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Class 8 Ganita Prakash · Chapter 5
Class 8 Ganita Prakash · Chapter 5
Number Play
Page-wise textbook solutions (pages 112–135) · Every step · Figures · Common mistakes · Tips
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✔ Understood 0/59
๐️ Key ideas of this chapter
- Parity: odd ± odd = even, even ± even = even, odd ± even = odd. All a ± b ± c ± … have the same parity.
- If a divides M and N, it divides M + N and M − N. If A is divisible by k, so are all multiples of A; and A is divisible by all factors of k.
- Divisible by k and by m ⇒ divisible by LCM(k, m) (not always by k × m).
- Numbers leaving remainder r on dividing by n: \(nk + r\).
- Divisibility: 9 and 3 → digit sum; 11 → alternating sum; digital root = remainder on division by 9 (9 if divisible).
๐ Textbook Page 112
๐ Page 112Math Talk
Can every natural number be written as a sum of consecutive numbers (2 or more)? Which numbers can be written in more than one way? Can every even number be written this way? Can 0?
1Odd numbers: \(2n + 1 = n + (n+1)\), e.g. 9 = 4 + 5.
2Try small numbers: 3 = 1+2, 5 = 2+3, 6 = 1+2+3, 7 = 3+4, 9 = 4+5 = 2+3+4, 10 = 1+2+3+4 …
31, 2, 4, 8, 16, 32 … (powers of 2) can NEVER be written this way (with positive numbers).
4Numbers with more odd factors have more ways: 15 = 7+8 = 4+5+6 = 1+2+3+4+5 (3 ways).
5Even numbers that are not powers of 2 can be written: 6 = 1+2+3, 12 = 3+4+5, 20 = 2+3+4+5+6.
60 = (−1) + 0 + 1 = (−2) + (−1) + 0 + 1 + 2 using negative numbers.
✅ Answer: All numbers except powers of 2 (1, 2, 4, 8, 16 …). Numbers like 9, 15, 18, 21 have more than one way. 0 = −1 + 0 + 1.
๐ก Tip: Number of ways = (number of odd factors) − 1. 15 has odd factors 1, 3, 5, 15 → 3 ways.
๐ Page 112Explore
Put + and − signs between 3, 4, 5, 6 in all possible ways. How many are there? Find each value.
1There are 3 gaps, each can be + or −: 2 × 2 × 2 = 8 ways.
2Values: 18, 6, 8, −4, 10, −2, 0, −12.
✅ Answer: 8 expressions; all the answers are even.
๐ Textbook Page 113
๐ Page 113Math Talk
Repeat with other sets of 4 consecutive numbers (e.g. 5, 6, 7, 8). What do you notice? Explain with algebra.
1Let the numbers be a, a+1, a+2, a+3.
2a + (a+1) − (a+2) − (a+3) = −4 always.
3a − (a+1) + (a+2) − (a+3) = −2 always.
4a − (a+1) − (a+2) + (a+3) = 0 always.
5Other values change with a, but all values are even.
✅ Answer: The values 0, −2 and −4 always appear, and every value is even.
๐ Textbook Page 114
๐ Page 114Math Talk
Take ANY 4 numbers (not consecutive). Do all 8 sign-expressions have the same parity? Why?
1Example 2, 7, 9, 4: 2+7+9+4 = 22, 2−7+9−4 = 0, 2+7−9−4 = −4 … all even.
2Switching −b to +b changes the value by 2b (even). Switching +c to −c changes it by −2c (even).
3A change by an even number never changes parity, so all 8 values have the same parity.
✅ Answer: Yes — all of them always have the same parity.
๐ก Tip: This works for any count of numbers, not just 4: a ± b ± c ± … all have the same parity.
๐ Textbook Page 115
๐ Page 115Breaking Even
Without calculating, which are even? 43 + 37, 672 − 348, 4 × 347 × 3, 708 − 477, 809 + 214, 119 × 303, 543 − 479, 513³
1odd ± odd = even; even ± even = even; odd ± even = odd.
2A product is even if any factor is even; odd × odd = odd.
