NCERT Solutions for Class 8 Mathematics Ganita Prakash I Chapter 5
NCERT solutions for Class 8 Maths Ganita Prakash Part 1 Chapter 5 Number Play – complete answers & explanations
Class 8 Maths is an exciting stage where students begin to think like mathematicians, and Chapter 5 of Ganita Prakash Part 1 — Number Play — is one of the most thought-provoking chapters in the book. This chapter moves beyond simple calculations and invites students to explore patterns, divisibility rules, algebraic reasoning, and number puzzles in a deep and enjoyable way. From understanding whether every natural number can be written as a sum of consecutive numbers, to cracking clever cryptarithms, this chapter builds genuine mathematical curiosity. Parents and students often find this chapter challenging because the questions require reasoning, not just formulas. This blog brings you clear and reliable NCERT solutions for every question and activity in Chapter 5, organized exactly as they appear in the worksheet, so students can practice with confidence. Download the worksheet and practice alongside solutions for better clarity. If you'd like personalized expert support, book a free trial now to get expert guidance.
What this NCERT chapter covers?
1. The chapter opens with an
exploration of consecutive numbers — students investigate whether every natural number can be expressed as a sum of consecutive natural numbers and discover that powers of 2 are the only numbers that cannot be.
2. Students explore odd and even numbers in the context of consecutive sums — for example, every odd number can be written as a sum of two consecutive numbers, while most even numbers (except powers of 2) can also be expressed as sums of consecutive numbers.
3. A key concept is parity — students learn how placing '+' and '–' signs between four consecutive numbers always produces even results, and they explore why this happens through algebraic reasoning.
4. The chapter extends the parity discussion to any four numbers (not just consecutive), showing that all eight expressions formed always share the same parity, which is a powerful general result.
5. Students work on the "Breaking Even" section to identify — without calculating — which arithmetic and algebraic expressions always give even numbers.
6. The concept of "Pairs to Make Fours" introduces students to divisibility by 4, exploring when the sum of two even numbers is divisible by 4 and when it is not, using both algebra and visual models.
7. The "Always, Sometimes, or Never" strand develops critical mathematical reasoning — students examine divisibility statements and decide whether they are always true, sometimes true, or never true, justifying with algebra and examples.
8. The "What Remains?" section introduces modular thinking — finding numbers with specific remainders when divided by a given number, and representing all such numbers algebraically (e.g., 5k + 3 or 5k – 2).
9. Section 5.2, Checking Divisibility Quickly, explains algebraically why the divisibility shortcuts for 2, 4, 5, 8, 9, 3, and 11 work, connecting place value to remainders.
10. The Digital Roots concept is explored — students discover that the digital root of a number equals its remainder when divided by 9, and observe cycling patterns in digital roots of consecutive numbers and multiples.
11. Section 5.3, Digits in Disguise, covers cryptarithms — puzzles where letters represent digits in addition and multiplication problems — which sharpen logical deduction skills.
12. The chapter ends with the traditional Indian board game Navakankari (also known as Nine Men's Morris), connecting mathematical strategy to cultural heritage.
How to use these NCERT solutions?
1. Before reading any solution, attempt
every question on your own first — even if you are unsure, writing down your thinking helps you learn much more than reading an answer directly.
2. Use these solutions strictly for checking your work after attempting the question, not as a shortcut to copy answers. The process of arriving at the answer matters more than the answer itself in this chapter.
3. For exploration-based questions (marked "Math Talk" in the textbook), discuss your observations with a classmate or parent before reading the solution here — the goal is to notice patterns, and your own discoveries are more valuable.
4. The solutions follow the worksheet order exactly, section by section and question by question, so you can easily find what you're looking for without confusion.
5. For algebraic justification questions, pay special attention to how the solution is set up — note how variables are assigned and how the reasoning flows, as this pattern will appear in exams.
6. Parents and teachers can use this blog to review whether a student's attempt is on the right track — the answers here are aligned with NCERT expectations and help confirm the correct approach to answering.
7. For cryptarithm problems (Digits in Disguise), try to solve by trial and logic before reading — these puzzles are designed to be solved by reasoning, not calculation alone.
Important tips & tricks for students
1. In parity questions, always start by identifying whether each number is odd or even before performing any operation — the rules odd + odd = even, even + even = even, and odd + even = odd are your most important tools.
2. For "Always, Sometimes, or Never" questions, never rely on just one example. To prove "always true," you need algebra. To prove "sometimes true" or "never true," you need both an example that works and one that does not.
3. When writing algebraic expressions for numbers with a given remainder, remember the general form: if a number leaves remainder r when divided by n, it can be written as nk + r. Practice recognizing this structure in every question.
4. For divisibility by 9 and 3, always check the digit sum first. If the digit sum is large, find its digit sum again until you reach a single digit — this is the digital root and equals the remainder when the number is divided by 9.
5. For divisibility by 11, apply the alternating digit sum rule: subtract and add digits alternating from the right. If the result is 0 or a multiple of 11, the number is divisible by 11.
6. In cryptarithm puzzles, start with the column that gives you the most constraint — usually the leftmost digit or the units digit — and use the constraints to narrow down digit choices systematically.
7. Never assume a divisibility rule works for a composite number without verifying — for example, checking divisibility by 4 and 6 does NOT confirm divisibility by 24. Always use coprime factor pairs like 3 and 8 for 24.
8. When the question asks to "explain using algebra," write a complete algebraic argument — assign variable names, set up the expression, simplify, and state your conclusion clearly. Partial algebraic work will not earn full marks.
NCERT solutions – complete answer key
Sum of consecutive numbers
Explore – Can every natural number be written as a sum of consecutive numbers?
