Figure it Out — Multiples, Remainders and Consecutive Numbers — Number Play

1. 12 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 122

   The sum of four consecutive numbers is 34. What are these numbers?

   Hint. Call the smallest one n and write the other three in terms of it.

   Step 1 — Name the numbers algebraically. Let the smallest be n. Consecutive numbers go up in steps of 1, so the four are n, n + 1, n + 2, n + 3.

   Step 2 — Form the equation. n + (n + 1) + (n + 2) + (n + 3) = 34 4n + 6 = 34

   Step 3 — Solve. 4n = 28, so n = 7.

   Step 4 — Write out the four numbers. 7, 8, 9, 10.

   Check: 7 + 8 + 9 + 10 = 34 ✓

   A shortcut worth knowing. The sum of four consecutive numbers is always 4n + 6, so it is always even but never a multiple of 4 — because 4n is a multiple of 4 and 6 is not. Dividing 34 by 4 gives 8.5, and the four numbers straddle that value: 7, 8, 9, 10.

   ✦ The numbers are 7, 8, 9 and 10.

2. 22 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 122

   Suppose p is the greatest of five consecutive numbers. Describe the other four numbers in terms of p.

   Hint. If p is the greatest, every other number lies below it.

   Since p is the greatest of the five, all the others are smaller, so each is obtained by subtracting from p rather than adding.

   Counting down one at a time from p: • the number just below p is p − 1 • the next is p − 2 • then p − 3 • and the smallest is p − 4

   So the five consecutive numbers, written in order, are p − 4, p − 3, p − 2, p − 1, p.

   Check with an example: if p = 20, the five numbers are 16, 17, 18, 19, 20 ✓ — five numbers, consecutive, with 20 the greatest.

   A useful extra. Their sum is (p − 4) + (p − 3) + (p − 2) + (p − 1) + p = 5p − 10 = 5(p − 2), which is always a multiple of 5. That happens because the middle number is p − 2, and the sum of any odd count of consecutive numbers equals the count times the middle one.

   ✦ The other four numbers are p − 1, p − 2, p − 3 and p − 4.

3. 35 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 122

   For each statement below, determine whether it is always true, sometimes true, or never true. Justify your claim using algebra, with examples and non-examples. (i) The sum of two even numbers is a multiple of 3. (ii) If a number is not divisible by 18, then it is also not divisible by 9. (iii) If two numbers are not divisible by 6, then their sum is not divisible by 6. (iv) The sum of a multiple of 6 and a multiple of 9 is a multiple of 3. (v) The sum of a multiple of 6 and a multiple of 3 is a multiple of 9.

   Hint. To show 'sometimes true' you need one example AND one non-example. To show 'always true' you need algebra, not examples.

   The rule of the game: one example is never enough to prove 'always', but a single non-example is enough to destroy it. So 'always true' claims need algebra, while 'sometimes true' claims need a matching pair.

   (i) The sum of two even numbers is a multiple of 3 — SOMETIMES TRUE. Examples: 2 + 4 = 6 ✓ and 4 + 8 = 12 ✓ are multiples of 3. Non-examples: 2 + 6 = 8 ✗ and 6 + 8 = 14 ✗ are not. Algebraically the sum is 2a + 2b = 2(a + b), which is guaranteed even but says nothing about divisibility by 3, since 3 does not divide 2 and may or may not divide (a + b).

   (ii) Not divisible by 18 ⇒ not divisible by 9 — SOMETIMES TRUE. Example where it holds: 30 is divisible by neither 18 nor 9 ✓ Non-example: 27 is not divisible by 18, yet it is divisible by 9 ✗ The reasoning: 18 = 2 × 9, so a multiple of 18 must be a multiple of 9, but the converse fails for odd multiples of 9 like 9, 27, 45, 63.

   (iii) Two numbers not divisible by 6 ⇒ their sum not divisible by 6 — SOMETIMES TRUE. Example where it holds: 9 and 11 are not divisible by 6, and neither is 20 ✓ Non-example: 8 and 10 are not divisible by 6, but 8 + 10 = 18 is ✗ The reason is that the remainders can add up to 6: 8 leaves 2 and 10 leaves 4, and 2 + 4 = 6.

   (iv) Multiple of 6 + multiple of 9 is a multiple of 3 — ALWAYS TRUE. Let the numbers be 6x and 9y. 6x + 9y = 3(2x + 3y) Since the sum is 3 times a whole number, it is a multiple of 3 for every choice of x and y. This works because 3 divides both 6 and 9, so it divides any combination of them.

   (v) Multiple of 6 + multiple of 3 is a multiple of 9 — SOMETIMES TRUE. Example: 18 + 9 = 27 ✓ is a multiple of 9. Non-example: 12 + 9 = 21 ✗ is not. Algebraically 6x + 3y = 3(2x + y), which guarantees a multiple of 3 but only gives a multiple of 9 when (2x + y) happens to be a multiple of 3.

   ✦ (i) Sometimes true. (ii) Sometimes true. (iii) Sometimes true. (iv) Always true, since 6x + 9y = 3(2x + 3y). (v) Sometimes true.

4. 43 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 122

   Find a few numbers that leave a remainder of 2 when divided by 3 and a remainder of 2 when divided by 4. Write an algebraic expression to describe all such numbers.

   Hint. If the remainder is the same both times, what can you say about the number minus that remainder?

   Step 1 — Strip off the common remainder. Let the number be x. We are told x = 3a + 2 and x = 4b + 2. Subtracting 2 from both: x − 2 = 3a and x − 2 = 4b.

   So x − 2 is a common multiple of 3 and 4. This is the key move — because the two remainders happen to be equal, the same shift of 2 clears both divisions at once.