343 + 37: odd + odd = even; 672 − 348: even; 4 × 347 × 3: even (factor 4)
4708 − 477: even − odd = odd; 809 + 214: odd; 119 × 303: odd
5543 − 479: odd − odd = even; 513³ = 513 × 513 × 513: odd
✅ Answer: Even: 43 + 37, 672 − 348, 4 × 347 × 3, 543 − 479.
๐ Page 115Breaking Even
Which expressions are even for ALL integers? 2a + 2b, 3g + 5h, 4m + 2n, 2u − 4v, 13k − 5k, 6m − 3n, x² + 2, b² + 1, 4k × 3j
12a + 2b = 2(a + b) ✔
23g + 5h: g = 1, h = 0 gives 3 ✘ (not always)
34m + 2n = 2(2m + n) ✔
42u − 4v = 2(u − 2v) ✔
513k − 5k = 8k ✔
66m − 3n: m = 0, n = 1 gives −3 ✘
7x² + 2: x = 3 gives 11 ✘
8b² + 1: b = 2 gives 5 ✘
94k × 3j = 12kj ✔
✅ Answer: Always even: 2a + 2b, 4m + 2n, 2u − 4v, 13k − 5k, 4k × 3j.
๐ก Tip: An expression is always even if you can take out a factor 2.
๐ Textbook Page 116
๐ Page 116Practice
Write a few algebraic expressions that always give an even number.
1Any expression with a common factor 2 works.
✅ Answer: e.g. 2n, 6a + 4b, 10x − 2y, n(n + 1), 2(a + b + c).
๐ก n(n + 1) is always even because one of two consecutive numbers is even.
๐ Page 116Pairs to Make Fours
When is the sum of two even numbers a multiple of 4?
1Even numbers are either 4p (multiple of 4) or 4p + 2.
24p + 4q = 4(p + q) → multiple of 4 ✔
3(4p + 2) + (4q + 2) = 4(p + q + 1) → multiple of 4 ✔
44p + (4q + 2) = 4(p + q) + 2 → NOT a multiple of 4
✅ Answer: When both are multiples of 4, or both are not. If only one is a multiple of 4, the sum is not.
Examples: 12 + 16 = 28 ✔, 6 + 10 = 16 ✔, 8 + 6 = 14 ✘
๐ Textbook Page 118
๐ Page 118Always/Sometimes/Never
Statement 1 is true for addition. Is it true for subtraction: if 8 divides two numbers, does it divide their difference?
18a − 8b = 8(a − b)
✅ Answer: Yes, always true. Example: 56 − 16 = 40 = 8 × 5.
๐ Textbook Page 121
๐ Page 121Math Talk
Always, sometimes or never? (6) A number divisible by both 9 and 4 must be divisible by 36. (7) A number divisible by both 6 and 4 must be divisible by 24.
1(6) 9 and 4 have no common factor, so LCM(9, 4) = 36 → always true.
2(7) LCM(6, 4) = 12, not 24. 12 is divisible by 6 and 4 but not by 24 → sometimes true (true for 24, 48 …).
✅ Answer: (6) Always true (7) Sometimes true
๐ก Tip: Divisible by k and by m → divisible by LCM(k, m), not always by k × m.
๐ Page 121What Remains?
Find numbers that leave remainder 3 when divided by 5. Which expressions describe all of them: 3k + 5, 3k − 5, 3k/5, 5k + 3, 5k − 2, 5k − 3?
1Examples: 3, 8, 13, 18, 23, 28 …
2They are 3 more than a multiple of 5 → 5k + 3 (k = 0, 1, 2 …).
3They are also 2 less than a multiple of 5 → 5k − 2 (k = 1, 2, 3 …).
4Others like 5k + 8 or 5k − 7 also work (with suitable k).
✅ Answer: (iv) 5k + 3 and (v) 5k − 2.
๐ Textbook Page 122
๐ Page 122Figure it Out · Q1
The sum of four consecutive numbers is 34. Find them.
1Let them be n, n+1, n+2, n+3.
24n + 6 = 34 → 4n = 28 → n = 7
✅ Answer: 7, 8, 9, 10
๐ Page 122Figure it Out · Q2
p is the greatest of five consecutive numbers. Write the other four in terms of p.