Not every natural number can be written as a sum of consecutive natural numbers.
Powers of 2 (like 1, 2, 4, 8, 16, 32 ...) CANNOT be written as a sum of consecutive natural numbers.
All odd numbers can be written as a sum of two consecutive numbers: e.g., 7 = 3 + 4.
Most even numbers (except powers of 2) can also be written as sums of consecutive numbers: e.g., 12 = 3 + 4 + 5.
Which numbers can be written as a sum of consecutive numbers in more than one way?
Numbers with more than one odd factor (other than 1) can be expressed in more than one way.
Example: 15 = 7 + 8 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5.
Can all even numbers be written as a sum of consecutive numbers?
No. Powers of 2 (2, 4, 8, 16, 32, ...) cannot be written as a sum of consecutive natural numbers.
Other even numbers like 6, 10, 12 can be expressed as such sums.
Take any 4 consecutive numbers (e.g., 3, 4, 5, 6). Place '+' and '–' signs. Write all 8 expressions and evaluate.
Using 3, 4, 5, 6:
1. 3 + 4 + 5 + 6 = 18
2. 3 + 4 + 5 – 6 = 6
3. 3 + 4 – 5 + 6 = 8
4. 3 + 4 – 5 – 6 = –4
5. 3 – 4 + 5 + 6 = 10
6. 3 – 4 + 5 – 6 = –2
7. 3 – 4 – 5 + 6 = 0
8. 3 – 4 – 5 – 6 = –12
Observation: All results are even numbers.
Now take 4 other consecutive numbers (e.g., 5, 6, 7, 8):
1. 5 + 6 + 7 + 8 = 26
2. 5 + 6 + 7 – 8 = 10
3. 5 + 6 – 7 + 8 = 12
4. 5 + 6 – 7 – 8 = –4
5. 5 – 6 + 7 + 8 = 14
6. 5 – 6 + 7 – 8 = –2
7. 5 – 6 – 7 + 8 = 0
8. 5 – 6 – 7 – 8 = –16
Observation: All results are even numbers again.
Is there a way to explain why all expressions with 4 consecutive numbers give even results?
Let the 4 consecutive numbers be n, n+1, n+2, n+3.
When any '+' sign is changed to '–', the value changes by 2 × (that number), which is always even.
So all 8 expressions have the same parity.
Starting from n + (n+1) + (n+2) + (n+3) = 4n + 6 (always even), all expressions remain even.
Take any 4 numbers (not necessarily consecutive). Place '+' and '–' signs. What is the parity?
The parity of all 8 expressions formed from any 4 numbers a, b, c, d is always the same.
This is because switching any sign changes the value by 2 × (that number), which is always even, so parity never changes.
However, unlike consecutive numbers, the results need not always be even — parity depends on the specific numbers chosen.
Is the same-parity phenomenon limited to 4 numbers?
No. It applies to any number of terms. For any set of numbers a ± b ± c ± d ± e ± ..., all expressions formed by changing signs always have the same parity.
Breaking Even
Activity: Without computing, find which arithmetic expressions are even:
1. 43 + 37 → 43 (odd) + 37 (odd) = even ✓ EVEN
2. 672 – 348 → even – even = even ✓ EVEN
3. 4 × 347 × 3 → even × any = even ✓ EVEN
4. 708 – 477 → even – odd = odd ✗ ODD
5. 809 + 214 → odd + even = odd ✗ ODD
6. 119 × 303 → odd × odd = odd ✗ ODD
7. 543 – 479 → odd – odd = even ✓ EVEN
8. 513³ → odd³ = odd ✗ ODD
Activity Q: Identify which algebraic expressions always give even numbers:
1. 2a + 2b → 2(a + b) — always even ✓ ALWAYS EVEN
2. 3g + 5h → odd×g + odd×h — depends on g, h ✗ NOT ALWAYS EVEN
3. 4m + 2n → 2(2m + n) — always even ✓ ALWAYS EVEN
4. 2u – 4v → 2(u – 2v) — always even ✓ ALWAYS EVEN
5. 13k – 5k → 8k — always even ✓ ALWAYS EVEN
6. 6m – 3n → 3(2m – n) — odd when n is odd ✗ NOT ALWAYS EVEN
7. x² + 2 → if x is odd, x² is odd, x²+2=odd ✗ NOT ALWAYS EVEN
8. b² + 1 → if b is odd, b²+1 = even; if b is even, b²+1 = odd ✗ NOT ALWAYS EVEN
9. 4k × 3j → 12kj — always even ✓ ALWAYS EVEN
Pairs to Make Fours
Take a pair of even numbers. When is their sum divisible by 4?
Even numbers are of two types:
Type A: Multiples of 4 (remainder 0 when divided by 4) — e.g., 4, 8, 12, 16
Type B: Non-multiples of 4 (remainder 2 when divided by 4) — e.g., 2, 6, 10, 14
Rules:
Type A + Type A = Multiple of 4 (e.g., 4 + 12 = 16 ✓)
Type B + Type B = Multiple of 4 (e.g., 2 + 6 = 8 ✓, remainders 2+2=4)
Type A + Type B = NOT multiple of 4 (e.g., 4 + 6 = 10 ✗, remainder = 2)
(Table — Adding multiple of 4 to non-multiple of 4):
Explanation with Algebra:
Let the numbers be 4p and (4q + 2).
4p + (4q + 2) = 4p + 4q + 2 = 4(p + q) + 2.
This leaves a remainder of 2 when divided by 4, so it is NOT a multiple of 4.
Example: 4 + 6 = 10 (not a multiple of 4); 8 + 10 = 18 (not a multiple of 4).
