NCERT solutions for Class 8 Maths Chapter 5 Number Play

This worksheet provides complete and accurate NCERT solutions for Class 8 Maths Chapter 5 Number Play from Ganita Prakash Part 1. Number Play is one of the most engaging and thought-provoking chapters in the Class 8 Maths syllabus. It takes students beyond basic arithmetic and into the fascinating world of number patterns, divisibility rules, parity, remainders, digital roots, and cryptarithms. The chapter encourages students to explore, observe, and reason mathematically rather than simply memorise formulas. This makes it especially important for building strong analytical thinking and problem-solving skills that are essential for higher classes and competitive examinations.

Chapter summary: themes and concepts covered

Chapter 5 Number Play is not built around a story or poem — it is an exploration-based and activity-driven chapter. The chapter begins by asking students to explore whether every natural number can be written as a sum of consecutive numbers. Through guided discovery, students learn that powers of 2 cannot be expressed as such sums, while most other numbers can be. The chapter then moves through several interconnected mathematical themes — parity of arithmetic expressions, divisibility rules for 2, 3, 4, 5, 8, 9, 11, and 24, the concept of digital roots, solving cryptarithms (digit puzzles), and understanding remainders using algebra. A puzzle activity called Navakankari, a traditional two-player strategy board game, is also included at the end of the chapter.

What this NCERT chapter covers?

The key learning areas explored in this chapter include the following. Students investigate the sum of consecutive natural numbers and identify which numbers can or cannot be expressed as such sums. They study the parity (odd or even nature) of arithmetic and algebraic expressions, learning to determine parity without actual computation. The chapter covers divisibility rules for 2, 3, 4, 5, 8, 9, 11, and 24, with algebraic explanations for why each rule works. Students learn about digital roots — what they are, how they relate to divisibility by 9, and what patterns they form for consecutive numbers and multiples. Cryptarithms (also called digit-in-disguise puzzles) are introduced, requiring students to find unknown digits using logical reasoning and algebraic thinking. Concepts like LCM, remainders, modular thinking, and Venn diagrams for sets of multiples are also woven into the chapter.

How to use these NCERT solutions?

Students should first attempt every question in the worksheet on their own before referring to the solutions provided here. For exploration-type questions, they should try to observe patterns and form their own conclusions before checking the answers. Parents can use this worksheet to guide their child through the chapter section by section, verifying answers and discussing the reasoning behind each solution. Teachers will find the step-by-step algebraic explanations particularly useful for classroom discussions and concept reinforcement. Since all solutions follow the exact order and structure of the NCERT chapter, this worksheet is equally helpful for first-time learning and for revision before exams and unit tests.

Student tips and learning tricks

For the parity section, remember that switching any sign from + to − changes the value by an even amount, so the parity of all expressions formed from the same set of numbers is always the same. When checking divisibility, always apply the specific rule — for 9, add all digits; for 11, use the alternating digit sum from the units place. For cryptarithms, always start with the units digit and work column by column, tracking carries carefully. When a letter appears as both the multiplier and part of the product (such as in GH × H), use trial and error with small values and verify each step. A common mistake is assuming that divisibility by two factors automatically means divisibility by their product — remember that this only works when the two factors are coprime (for example, 3 and 8 for 24, not 4 and 6). For digital roots, remember that the digital root always equals the remainder when the number is divided by 9, with the digital root being 9 when the remainder is 0.

Why NCERT solutions are important?

NCERT solutions aligned to the Ganita Prakash textbook ensure that students receive answers that are fully consistent with what is expected in school assessments and board examinations. For a concept-heavy chapter like Number Play, having access to clear, step-by-step algebraic reasoning helps students understand not just the answer but also the method. This builds mathematical confidence and reduces dependence on rote learning. Strong foundational understanding of divisibility, parity, and number properties directly supports performance in topics like algebra, number theory, and reasoning in later classes. These solutions also help parents and teachers identify where a student may need additional practice or clarification.

Complete answer key – NCERT solutions

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.

Four consecutive numbers – expressions with + and − signs

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.

Using 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: Identify which algebraic expressions always give even numbers:
1. 2a + 2b → 2(a + b) — always even ✓ ALWAYS EVEN
2. 3g + 5h → 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 ✓)
Type A + Type B = NOT a multiple of 4 (e.g., 4 + 6 = 10 ✗, remainder = 2)
Algebraic explanation: Let the numbers be 4p and (4q + 2). Then 4p + (4q + 2) = 4(p + q) + 2. This leaves remainder 2 when divided by 4, so it is not a multiple of 4. Example: 4 + 6 = 10; 8 + 10 = 18 — neither is 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 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 algebraic structure of place values.)

(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.
1 ÷ 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?
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.

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:
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. 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: 608, 617, 626, 635, 644, 653, 662, 671, 680, 689, 698.
(ii) 7: 610, 619, 628, 637, 646, 655, 664, 673, 682, 691, 700.
(iii) 3: 606, 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. They form a repeating pattern: 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 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. The digital roots increase by 2 each step: 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, …). 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 ✓
Answer: A = 7, B = 9.

ii. AB + 37 = 6A
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 (2-digit result)
Try ON = 15: 3×15 = 45. O=1, N=5, P=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: 3×50 = 150. P=1, Q=5, R=0. All distinct ✓.
If R = 5: 3×85 = 255. Q=8, R=5, P=2. All distinct ✓.
Most common answer: Q=8, R=5, P=2 (QR=85, PRR=255).
Answer: Q=8, R=5, P=2.