1Each is 1 less than the next.
✅ Answer: p − 1, p − 2, p − 3, p − 4
๐ Page 122Figure it Out · Q3
Always, sometimes or never? (i) The sum of two even numbers is a multiple of 3. (ii) If a number is not divisible by 18, it is not divisible by 9. (iii) If two numbers are not divisible by 6, their sum is not divisible by 6. (iv) A multiple of 6 + a multiple of 9 is a multiple of 3. (v) A multiple of 6 + a multiple of 3 is a multiple of 9.
1(i) Sometimes: 2 + 4 = 6 ✔, 2 + 6 = 8 ✘
2(ii) Sometimes: 20 — not divisible by 18 or 9 ✔; 27 — not divisible by 18 but divisible by 9 ✘
3(iii) Sometimes: 7 + 8 = 15 ✔ (not divisible); 4 + 8 = 12 ✘ (divisible)
4(iv) Always: 6x + 9y = 3(2x + 3y)
5(v) Sometimes: 18 + 9 = 27 ✔, 12 + 9 = 21 ✘
✅ Answer: (i) S (ii) S (iii) S (iv) A (v) S
๐ Page 122Figure it Out · Q4
Find numbers that leave remainder 2 when divided by 3 and also when divided by 4. Write an expression for all of them.
1The number minus 2 must be a multiple of both 3 and 4, i.e. of 12.
2Numbers: 2, 14, 26, 38, 50 …
✅ Answer: 12n + 2 (n = 0, 1, 2, …)
๐ Page 122Figure it Out · Q5
Pebble riddle: groups of 3 leave 1, pairs leave 1 (odd), groups of 5 leave 1, groups of 7 leave none, fewer than 100. How many pebbles?
1Leaves 1 with 2, 3 and 5 → the number is 1 more than a multiple of LCM(2, 3, 5) = 30.
2Candidates: 31, 61, 91.
3Only 91 = 7 × 13 is a multiple of 7.
✅ Answer: 91 pebbles
๐ Page 122Figure it Out · Q6
Numbers leave remainder 2 when divided by 6. Is the sum of any three of them always a multiple of 6?
1(6a + 2) + (6b + 2) + (6c + 2) = 6(a + b + c) + 6 = 6(a + b + c + 1)
✅ Answer: Yes, Tathagat is right. Example: 8 + 14 + 20 = 42.
๐ Textbook Page 123
๐ Page 123Figure it Out · Q7
661 leaves remainder 3 and 4779 leaves remainder 5 when divided by 7. Find the remainders of (i) 4779 + 661 (ii) 4779 − 661.
14779 = 7p + 5, 661 = 7q + 3
2(i) Sum = 7(p + q) + 8 = 7(p + q + 1) + 1 → remainder 1
3(ii) Difference = 7(p − q) + 2 → remainder 2
4Visual: the leftover dots 5 + 3 = 8 make one more row of 7 with 1 left; 5 − 3 leaves 2.
✅ Answer: (i) 1 (ii) 2
๐ Page 123Figure it Out · Q8
Find the smallest number that leaves remainder 2 when divided by 3, 3 when divided by 4 and 4 when divided by 5.
1Each remainder is 1 less than the divisor → the number + 1 is a multiple of 3, 4 and 5.
2LCM(3, 4, 5) = 60 → number = 60 − 1 = 59.
3It is the smallest because 60 is the smallest common multiple.
✅ Answer: 59 (others: 119, 179 …)
๐ Page 123Explain
Using algebra, explain the divisibility shortcuts for 5, 2, 4 and 8.
1Number = … + 1000d + 100c + 10b + a.
210b, 100c, 1000d … are multiples of 10, so of 2 and of 5 → only a decides divisibility by 2 and 5.
3100c, 1000d … are multiples of 4 → only the last two digits (10b + a) decide divisibility by 4.
41000d, 10000e … are multiples of 8 → only the last three digits decide divisibility by 8.
✅ Answer: Higher place values are already multiples, so only the last 1, 2 or 3 digits matter.