What Remains?
Find numbers with remainder 3 when divided by 5.
Such numbers: 3, 8, 13, 18, 23, 28, ...
These are 3 more than multiples of 5, i.e., of the form 5k + 3 (where k = 0, 1, 2, 3, ...)
Which algebraic expressions capture all such numbers?
(iv) 5k + 3 (for k = 0, 1, 2, 3, ...) → gives 3, 8, 13, 18, 23 ✓
(v) 5k – 2 (for k = 1, 2, 3, ...) → gives 3, 8, 13, 18, 23 ✓
Both (iv) and (v) are correct.
Are there other expressions that give numbers 3 more than a multiple of 5?
Yes. Any expression of the form 5k + 3 or equivalently 5k – 2 (for appropriate values of k) works.
Also: 5(k+1) – 2 = 5k + 3 — they are the same family of numbers.
Figure it Out
1. Let the four consecutive numbers be n, n+1, n+2, n+3.
n + (n+1) + (n+2) + (n+3) = 34
4n + 6 = 34
4n = 28
n = 7
The four numbers are: 7, 8, 9, 10.
2. If p is the greatest, the five consecutive numbers in increasing order are:
p–4, p–3, p–2, p–1, p.
3. (i) The sum of two even numbers is a multiple of 3.
Answer: SOMETIMES TRUE
Example (True): 6 + 12 = 18 = 3 × 6 ✓
Example (False): 2 + 8 = 10, not a multiple of 3 ✓
Algebra: 2a + 2b = 2(a+b). This is a multiple of 3 only when (a+b) is a multiple of 3 — not always.
(ii) If a number is not divisible by 18, it is also not divisible by 9.
Answer: SOMETIMES TRUE
Example (True): 7 is not divisible by 18 and not divisible by 9 ✓
Example (False): 9 is not divisible by 18, but IS divisible by 9 ✗
(iii) If two numbers are not divisible by 6, their sum is not divisible by 6.
Answer: SOMETIMES TRUE
Example (True): 4 and 5 — neither divisible by 6; 4+5=9, not divisible by 6 ✓
Example (False): 4 and 8 — neither divisible by 6; 4+8=12, which IS divisible by 6 ✗
(iv) The sum of a multiple of 6 and a multiple of 9 is a multiple of 3.
Answer: ALWAYS TRUE
A multiple of 6 = 6a = 3(2a) — divisible by 3.
A multiple of 9 = 9b = 3(3b) — divisible by 3.
Sum = 3(2a + 3b) — always divisible by 3.
(v) The sum of a multiple of 6 and a multiple of 3 is a multiple of 9.
Answer: SOMETIMES TRUE
Example (True): 18 + 9 = 27 = 9 × 3 ✓
Example (False): 6 + 6 = 12, not a multiple of 9 ✗
4. Numbers with remainder 2 when divided by 3: 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, ...
Numbers with remainder 2 when divided by 4: 2, 6, 10, 14, 18, 22, 26, 30, ...
Common numbers: 2, 14, 26, 38, ... (differ by 12, which is LCM of 3 and 4)
Algebraic expression: 12k + 2 (for k = 0, 1, 2, 3, ...)
5. We need: number ≡ 1 (mod 3), ≡ 1 (mod 2), ≡ 1 (mod 5), ≡ 0 (mod 7)
LCM(2, 3, 5) = 30. Numbers that are 1 more than a multiple of 30: 31, 61, 91, ...
Check divisibility by 7:
31 ÷ 7 = 4 remainder 3 ✗
61 ÷ 7 = 8 remainder 5 ✗
91 ÷ 7 = 13 exactly ✓
The number of pebbles is 91.
6. TRUE
Each number = 6k + 2 (for some integer k).
Sum of three such numbers = (6a+2) + (6b+2) + (6c+2) = 6(a+b+c) + 6 = 6(a+b+c+1).
This is always a multiple of 6. So Tathagat's claim is correct.
7. (i) 4779 + 661:
Algebraically: 661 = 7m + 3 and 4779 = 7n + 5.
Sum = 7(m+n) + 8 = 7(m+n) + 7 + 1 = 7(m+n+1) + 1.
Remainder = 1.
(ii) 4779 – 661:
Difference = 7(n–m) + (5–3) = 7(n–m) + 2.
Remainder = 2.
8. Notice: each number is 1 less than a multiple of 3, 4, and 5 respectively.
i.e., the number + 1 is divisible by 3, 4, and 5.
LCM(3, 4, 5) = 60.
So number + 1 = 60, meaning number = 59.
Check: 59 ÷ 3 = 19 remainder 2 ✓; 59 ÷ 4 = 14 remainder 3 ✓; 59 ÷ 5 = 11 remainder 4 ✓.
The smallest such number is 59.
Checking divisibility quickly
Explain using algebra why divisibility shortcuts for 5, 2, 4, and 8 work.
Any number = ...1000d + 100c + 10b + a (where a = units digit)
Divisibility by 5:
All place values except units (10, 100, 1000, ...) are multiples of 5.
So the number is divisible by 5 if and only if the units digit a is 0 or 5.
Divisibility by 2:
All place values except units are multiples of 2 (since 10, 100, ... are all even).
So the number is divisible by 2 if and only if the units digit a is even (0, 2, 4, 6, 8).
Divisibility by 4:
10 is not divisible by 4, but 100, 1000, ... all are (since 100 = 4 × 25).
So divisibility by 4 depends only on the last two digits (10b + a).
A number is divisible by 4 if and only if its last two digits form a number divisible by 4.
Divisibility by 8:
100 is not divisible by 8, but 1000, 10000, ... all are (since 1000 = 8 × 125).
