 9×□ + 1.
3️⃣ Divisibility by 3
Sum of digits divisible by 3 → number divisible by 3. (Same logic as for 9, but weaker condition.)
Sum of digits divisible by 3 → number divisible by 3. (Same logic as for 9, but weaker condition.)
1️⃣1️⃣ Divisibility by 11
Alternating digit sum (from units) = 0 or multiple of 11. Place values alternate as 1 more/less than a multiple of 11.
Alternating digit sum (from units) = 0 or multiple of 11. Place values alternate as 1 more/less than a multiple of 11.
🌱 Digital Roots
Repeatedly add digits until single digit. Equals remainder ÷ 9 (digital root of multiples of 9 is always 9).
Repeatedly add digits until single digit. Equals remainder ÷ 9 (digital root of multiples of 9 is always 9).
🗒️ Key Formulas at a Glance
Numbers leaving remainder r when divided by d → dk + r (k = 0, 1, 2, …)
Algebraic expression always even → can be written as 2 × (integer expression)
Sum of two even numbers is divisible by 4 → both are multiples of 4, OR neither is a multiple of 4
Important Exam Questions with Answers
Q1. The sum of four consecutive numbers is 34. What are the numbers?
Ans: Let the numbers be n, n+1, n+2, n+3. Then 4n+6 = 34 → 4n = 28 → n = 7. Numbers are 7, 8, 9, 10.
Q2. Without dividing, check if 358095 is divisible by 9.
Ans: Digit sum = 3+5+8+0+9+5 = 30 → 3+0 = 3. Since 3 is NOT divisible by 9, 358095 is NOT divisible by 9.
Q3. Is the expression 4m + 2n always even? Justify.
Ans: 4m + 2n = 2(2m + n). Since 2 is a factor, this expression is always even for any integers m and n.
Q4. If 31z5 is a multiple of 9, find z.
Ans: Digit sum = 3+1+z+5 = 9+z. For divisibility by 9: 9+z = 9 → z = 0, OR 9+z = 18 → z = 9. So z = 0 or z = 9.
Q5. Check whether 90904 is divisible by 11.
Ans: Alternating sum from units: 4 − 0 + 9 − 0 + 9 = 22 = 2 × 11. Since 22 is a multiple of 11, 90904 IS divisible by 11.
Q6. Find the digital root of 489710.
Ans: 4+8+9+7+1+0 = 29 → 2+9 = 11 → 1+1 = 2. Digital root = 2.
Q7. Is this Always True, Sometimes True, or Never True: “If a number is divisible by both 9 and 4, it must be divisible by 36.”
Ans: Always True. LCM(9, 4) = 36 (since 9 and 4 share no common factors). By the LCM rule, divisibility by both 9 and 4 implies divisibility by 36.
Q8. A number leaves remainder 2 when divided by 3 AND remainder 2 when divided by 4. Write the algebraic expression and find such numbers.
Ans: Numbers are 2 more than multiples of both 3 and 4 = LCM(3,4) = 12. So the expression is 12k + 2. Numbers: 2, 14, 26, 38, 50…
Q9. 661 leaves remainder 3 when divided by 7, and 4779 leaves remainder 5. Find the remainder when 4779 + 661 is divided by 7.
Ans: 661 = 7a + 3, 4779 = 7b + 5. Sum = 7(a+b) + 8 = 7(a+b+1) + 1. Remainder = 1.
Q10. Sreelatha says a number divisible by 9, when its digits are reversed, is still divisible by 9. Is she right?
Ans: Yes! Reversing digits doesn’t change the digit sum. Since divisibility by 9 depends only on digit sum, a number divisible by 9 remains divisible by 9 after any digit rearrangement. Always True.
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