๐ Textbook Page 124
๐ Page 124Think
Is 10 divisible by 9? What remainder do multiples of 10 and 100 leave when divided by 9? Use it to find the remainder of 427 ÷ 9.
110 = 9 + 1 → remainder 1. 30 = 3 × 9 + 3 → remainder 3 (number of tens).
2100 = 99 + 1 → each hundred leaves 1. 400 leaves 4.
3427: 4 + 2 + 7 = 13 → 13 = 9 + 4 → remainder 4.
✅ Answer: 427 ÷ 9 leaves remainder 4.
๐ Textbook Page 125
๐ Page 125Check
Which are correct? (i) Divisible by 9 → digit sum divisible by 9. (ii) Digit sum divisible by 9 → divisible by 9. (iii) Not divisible by 9 → digit sum not divisible by 9. (iv) Digit sum not divisible by 9 → number not divisible by 9.
1The number and its digit sum always leave the same remainder when divided by 9.
2So one is divisible by 9 exactly when the other is.
✅ Answer: All four are correct.
๐ Textbook Page 126
๐ Page 126Figure it Out · Q1
Without dividing, which are divisible by 9? (i) 123 (ii) 405 (iii) 8888 (iv) 93547 (v) 358095
1Digit sums: 6, 9, 32, 28, 30
2Only 9 is a multiple of 9.
✅ Answer: Only (ii) 405.
๐ Page 126Figure it Out · Q2
Find the smallest multiple of 9 with no odd digits.
1Digits must be 0, 2, 4, 6, 8 and the digit sum a multiple of 9.
2The sum of even digits is even, so it must be 18 (not 9).
3Two digits give at most 16, so we need 3 digits: 2 + 8 + 8 = 18 → 288.
✅ Answer: 288
๐ Page 126Figure it Out · Q3
Find the multiple of 9 closest to 6000.
16000: digit sum 6 → 6000 = multiple of 9 + 6.
26000 − 6 = 5994 (6 away); 5994 + 9 = 6003 (3 away).
✅ Answer: 6003
๐ Page 126Figure it Out · Q4
How many multiples of 9 are there between 4300 and 4400?
1First: 4302 (digit sum 9). Last: 4392.
2(4392 − 4302) ÷ 9 + 1 = 10 + 1 = 11
✅ Answer: 11
๐ Page 126Explain
Why does the digit-sum rule work for 3?
110 = 9 + 1, 100 = 99 + 1 … every place value is 1 more than a multiple of 9, so also of 3.
2So the number leaves the same remainder as its digit sum when divided by 3.
✅ Answer: Because every power of 10 leaves remainder 1 when divided by 3.
๐ Textbook Page 127
๐ Page 127Math Talk
Is 462 divisible by 11? What is a general shortcut?
1Place values 1, 100, 10000 … are 1 MORE than a multiple of 11; 10, 1000 … are 1 LESS.
2462: 4 (hundreds, +4) − 6 (tens) + 2 (units) = 0
30 → divisible. 462 = 11 × 42.
✅ Answer: Yes. Rule: (sum of digits in odd places from the right) − (sum of digits in even places) must be 0 or a multiple of 11.
๐ Textbook Page 128
๐ Page 128Practice
Using the shortcut, check divisibility by 11 and find the remainder: (i) 158 (ii) 841 (iii) 481 (iv) 5529 (v) 90904 (vi) 857076
1(i) 8 − 5 + 1 = 4 → remainder 4
2(ii) 1 − 4 + 8 = 5 → remainder 5
3(iii) 1 − 8 + 4 = −3 → 3 short, remainder 11 − 3 = 8
4(iv) 9 − 2 + 5 − 5 = 7 → remainder 7
5(v) 4 − 0 + 9 − 0 + 9 = 22 → multiple of 11 → divisible
6(vi) 6 − 7 + 0 − 7 + 5 − 8 = −11 → divisible
✅ Answer: (i) 4 (ii) 5 (iii) 8 (iv) 7 (v) divisible (vi) divisible
⚠️ Common mistake: For a negative result like −3, the remainder is 11 − 3 = 8, not 3.