So divisibility by 8 depends only on the last three digits (100c + 10b + a).
A number is divisible by 8 if and only if its last three digits form a number divisible by 8.
Can any number made of only digits 0 and 9 always be divisible by 9?
YES
Each digit is either 0 or 9, so each term in expanded form is 0×(place value) or 9×(place value), both multiples of 9. So the entire number is a multiple of 9.
Is 10 divisible by 9? What is the remainder?
10 = 9 × 1 + 1. Remainder = 1.
Check multiples of 10 divided by 9.
10 ÷ 9: remainder 1; 20 ÷ 9: remainder 2; 30 ÷ 9: remainder 3.
The remainder equals the number of tens (i.e., the tens digit).
Check multiples of 100 divided by 9.
100 ÷ 9: remainder 1; 200 ÷ 9: remainder 2; 300 ÷ 9: remainder 3.
The remainder equals the number of hundreds (i.e., the hundreds digit).
Find remainder when 427 is divided by 9.
4 (hundreds) + 2 (tens) + 7 (units) = 13. 13 = 9 + 4.
Remainder = 4.
Statements — which are correct and why?
(i) If a number is divisible by 9, then the sum of its digits is divisible by 9.
Answer: CORRECT. Since the number = (multiple of 9) + (digit sum), if the number is divisible by 9, the digit sum must also be divisible by 9.
(ii) If the sum of the digits of a number is divisible by 9, then the number is divisible by 9.
Answer: CORRECT. This follows directly from the same algebraic breakdown.
(iii) If a number is not divisible by 9, then the sum of its digits is not divisible by 9.
Answer: CORRECT. This is the contrapositive of statement (ii), which is true.
(iv) If the sum of the digits of a number is not divisible by 9, then the number is not divisible by 9.
Answer: CORRECT. This is the contrapositive of statement (i), which is true.
All four statements are correct.
Figure it Out
1. (i) 123: 1+2+3 = 6. Not divisible by 9. → NOT DIVISIBLE
(ii) 405: 4+0+5 = 9. Divisible by 9. → DIVISIBLE
(iii) 8888: 8+8+8+8 = 32; 3+2 = 5. Not divisible. → NOT DIVISIBLE
(iv) 93547: 9+3+5+4+7 = 28; 2+8 = 10; 1+0 = 1. Not divisible. → NOT DIVISIBLE
(v) 358095: 3+5+8+0+9+5 = 30; 3+0 = 3. Not divisible. → NOT DIVISIBLE
2. - 288: digits 2, 8, 8 → sum 18 ✓
- Any smaller? 200–288: need digit sum = 9 or 18 with all even digits.
Smallest with digit sum 18: 288 (since 2+8+8=18). ✓
The smallest multiple of 9 with no odd digits is 288.
3. 6000 ÷ 9 = 666.67
666 × 9 = 5994
667 × 9 = 6003
|6000 – 5994| = 6
|6003 – 6000| = 3
Closest multiple of 9 to 6000 is 6003.
4. 4300 ÷ 9 = 477.8 → next multiple = 478 × 9 = 4302
4400 ÷ 9 = 488.9 → last multiple = 488 × 9 = 4392
Multiples: 4302, 4311, 4320, 4329, 4338, 4347, 4356, 4365, 4374, 4383, 4392
Count = 488 – 478 + 1 = 11 multiples.
A shortcut for divisibility by 3
Explain why divisibility by 3 works using remainders when powers of 10 are divided by 3.
÷ 3: remainder 1
10 ÷ 3: remainder 1 (since 10 = 9 + 1)
100 ÷ 3: remainder 1 (since 100 = 99 + 1)
1000 ÷ 3: remainder 1 (since 1000 = 999 + 1)
Every power of 10 leaves remainder 1 when divided by 3.
So for any number = ...1000d + 100c + 10b + a:
Remainder = d×1 + c×1 + b×1 + a×1 = d + c + b + a = digit sum.
Therefore, a number is divisible by 3 if and only if its digit sum is divisible by 3.
A shortcut for divisibility by 11
Can you tell whether 462 is divisible by 11?
Digits from units place alternating excess/short:
Units (2): 2 more than a multiple of 11
Tens (6): 6 less than a multiple of 11
Hundreds (4): 4 more than a multiple of 11
Excess = 2 + 4 = 6
Short = 6
Difference = 6 – 6 = 0
Since the difference is 0, 462 IS divisible by 11.
Check: 462 ÷ 11 = 42 ✓
What is the general shortcut for divisibility by 11?
Add the digits at odd positions from the right (units, hundreds, ten-thousands, ...)
Subtract the digits at even positions from the right (tens, thousands, hundred-thousands, ...)
If the result is 0 or a multiple of 11, the number is divisible by 11.
This is equivalent to: (sum of odd-position digits) – (sum of even-position digits) = 0 or multiple of 11.
Is this method similar to the previous one?
Steps: Place alternating '+' and '–' starting from units digit, then evaluate.
For 328105: –3 + 2 – 8 + 1 – 0 + 5 = –3.
This gives the same result as the first method — both find the same alternating digit sum.
The two methods ARE the same; just presented differently.
Fill in the divisibility table.