Cryptarithms (Multiplication)

v. 12 × 8 = 96. P=1, Q=2, R=9, S=6. All distinct ✓.

vi. Valid answers: 12×8=96 (G=1, H=2, K=6) and 24×4=96 (G=2, H=4, K=6).

vii. Try BYE = 142: 142 × 6 = 852. B=1, Y=4, E=2, R=8, A=5. 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
20U + 2T = 100P
10U + T = 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. 5 × AB = BC
AB = 10A + B, then 5(10A+B) = 10B + C.
50A + 5B = 10B + C → 50A = 5B + C.
A=1: 50 = 5B + C. B=9, C=5: 5×9+5=50 ✓. AB=19, BC=95. 19×5=95 ✓.
Answer: A=1, B=9, C=5. (19×5=95)

iii. 2 × L2N = 2NP
L=1, N=4, P=8. Check: L2N = 124. 124×2 = 248. 2NP = 248 ✓.
Answer: L=1, N=4, P=8. (124×2=248)

iv. 4 × XY = ZX
4(10X+Y) = 10Z + X → 39X + 4Y = 10Z.
Try X=2: 78 + 4Y = 10Z. Y=3: 78+12=90 → Z=9. 23×4=92 ✓.
Answer: X=2, Y=3, Z=9. (23×4=92)

v. PP × QQ = PRP
P=2, Q=1, R=4. PRP=242. 22×11=242 ✓. All distinct ✓.
Answer: P=2, Q=1, R=4. (22×11=242)

vi. JK × 6 = KKK
KKK = 111K = 3×37×K. JK = 37K/2. K must be even.
K=4: JK = 74. Units of 74 = 4 = K ✓. J=7, K=4. Check: 74×6=444=KKK ✓.
Answer: J=7, K=4. (74×6=444)

Further cryptarithms

(i) EF × E = GGG
GGG = 111G = 3×37×G. Try E=3: 3×37=111. F=7, G=1. 37×3=111 ✓.
Answer: E=3, F=7, G=1. (37×3=111)

(ii) WOW × 5 = MEOW
W=5 (since units of 5×5=25 → W=5).
Try O=3: 5×535=2675. M=2, E=6, O=3, W=5 all distinct ✓.
Answer: W=5, O=3, M=2, E=6. (535×5=2675)

Figure it Out

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. Both 3105 (digit sum=9) and 3195 (digit sum=18) are multiples of 9.

2. First number: 12a + 8. Second number: 12b − 4.
Sum = 12a + 8 + 12b − 4 = 12(a+b) + 4 = 4(3(a+b) + 1).
This is divisible by 4 but NOT necessarily by 8.
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 multiple of 6 (even multiple of 3) or 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: Sum = 3(2a+1) + 3(2b+1) = 6(a+b+1) → multiple of 6 ✓.
Case 3: One is a multiple of 6, other is an odd multiple of 3: Sum = 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. 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.

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.
All possible pairs (a, b): (1,0), (8,2), (6,4), (4,6), (2,8).

6. 44 = 4 × 11.
Divisibility by 4: q must be even → q ∈ {0, 2, 4, 6, 8}.
Alternating digit sum for 3p7q8: 8 − q + 7 − p + 3 = 18 − p − q.
For divisibility by 11: 18 − p − q = 11 → p+q = 7.
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 ≡ 2 (mod 4), so n = 4k+2. Then n+1 = 4k+3 must be divisible by 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. They occur every 12 numbers.

8. 45,000 ÷ 36 = 1250. So 1250×36 = 45,000.
Five multiples of 36 after 45,000: 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, the number must be divisible by 6 (even ending and digit sum divisible by 3). Original must start with an even digit and end in 5 (to be divisible by 15). Example: 213045. Reversed: 540312 — ends in 2 (even) ✓, digit sum=15 ✓ → 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.

12.
(i) Product of a multiple of 6 and a multiple of 3 is a multiple of 9.
6a × 3b = 18ab = 9(2ab) — 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.
Both have the same set of digits — same digit sum (divisibility by 3) and same last digit f (divisibility by 2). Answer: ALWAYS TRUE.

(iv) 4(3b−7) divisible by 12?
3b−7 ≡ 2 (mod 3). So 4(3b−7) is NOT always divisible by 3, hence not always 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, or all three have different remainders mod 3 (0+1+2=3). Sum is NOT divisible by 3 in all other cases.

14. Product of 2 consecutive integers always a multiple of 2? YES. Among any 2 consecutive integers, one is even.
Product of 3 consecutive integers always a multiple of 6? 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? Always a multiple of 24. Among 4 consecutive integers, there is at least one multiple of 4, at least one multiple of 3. Product = 4! × C(n,4), always divisible by 4! = 24.
Product of 5 consecutive integers? Always a multiple of 120 (= 5!). The product of 5 consecutive integers is always divisible by 5! = 120.

15.
(i) EF × E = GGG: E=3, F=7, G=1. (37×3=111)
(ii) WOW × 5 = MEOW: 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. 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 where each player has 9 pawns. Phase 1 (Placing): Players take turns placing pawns on the 25 intersections of the board, with 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.

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