๐ Textbook Page 129
๐ Page 129Table
Fill the table: is each number divisible by 2, 3, 4, 5, 6, 8, 9, 10, 11?
12, 5, 10: last digit. 4: last two digits. 8: last three digits.
23, 9: digit sum. 6: both 2 and 3. 11: alternating sum.
3Note: 128 is divisible by 4 (28 = 4 × 7).
| Number | 2 | 3 | 4 | 5 | 6 | 8 | 9 | 10 | 11 |
|---|---|---|---|---|---|---|---|---|---|
| 128 | ✔ | ✘ | ✔ | ✘ | ✘ | ✔ | ✘ | ✘ | ✘ |
| 990 | ✔ | ✔ | ✘ | ✔ | ✔ | ✘ | ✔ | ✔ | ✔ |
| 1586 | ✔ | ✘ | ✘ | ✘ | ✘ | ✘ | ✘ | ✘ | ✘ |
| 275 | ✘ | ✘ | ✘ | ✔ | ✘ | ✘ | ✘ | ✘ | ✔ |
| 6686 | ✔ | ✘ | ✘ | ✘ | ✘ | ✘ | ✘ | ✘ | ✘ |
| 639210 | ✔ | ✔ | ✘ | ✔ | ✔ | ✘ | ✘ | ✔ | ✔ |
| 429714 | ✔ | ✔ | ✘ | ✘ | ✔ | ✘ | ✔ | ✘ | ✘ |
| 2856 | ✔ | ✔ | ✔ | ✘ | ✔ | ✔ | ✘ | ✘ | ✘ |
| 3060 | ✔ | ✔ | ✔ | ✔ | ✔ | ✘ | ✔ | ✔ | ✘ |
| 406839 | ✘ | ✔ | ✘ | ✘ | ✘ | ✘ | ✘ | ✘ | ✘ |
✅ Answer: See the table (Y = yes, N = no).
๐ Page 129Check
Does checking 2 and 3 work for 6? Test 38, 225, 186, 64.
138: even, digit sum 11 ✘ → no
2225: odd → no
3186: even, digit sum 15 ✔ → yes (6 × 31)
464: digit sum 10 ✘ → no
✅ Answer: Yes — only 186 is divisible by 6.
๐ Textbook Page 130
๐ Page 130Explain
Why does checking 4 and 6 NOT work for 24, while checking 3 and 8 does?
124 = 2³ × 3.
24 = 2² and 6 = 2 × 3 → together they only guarantee 2² × 3 = 12. Example: 12 passes both but is not divisible by 24.
38 = 2³ and 3 have no common factor → together they guarantee 2³ × 3 = 24.
✅ Answer: Because LCM(4, 6) = 12, but LCM(3, 8) = 24.
๐ Page 130Digital Roots
Between 600 and 700, which numbers have digital root (i) 5 (ii) 7 (iii) 3?
1Digital root = remainder on dividing by 9 (9 if divisible).
2(i) 608, 617, 626, 635, 644, 653, 662, 671, 680, 689, 698
3(ii) 601, 610, 619, 628, 637, 646, 655, 664, 673, 682, 691 (and 700)
4(iii) 606, 615, 624, 633, 642, 651, 660, 669, 678, 687, 696
✅ Answer: See the lists — numbers with the same digital root are 9 apart.
๐ Page 130Math Talk
Find digital roots of 12 consecutive numbers, and of consecutive multiples of 3, 4 and 6. What about numbers 1 more than a multiple of 6?
1Consecutive numbers: 1, 2, …, 9, 1, 2, 3 — the roots go round 1 to 9.
2Multiples of 3: 3, 6, 9, 3, 6, 9 …
3Multiples of 4: 4, 8, 3, 7, 2, 6, 1, 5, 9, then repeat.
4Multiples of 6: 6, 3, 9, 6, 3, 9 …
51 more than a multiple of 6 (7, 13, 19, 25 …): 7, 4, 1, 7, 4, 1 …
✅ Answer: Digital roots repeat in cycles because they are remainders on division by 9.
๐ Page 130Riddle
Digits each the tiniest odd digit; digits count, digit sum and root all equal the largest odd single digit. What is the number?