Number | 2 | 3 | 4 | 5 | 6 | 8 | 9 | 10 | 11
-------|-----|-----|-----|-----|-----|-----|-----|-----|-----
128 | Yes | No | No | No | No | Yes | No | No | No
990 | Yes | Yes | No | Yes | Yes | No | Yes | Yes | Yes
1586 | Yes | No | No | No | No | No | No | No | No
275 | No | No | No | Yes | No | No | No | No | No
6686 | Yes | No | No | No | No | No | No | No | Yes
639210 | Yes | Yes | No | Yes | Yes | No | Yes | Yes | Yes
429714 | Yes | Yes | Yes | No | Yes | No | Yes | No | No
2856 | Yes | Yes | Yes | No | Yes | Yes | Yes | No | No
3060 | Yes | Yes | Yes | Yes | Yes | No | Yes | Yes | No
406839 | No | Yes | No | No | No | No | No | No | No
More on divisibility shortcuts
How to check divisibility by 6?
Check divisibility by both 2 AND 3 (since 6 = 2 × 3, and 2 and 3 are coprime).
If a number is divisible by both 2 and 3, it is divisible by 6.
Verification with given numbers:
38: even ✓, digit sum=11 (not ÷3) → Not divisible by 6.
225: odd → Not divisible by 6.
186: even ✓, digit sum=15 (÷3) ✓ → Divisible by 6. (186÷6=31 ✓)
64: even ✓, digit sum=10 (not ÷3) → Not divisible by 6.
Can we check divisibility by 24 using factors 4 and 6?
NO. This does not work.
Counterexample: 12 is divisible by 4 and by 6, but 12 ÷ 24 is not a whole number.
To check for 24, check divisibility by 3 and 8 instead.
Explain using prime factorisation why 3 and 8 work for 24.
24 = 2³ × 3. So its prime factors are 2³ and 3.
8 = 2³ and 3 = 3. Since gcd(8, 3) = 1 (coprime), checking both 8 and 3 captures the full prime factorisation of 24.
For 4 and 6: 4 = 2² and 6 = 2 × 3. Their LCM = 12, not 24. The factor 2³ is not fully captured (4 only checks for 2², not 2³). Hence insufficient.
Digital Roots
What property does the digital root have?
The digital root of a number equals the remainder when the number is divided by 9 (with digital root = 9 when the remainder is 0, i.e., for multiples of 9).
Between 600 and 700, which numbers have digital root:
(i) 5: Need numbers ≡ 5 (mod 9). Starting from 600+?: 600 has digital root 6.
601→7, 602→8, 603→9, 604→1, 605→2, 606→3, 607→4, 608→5
Subsequent: 617, 626, 635, 644, 653, 662, 671, 680, 689, 698.
(ii) 7:
610→7 ✓, 619, 628, 637, 646, 655, 664, 673, 682, 691, 700.
(iii) 3:
606→3 ✓, 615, 624, 633, 642, 651, 660, 669, 678, 687, 696.
Write digital roots of any 12 consecutive numbers. What do you observe?
The digital roots of 12 consecutive numbers cycle through all 9 values (1–9) and then repeat. The digital roots form a repeating pattern: they are 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3 (or starting from wherever the sequence begins). They cycle with period 9.
Digital roots of consecutive multiples of 3, 4, and 6:
(i) Multiples of 3: Digital roots cycle as 3, 6, 9, 3, 6, 9, ...
(ii) Multiples of 4: Digital roots cycle as 4, 8, 3, 7, 2, 6, 1, 5, 9, 4, 8, 3, ... (period 9)
(iii) Multiples of 6: Digital roots cycle as 6, 3, 9, 6, 3, 9, ...
Digital roots of numbers that are 1 more than a multiple of 6:
Numbers: 1, 7, 13, 19, 25, 31, 37, ...
Digital roots: 1, 7, 4, 1, 7, 4, 1, 7, 4, ...
They cycle as 1, 7, 4 with period 3.
Riddle: "I'm made of digits, each tiniest and odd... digital root = 9, digits all odd, not coprime with 9... largest odd single-digit."
The number is 999 (digits: 9, 9, 9 — all odd; digit count = 3; sum = 27; root = 9; largest odd single-digit digit = 9).
Figure it Out
1. Adding 10 to the number changes the digit sum by 1+0 = 1 (adds 1 to the tens digit).
New digit sum → digital root increases by 1.
Original digital root = 5.
New digital root = 6.
2. Adding 11 adds 1+1=2 to the digit sum each time (approximately).
The digital roots increase by 2 each step: e.g., if starting number has root r, the sequence has roots r, r+2, r+4, ... cycling mod 9.
The pattern of digital roots repeats with period 9 (they increase by 2 each time, cycling through all 9 values).
3. 9a is divisible by 9 → digital root contribution = 0.
36b is divisible by 9 → digital root contribution = 0.
13 → digital root = 1+3 = 4.
So digital root of 9a + 36b + 13 = 4.
4. (i) Parity and digital root:
No fixed relationship. Even numbers can have any digital root (2→2, 4→4, 6→6, 8→8, 18→9, 20→2, ...). Odd numbers similarly.
Parity and digital root are independent.
(ii) Digital root and remainder when divided by 3 or 9:
The digital root IS the remainder when divided by 9 (with 9 representing remainder 0).
For divisibility by 3: if digital root is 3, 6, or 9, the number is divisible by 3. The remainder when divided by 3 equals (digital root mod 3).
Digits in Disguise
i. A1 + 1B = B0
Units: 1 + B = 0 (mod 10), so B = 9 and carry = 1.
Tens: A + 1 + 1(carry) = B → A + 2 = 9 → A = 7.
Check: 71 + 19 = 90 ✓ (B=9, B0=90)
Answer: A = 7, B = 9.
ii. AB + 37 = 6A
Units: B + 7 = A (if no carry) or B + 7 = A + 10 (with carry).
Tens: A + 3 = 6 (no carry) → A = 3; or A + 3 + 1 = 6 → A = 2.