1Tiniest odd digit: 1 → all digits are 1.
2Largest odd single digit: 9 → nine digits, digit sum 9, root 9.
✅ Answer: 111111111 (eleven crore eleven lakh eleven thousand one hundred eleven)
๐ Textbook Page 131
๐ Page 131Figure it Out · Q1
The digital root of an 8-digit number is 5. What is the digital root of 10 more than it?
1Adding 10 = adding 9 + 1; adding 9 does not change the root.
2So the root goes up by 1.
✅ Answer: 6
๐ Page 131Figure it Out · Q2
Start with any number and keep adding 11. What happens to the digital roots?
1Adding 11 = adding 9 + 2 → the root goes up by 2 each time (after 9 comes back round).
2Start 10: roots 1, 3, 5, 7, 9, 2, 4, 6, 8, 1 …
✅ Answer: Roots increase by 2 each step and cycle through all of 1–9.
๐ Page 131Figure it Out · Q3
What is the digital root of 9a + 36b + 13?
19a + 36b + 13 = 9(a + 4b + 1) + 4
✅ Answer: 4
๐ Page 131Figure it Out · Q4
Conjectures: (i) parity of a number vs its digital root (ii) digital root vs remainder on dividing by 3 or 9.
1(i) 12 (even) root 3; 13 (odd) root 4; 21 (odd) root 3 → no link.
2(ii) root 9 → remainder 0 by 9; otherwise remainder by 9 = root.
3Root 3, 6, 9 → divisible by 3; root 1, 4, 7 → remainder 1; root 2, 5, 8 → remainder 2.
✅ Answer: (i) No pattern (ii) The root gives the remainder by 9 (9 means 0) and by 3.
๐ Page 131Cryptarithms
Solve: (i) A1 + 1B = B0 (ii) AB + 37 = 6A (iii) ON + ON + ON = PO (iv) QR + QR + QR = PRR
1(i) Units: 1 + B ends in 0 → B = 9 (carry 1). Tens: A + 1 + 1 = 9 → A = 7. 71 + 19 = 90 ✔
2(ii) Units: B + 7 ends in A, tens: A + 3 (+carry) = 6 → A = 2, B = 5. 25 + 37 = 62 ✔
3(iii) 3 × ON = PO, so ON is at most 33 and 3 × N must end in O.
413 × 3 = 39 → O = 3, N = 1, P = 9. (17 × 3 = 51 and 24 × 3 = 72 also work.)
5(iv) 3 × QR = PRR → 85 × 3 = 255 (Q = 8, R = 5, P = 2)
✅ Answer: (i) A = 7, B = 9 (ii) A = 2, B = 5 (iii) O = 3, N = 1, P = 9 (other answers: 17 × 3 = 51, 24 × 3 = 72) (iv) Q = 8, R = 5, P = 2
๐ Page 131Cryptarithms
(v) PQ × 8 = RS — can PQ be 13? (vi) GH × H = 9K (vii) BYE × 6 = RAY
1(v) 13 × 8 = 104 has 3 digits, so no. Only 12 × 8 = 96 works.
2(vi) H × H must end in K and units of GH is H: from the list only 24 × 4 = 96 fits (G = 2, H = 4, K = 6).
3(vii) B = 1. Y is even and less than 7. E × 6 must end in Y; try: 105 × 6 = 630 (Y = 0, E = 5, R = 6, A = 3).
✅ Answer: (v) No; PQ = 12 (vi) 24 × 4 = 96 (vii) 105 × 6 = 630
๐ Textbook Page 132
๐ Page 132Cryptarithms
Solve: (i) UT × 3 = PUT (ii) AB × 5 = BC (iii) L2N × 2 = 2NP (iv) XY × 4 = ZX (v) PP × QQ = PRP (vi) JK × 6 = KKK
1(i) 50 × 3 = 150 → U = 5, T = 0, P = 1
2(ii) 19 × 5 = 95 → A = 1, B = 9, C = 5
3(iii) 125 × 2 = 250 → L = 1, N = 5, P = 0 (also 124 × 2 = 248)
4(iv) 23 × 4 = 92 → X = 2, Y = 3, Z = 9
5(v) 22 × 11 = 242 → P = 2, Q = 1, R = 4 (also 33 × 11 = 363, 44 × 11 = 484)
6(vi) 74 × 6 = 444 → J = 7, K = 4
✅ Answer: (i) 50 × 3 = 150 (ii) 19 × 5 = 95 (iii) 125 × 2 = 250 (iv) 23 × 4 = 92 (v) 22 × 11 = 242 (vi) 74 × 6 = 444
๐ก Tip: Start with the units digit and with the size of the answer.
๐ Page 132Figure it Out · Q1
31z5 is a multiple of 9. Find z. Why are there two answers?
13 + 1 + z + 5 = 9 + z must be a multiple of 9.
2z = 0 (sum 9) or z = 9 (sum 18).
✅ Answer: z = 0 or 9 — both 0 and 9 keep the sum a multiple of 9.
๐ Page 132Figure it Out · Q2
A number leaves remainder 8 on dividing by 12; another is 4 short of a multiple of 12. Snehal says their sum is always a multiple of 8. Check.
1(12n + 8) + (12m − 4) = 12(n + m) + 4
2n + m = 1: 16 ✔; n + m = 2: 28 ✘
✅ Answer: Not always — only sometimes (when n + m is odd).
๐ Page 132Figure it Out · Q3
When is the sum of two multiples of 3 a multiple of 6?
13m + 3n = 3(m + n)
2It is a multiple of 6 when m + n is even: both m, n even or both odd.
36 + 12 = 18 ✔ (m = 2, n = 4); 3 + 9 = 12 ✔ (1, 3); 3 + 6 = 9 ✘ (1, 2)
✅ Answer: When both multiples are even, or both are odd (i.e. m + n even).
๐ Page 132Figure it Out · Q4
Sreelatha says reversing the digits of a multiple of 9 gives another multiple of 9. (i) Is it always true? (ii) What other shuffles work?
1Reversing or shuffling digits does not change the digit sum.
2So the digit sum stays a multiple of 9.
✅ Answer: (i) Always true (ii) Every rearrangement of the digits is still a multiple of 9.
๐ Page 132Figure it Out · Q5
48a23b is a multiple of 18. List all pairs (a, b).
118 = 2 × 9 → b is even and 17 + a + b is a multiple of 9.
2a + b = 1 or 10: (1, 0), (8, 2), (6, 4), (4, 6), (2, 8)
✅ Answer: (1, 0), (8, 2), (6, 4), (4, 6), (2, 8)
๐ Textbook Page 133
๐ Page 133Figure it Out · Q6
3p7q8 is divisible by 44. List all pairs (p, q).
144 = 4 × 11. By 4: q8 must be divisible by 4 → q = 0, 2, 4, 6, 8.
2By 11: (8 + 7 + 3) − (p + q) = 18 − (p + q) must be 0 or 11.
3p + q = 18 needs p = q = 9, but q must be even → so p + q = 7.
4q = 0, p = 7; q = 2, p = 5; q = 4, p = 3; q = 6, p = 1
✅ Answer: (7, 0), (5, 2), (3, 4), (1, 6)
๐ Page 133Figure it Out · Q7
Find three consecutive numbers where the first is a multiple of 2, the second of 3 and the third of 4. How often do they occur?
12, 3, 4 works.
2Adding LCM(2, 3, 4) = 12 keeps all conditions: 14, 15, 16; 26, 27, 28 …
✅ Answer: 2, 3, 4 and then every 12: (12k + 2, 12k + 3, 12k + 4).
๐ Page 133Figure it Out · Q8
Write five multiples of 36 between 45,000 and 47,000.
145000 = 36 × 1250, so keep adding 36.
2Check: divisible by 4 (last two digits) and 9 (digit sum).
✅ Answer: 45036, 45072, 45108, 45144, 45180
๐ Page 133Figure it Out · Q9
The middle of 5 consecutive even numbers is 5p. Write the others.
1Consecutive even numbers differ by 2.