Case 1 (A=3): 3 + 3 = 6 ✓. Units: B + 7 = 3 → B = –4 (impossible).
So carry from units: B + 7 ≥ 10 → B + 7 = A + 10 → B = A + 3.
Tens with carry: A + 3 + 1 = 6 → A = 2. Then B = 2 + 3 = 5.
Check: 25 + 37 = 62. 6A = 62 ✓.
Answer: A = 2, B = 5.
iii. 3 × ON = PO where P and O are different digits.
Try N = 5: 3×5=15, so units digit of product = 5 ✓, carry = 1.
3 × ON = PO (2-digit result).
Try ON = 15: 3×15=45. O=1, N=5, P=4. All different ✓.
Check: O≠N≠P, and P≠O: 1,5,4 all different ✓.
Answer: O=1, N=5, P=4.
iv. 3 × QR = PRR (3-digit number where last two digits are same = R).
Units: 3×R ends in R → R = 0 or 5.
If R = 0: PRR = P00. 3×Q0 = P00 → Q=5: 150... P=1, R=0, Q=5.
But 3×50=150. QR=50, PRR=150. P=1,R=0,Q=5. All distinct ✓.
If R = 5: PRR = P55. 3×Q5 = P55. 3×85=255 ✓. Q=8, R=5, P=2. All distinct ✓.
Both work. Most common answer: Q=8, R=5, P=2 (QR=85, PRR=255).
Answer: Q=8, R=5, P=2.
v. 12 × 8 = 96. P=1, Q=2, R=9, S=6. All distinct.
vi. GH × H = 9K.
11×9=99: G=1,H=1 — G=H, not allowed (different letters must be different digits).
12×8=96: G=1,H=2, product=96=9K → K=6. G=1,H=2,K=6 all different ✓.
24×4=96: G=2,H=4,K=6. Check: units of H×H = 4×4=16, so K=6 ✓. GH×H=24×4=96=9K ✓.
Valid answers: 12×8=96 (G=1,H=2,K=6) and 24×4=96 (G=2,H=4,K=6).
vii. BYE × 6 = RAY.
B = 1 (as explained: product must be 3 digits, so B can only be 1).
Y is even and less than 7.
Try BYE = 162: 162 × 6 = 972. R=9, A=7, Y=2. B=1,Y=2,E=2? Y=E=2, not allowed.
Try BYE = 164: 164 × 6 = 984. R=9, A=8, Y=4. B=1,Y=4,E=4? Y=E=4, not allowed.
Try BYE = 132: 132 × 6 = 792. R=7, A=9, Y=3. But Y must be even... 3 is odd.
Try BYE = 142: 142 × 6 = 852. R=8, A=5, Y=4. B=1,Y=4,E=2: R=8,A=5,Y=4 ✓. Check all distinct: 1,4,2,8,5 ✓.
Answer: B=1, Y=4, E=2, R=8, A=5. (BYE=142, RAY=852).
Solve the following cryptarithms:
i. 3×UT = PUT → 3(10U+T) = 100P + 10U + T
30U + 3T = 100P + 10U + T
20U + 2T = 100P
10U + T = 50P
UT = 50P. P=1: UT=50. U=5,T=0,P=1. Check: 50×3=150=PUT ✓. All distinct: 5,0,1 ✓.
Answer: U=5, T=0, P=1. (50×3=150)
ii. AB = 10A + B, then 5(10A+B) = 10B + C.
50A + 5B = 10B + C → 50A = 5B + C. Since C is a single digit (0-9) and B is (0-9):
5B + C ≤ 45+9 = 54, so 50A ≤ 54 → A = 1.
50 = 5B + C. With B and C single digits: B=9, C=5: 5×9+5=50 ✓.
AB=19, BC=95. 19×5=95 ✓. B=9, C=5, A=1. All distinct ✓.
Answer: A=1, B=9, C=5. (19×5=95)
iii. 2L=1 (impossible). So L=1, carry from tens = 0.
Tens: 2×2 + carry_from_units = N (since carry_to_hundreds=0).
If carry from units = 0: N=4. Then P = 2×4 mod 10 = 8.
Check: L2N = 124. 124×2 = 248. 2NP = 248 → 2,N=4,P=8 ✓.
Answer: L=1, N=4, P=8. (124×2=248)
iv. 4×XY = ZX (2-digit result, units digit = X = tens digit of XY).
4(10X+Y) = 10Z + X
40X + 4Y = 10Z + X
39X + 4Y = 10Z
Try X=2: 78 + 4Y = 10Z. Y=3: 78+12=90 → Z=9. XY=23, ZX=92. 23×4=92 ✓. X=2,Y=3,Z=9 all distinct ✓.
Answer: X=2, Y=3, Z=9. (23×4=92)
v. P=1: 121Q = 101 + 10R. Q=1: 121=101+10R → R=2. PP=11,QQ=11 — P=Q=1 not allowed (different digits).
Try P=2: 242Q = 202 + 10R.
Q=1: 242 = 202+10R → 10R=40 → R=4. PP=22,QQ=11.
P=2,Q=1,R=4. PRP=242. 22×11=242 ✓. All distinct ✓.
Answer: P=2, Q=1, R=4. (22×11=242)
vi. KKK = 111K = 3×37×K. So JK×6 = 111K → JK = 111K/6 = 37K/2.
For JK to be integer, K must be even.
K=2: JK = 37. But then units of JK = 7 ≠ K=2. Contradiction.
K=4: JK = 74. Units of 74 = 4 = K ✓. J=7, K=4. Check: 74×6=444=KKK ✓. All distinct ✓.