✅ Answer: 5p − 4, 5p − 2, 5p + 2, 5p + 4
๐ Page 133Figure it Out · Q10
Write a 6-digit number divisible by 15 whose reverse is divisible by 6.
1Divisible by 15 → by 3 and 5 → ends in 5 (ending in 0 would make the reverse start with 0).
2Reverse divisible by 6 → reverse is even → the number starts with an even digit.
3Digit sum a multiple of 3: e.g. 200025 (sum 9).
✅ Answer: e.g. 200025 (reverse 520002 = 6 × 86667)
๐ Page 133Figure it Out · Q11
Deepak says some multiples of 11 stay multiples of 11 when doubled, but others do not. True?
111k × 2 = 11 × (2k) — always a multiple of 11.
✅ Answer: False — every multiple of 11 stays a multiple of 11 when doubled.
๐ Page 133Figure it Out · Q12
Always, sometimes or never? (i) Multiple of 6 × multiple of 3 is a multiple of 9. (ii) Sum of three consecutive even numbers is divisible by 6. (iii) If abcdef is a multiple of 6, so is badcef. (iv) 8(7b − 3) − 4(11b + 1) is a multiple of 12.
1(i) 6a × 3b = 18ab → always
2(ii) (2n − 2) + 2n + (2n + 2) = 6n → always
3(iii) Same digits (same digit sum) and same last digit f → always
4(iv) 56b − 24 − 44b − 4 = 12b − 28 = 12(b − 3) + 8 → never
✅ Answer: (i) A (ii) A (iii) A (iv) N
๐ Page 133Figure it Out · Q13
When is the sum of 3 numbers divisible by 3? Explore all cases.
1Look at the remainders (0, 1 or 2) on dividing by 3.
2Divisible when the remainders add to 0, 3 or 6:
3all three remainders the same (0,0,0), (1,1,1), (2,2,2), or all different (0, 1, 2).
✅ Answer: When all three leave the same remainder, or all three leave different remainders.
๐ Page 133Figure it Out · Q14
Is the product of 2 consecutive integers always a multiple of 2? Of 3 consecutive a multiple of 6? What about 4 and 5 consecutive integers?
12 consecutive: one is even → multiple of 2.
23 consecutive: one is a multiple of 3 and at least one is even → multiple of 6.
34 consecutive: a multiple of 3, a multiple of 4 and another even number → multiple of 24.
45 consecutive: also a multiple of 5 → multiple of 120.
✅ Answer: Yes; yes; 4 consecutive → multiple of 24; 5 consecutive → multiple of 120.
๐ Page 133Figure it Out · Q15
Solve: (i) EF × E = GGG (ii) WOW × 5 = MEOW
1(i) GGG = 111 × G = 3 × 37 × G → 37 × 3 = 111 (E = 3, F = 7, G = 1)
2(ii) W × 5 ends in W → W = 5. 575 × 5 = 2875 → O = 7, M = 2, E = 8
✅ Answer: (i) 37 × 3 = 111 (ii) 575 × 5 = 2875
๐ Textbook Page 134
๐ Page 134Figure it Out · Q16
Which Venn diagram shows multiples of 4, 8 and 32?
1Every multiple of 32 is a multiple of 8, and every multiple of 8 is a multiple of 4.
2So 32 inside 8 inside 4.
✅ Answer: (iv) — three circles one inside another, with multiples of 4 outside.
๐ Textbook Page 135
๐ Page 135Game
Navakankari (Nine Men's Morris): what strategy helps you win?
1Each player has 9 pawns; make lines of 3 to remove the opponent's pawns.
2Place pawns at the middle points of lines (they join more lines).
3Try to build two possible lines at once so the opponent can block only one.
✅ Answer: Plan lines ahead, block the opponent's two-in-a-row, and keep your pawns able to move.
๐ Where do we use this?
- Checking calculations quickly (digital roots were used by Aryabhata II for this).
- Barcodes, ISBN numbers and bank account checks use remainders to catch typing mistakes.
๐ Link to higher classes
- Class 9–10: Euclid's division lemma, HCF/LCM proofs, number theory; Class 11: mathematical induction.
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