Answer: J=7, K=4. (74×6=444)
Figure it Out (Final Section)
1. Digit sum = 3+1+z+5 = 9+z. For divisibility by 9: 9+z must be divisible by 9.
9+z = 9 → z=0, or 9+z = 18 → z=9.
Two answers: z = 0 or z = 9.
There are two answers because both 3105 (digit sum=9) and 3195 (digit sum=18) are multiples of 9.
2. First number: 12a + 8 (for some integer a).
Second number: 12b – 4 (for some integer b).
Sum = 12a + 8 + 12b – 4 = 12(a+b) + 4.
12(a+b) + 4 = 4(3(a+b) + 1).
This is divisible by 4 but NOT necessarily by 8.
Example: a=0, b=1: Numbers are 8 and 8. Sum=16 ÷ 8 = 2 ✓.
Example: a=0, b=2: Numbers are 8 and 20. Sum=28. 28÷8 = 3.5 ✗.
Snehal's claim is FALSE. The sum is always divisible by 4, but not always by 8.
3. Any multiple of 3 is either: (a) a multiple of 6 (even multiple of 3), or (b) an odd multiple of 3 (not divisible by 6).
Case 1: Both are multiples of 6: Sum = 6a + 6b = 6(a+b) → multiple of 6 ✓.
Case 2: Both are odd multiples of 3 (i.e., 3×odd): Sum = 3(2a+1) + 3(2b+1) = 3(2a+2b+2) = 6(a+b+1) → multiple of 6 ✓.
Case 3: One is a multiple of 6, other is an odd multiple of 3: Sum = 6a + 3(2b+1) = 3(2a+2b+1) = 3×odd → NOT a multiple of 6 ✗.
Conclusion: Sum is a multiple of 6 when both multiples of 3 are of the same type (both even or both odd multiples of 3). Sum is NOT a multiple of 6 when one is an even and one is an odd multiple of 3.
4. (i) YES. The reversed number has the same digits (just in different order), so the digit sum is the same. Since the original digit sum is divisible by 9, the reversed number's digit sum is also divisible by 9. So the reversed number is also divisible by 9.
(ii) YES. Any rearrangement of the digits of a multiple of 9 gives another multiple of 9, because all rearrangements have the same digit sum. So any permutation of the digits works.
5. 18 = 2 × 9. The number must be divisible by both 2 and 9.
Divisibility by 2: b must be even → b ∈ {0, 2, 4, 6, 8}.
Divisibility by 9: digit sum = 4+8+a+2+3+b = 17+a+b must be divisible by 9.
17+a+b = 18 → a+b = 1, or 17+a+b = 27 → a+b = 10.
Case a+b = 1 with b even: b=0, a=1. → (a,b) = (1,0) ✓.
Case a+b = 10 with b even:
b=2, a=8 → (8,2) ✓
b=4, a=6 → (6,4) ✓
b=6, a=4 → (4,6) ✓
b=8, a=2 → (2,8) ✓
All possible pairs (a, b): (1,0), (8,2), (6,4), (4,6), (2,8).
6. 44 = 4 × 11.
Divisibility by 4: last two digits q8 must be divisible by 4. 10q+8 divisible by 4 → q must be even.
So q ∈ {0, 2, 4, 6, 8}.
Divisibility by 11 (alternating digit sum from units):
Digits of 3p7q8 from units: 8, q, 7, p, 3.
Alternating sum: 8 – q + 7 – p + 3 = 18 – p – q.
Must be divisible by 11: 18–p–q = 11 → p+q = 7.
So p+q = 7, with q even:
q=0, p=7 → (7,0) ✓
q=2, p=5 → (5,2) ✓
q=4, p=3 → (3,4) ✓
q=6, p=1 → (1,6) ✓
All possible pairs (p, q): (7,0), (5,2), (3,4), (1,6).
7. Let the three consecutive numbers be n, n+1, n+2.
n divisible by 2 → n is even.
n+1 divisible by 3.
n+2 divisible by 4.
From n+2 divisible by 4 and n divisible by 2: n ≡ 2 (mod 4).
So n = 4k+2 for some k. Then n+1 = 4k+3 must be divisible by 3:
4k+3 ≡ 0 (mod 3) → k ≡ 0 (mod 3).
So k = 3m, giving n = 12m+2.
Smallest (m=0): n=2. Numbers: 2, 3, 4. Check: 2÷2 ✓, 3÷3 ✓, 4÷4 ✓.
Next: m=1: n=14. Numbers: 14, 15, 16. Check: 14÷2 ✓, 15÷3 ✓, 16÷4 ✓.
They occur every 12 numbers.
8. 45,000 ÷ 36 = 1250. So 1250×36 = 45,000.
45,000 + 36 = 45,036
45,036 + 36 = 45,072
45,072 + 36 = 45,108
45,108 + 36 = 45,144
45,144 + 36 = 45,180
Five multiples: 45,036; 45,072; 45,108; 45,144; 45,180.
9. 5 consecutive even numbers: 5p–4, 5p–2, 5p, 5p+2, 5p+4.
10. When reversed, divisible by 6: reversed number must be even (ends in even digit, so original starts with even digit) and digit sum divisible by 3 (same as original).
If original ends in 0: reversed starts with 0 (invalid as leading digit). So original must end in 5.
Original ends in 5 → divisible by 5 ✓. For div by 15, also need div by 3 (digit sum ÷ 3).
Reversed must end in even digit → original must start with even digit.
Example: 213045 — digit sum = 2+1+3+0+4+5=15 ✓ (div by 3). Ends in 5 ✓. Starts with 2 (even).
Reversed: 540312 — ends in 2 (even) ✓, digit sum=15 ✓ (div by 3) → divisible by 6 ✓.
Answer: 213045 (or any similar valid number).
11. Any multiple of 11 = 11k. Doubled = 22k = 11(2k) — always a multiple of 11.
So ALL multiples of 11, when doubled, are STILL multiples of 11.
Deepak's conjecture is FALSE. Doubling always preserves divisibility by 11.
12. (i) The product of a multiple of 6 and a multiple of 3 is a multiple of 9.
Answer: ALWAYS TRUE.
6a × 3b = 18ab = 9(2ab) — this IS always a multiple of 9.
Answer: ALWAYS TRUE.
(ii) Sum of three consecutive even numbers is divisible by 6.
Three consecutive even numbers: 2k, 2k+2, 2k+4. Sum = 6k+6 = 6(k+1). Always divisible by 6!
Answer: ALWAYS TRUE.
(iii) If abcdef is a multiple of 6, then badcef is a multiple of 6.
Answer: ALWAYS TRUE.
Both abcdef and badcef have the same set of digits (just the first two swapped).
Divisibility by 2: both end in same digit f — same result.
Divisibility by 3: digit sum is the same (same digits) — same result.
So if abcdef is divisible by 6, badcef has same divisibility by 2 and 3, hence also divisible by 6.
(iv) 8(7b – 3) – 4(11b + 1) is a multiple of 12.
We need 3|(3b–7). 3b–7 ≡ –7 ≡ –1 ≡ 2 (mod 3). So 3∤(3b–7).
So 4(3b–7) is NOT always divisible by 12.
Answer: SOMETIMES TRUE. (Divisible by 4 always, by 12 only sometimes.)
13. Any integer falls into one of 3 residue classes mod 3: remainder 0, 1, or 2.
Sum of 3 numbers is divisible by 3 when:
- All three have the same remainder mod 3 (0+0+0=0, 1+1+1=3, 2+2+2=6 — all ÷3 ✓).
- All three have different remainders mod 3 (0+1+2=3 — divisible by 3 ✓).
Sum is NOT divisible by 3 in all other cases (e.g., remainders 0,0,1 → sum ≡ 1 mod 3).
14. Product of 2 consecutive integers always a multiple of 2?
Answer: YES. Among any 2 consecutive integers, one is even. So product is always even (multiple of 2).
Product of 3 consecutive integers always a multiple of 6?
Answer: YES. Among any 3 consecutive integers, one is divisible by 3, and at least one is even. So product is divisible by both 2 and 3, hence by 6.
Product of 4 consecutive integers?
Answer: Always a multiple of 24. Among 4 consecutive integers, there is at least one multiple of 4, at least one multiple of 3, and (combined) factors include 2×4×3 = 24. More precisely, the product = 4! × C(n,4) for some n, always divisible by 4! = 24.
Product of 5 consecutive integers?
Answer: Always a multiple of 120 (= 5!). By the same logic, the product of 5 consecutive integers is always divisible by 5! = 120.
15. (i) EF × E = GGG
GGG = 111G = 3 × 37 × G. EF × E = GGG.
E × EF = 111G. Try E=3: 3 × 3F = 111G. 3 × 37 = 111 (F=7, G=1): 37×3=111 ✓. E=3,F=7,G=1. All distinct ✓.
Answer: E=3, F=7, G=1. (37×3=111)
(ii) WOW × 5 = MEOW
WOW is 3-digit, MEOW is 4-digit. 5 × WOW = MEOW. WOW × 5 gives 4 digits → WOW ≥ 200 (5×200=1000).
WOW = 100W + 10O + W = 101W + 10O.
Product ends in W (units of 5×W = units of W) → W must be 0 or 5.
W=0: WOW starts with 0 — invalid.
W=5: units of 5×5=25 → W=5 and last digit of MEOW = W = 5 ✓.
Try O=3: 5×535=2675. M=2,E=6,O=3,W=5. E=6 ✓. MEOW=2675. M=2,E=6,O=3,W=5 all distinct ✓.
Answer: W=5, O=3, M=2, E=6. (535×5=2675)
16. Every multiple of 32 is a multiple of 8 (since 32 = 4×8).
Every multiple of 8 is a multiple of 4 (since 8 = 2×4).
So: Multiples of 32 ⊂ Multiples of 8 ⊂ Multiples of 4.
This is a nested/concentric relationship.
The correct Venn diagram is (iii) — three concentric circles with Multiples of 32 innermost, then Multiples of 8, then Multiples of 4 outermost.
Puzzle Time – Navakankari
This is a two-player strategy board game. Each player has 9 pawns.
Phase 1 (Placing): Players take turns placing pawns on the 25 intersections of the board. At most one pawn per intersection.
Phase 2 (Moving): Once all pawns are placed, players take turns moving a pawn to an adjacent empty intersection to form a line of 3 (horizontally or vertically).
Phase 3 (Removing): When a player forms a line of 3, they remove any one of the opponent's pawns that is not part of the opponent's own line of 3.
Winning: A player wins when the opponent has fewer than 3 pawns or cannot make any valid move.
Why NCERT solutions help students?
Chapter 5 of Class 8 Maths is not a chapter that can be mastered by memorizing formulas — it requires genuine understanding of why number properties work. Having clear and reliable NCERT solutions helps students verify their reasoning, identify gaps in their thinking, and build the kind of algebraic confidence that directly supports exam performance. When students understand not just what the answer is, but why it is correct, they are far better prepared for higher-level maths. This chapter's style of reasoning — proving things always, sometimes, or never — is exactly the kind of thinking that NCERT and school exams expect, and working through these solutions systematically builds that mathematical maturity and confidence that lasts well beyond Class 8.
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