   Step 2 — Use the LCM. The common multiples of 3 and 4 are exactly the multiples of their LCM: LCM(3, 4) = 12. So x − 2 = 12n for some whole number n.

   Step 3 — Write the general form. x = 12n + 2

   Step 4 — List a few. n = 1 → 14, n = 2 → 26, n = 3 → 38, n = 4 → 50, n = 5 → 62.

   Check 38: 38 ÷ 3 = 12 remainder 2 ✓ and 38 ÷ 4 = 9 remainder 2 ✓

   Watch out: this shortcut only works because both remainders are 2. If the remainders differed, x − 2 would not clear both divisions and you would have to search the residues directly.

   ✦ The numbers are 14, 26, 38, 50, 62, … and in general 12n + 2.

5. 54 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 122 (pebble riddle)

   "I hold some pebbles, not too many, When I group them in 3's, one stays with me. Try pairing them up — it simply won't do, A stubborn odd pebble remains in my view. Group them by 5, yet one's still around, But grouping by seven, perfection is found. More than one hundred would be far too bold, Can you tell me the number of pebbles I hold?"

   Hint. Translate each line into a remainder condition, then handle the three 'remainder 1' conditions together.

   Step 1 — Turn the verse into arithmetic. • "group them in 3's, one stays" → remainder 1 when divided by 3 • "pairing them up… a stubborn odd pebble" → remainder 1 when divided by 2 • "group them by 5, one's still around" → remainder 1 when divided by 5 • "grouping by seven, perfection" → remainder 0 when divided by 7 • "more than one hundred would be too bold" → the number is at most 100

   Step 2 — Combine the three 'remainder 1' conditions. Since the remainder is 1 in all three cases, the number minus 1 is divisible by 3, 2 and 5 alike. So p − 1 is a common multiple of 3, 2 and 5, and LCM(3, 2, 5) = 30. Therefore p = 30k + 1.

   Step 3 — List the candidates under 100. k = 1 → 31 k = 2 → 61 k = 3 → 91 (k = 4 gives 121, which is over 100.)

   Step 4 — Apply the last condition, divisibility by 7. 31 ÷ 7 = 4 remainder 3 ✗ 61 ÷ 7 = 8 remainder 5 ✗ 91 ÷ 7 = 13 exactly ✓

   Check 91 against every line: 91 = 3×30 + 1 ✓, odd ✓, 91 = 5×18 + 1 ✓, 91 = 7×13 ✓, and 91 ≤ 100 ✓

   ✦ There are 91 pebbles.

6. 63 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 122

   Tathagat has written several numbers that leave a remainder of 2 when divided by 6. He claims, "If you add any three such numbers, the sum will always be a multiple of 6." Is Tathagat's claim true?

   Hint. Write three such numbers algebraically and add them. What happens to the three remainders?

   Yes, Tathagat's claim is true.

   Step 1 — Write the three numbers algebraically. A number leaving remainder 2 on division by 6 has the form 6k + 2. So let the three numbers be 6a + 2, 6b + 2, 6c + 2.

   Step 2 — Add them. (6a + 2) + (6b + 2) + (6c + 2) = 6a + 6b + 6c + 6 = 6(a + b + c + 1)

   Step 3 — Read off the conclusion. The sum is 6 times a whole number, so it is a multiple of 6 for every choice of a, b and c. The claim holds always, not just sometimes.

   Why it works — the idea behind the algebra. Each number carries a leftover of 2. Three leftovers of 2 make 6, which is exactly one more full group of 6. So the three stray remainders combine into a complete group and nothing is left over.

   Check with numbers: 8 + 14 + 20 = 42 = 6 × 7 ✓ and 2 + 2 + 2 = 6 ✓

   Worth noticing: the claim would fail for two such numbers, since (6a + 2) + (6b + 2) = 6(a + b) + 4 leaves remainder 4. It is specifically three that works, because 3 × 2 = 6.

   ✦ True — three numbers of the form 6k + 2 add to 6(a + b + c + 1), because the three leftovers of 2 combine into exactly one more group of 6.

7. 73 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 123

   When divided by 7, the number 661 leaves a remainder of 3, and 4779 leaves a remainder of 5. Without calculating, what remainders will these leave when divided by 7? (i) 4779 + 661 (ii) 4779 − 661

   Hint. Write each number in the form 7 × something + remainder, then add or subtract the two expressions.

   The point of this question is that remainders can be added and subtracted on their own, without touching the large numbers at all.

   Write the two numbers in remainder form: 4779 = 7p + 5 661 = 7q + 3

   (i) 4779 + 661 Adding the two expressions: 4779 + 661 = (7p + 5) + (7q + 3) = 7p + 7q + 8

   But 8 is bigger than the divisor 7, so it is not yet a valid remainder — one more group of 7 can still be taken out: = 7p + 7q + 7 + 1 = 7(p + q + 1) + 1

   So the remainder is 1.

   (ii) 4779 − 661 Subtracting: 4779 − 661 = (7p + 5) − (7q + 3) = 7p − 7q + 2 = 7(p − q) + 2

   Here 2 is already less than 7, so no regrouping is needed and the remainder is 2.

   Visual version. Picture each number as rows of 7 with a short row left over — 661 has a leftover row of 3, and 4779 has a leftover row of 5. Putting them together gives a leftover of 8, which is one full row of 7 plus 1 spare. Taking the smaller away from the larger leaves 5 − 3 = 2 spare.

   Check: 4779 + 661 = 5440, and 5440 = 7 × 777 + 1 ✓. 4779 − 661 = 4118 = 7 × 588 + 2 ✓

   ✦ (i) remainder 1, since the leftovers 5 + 3 = 8 give one more full 7 with 1 spare. (ii) remainder 2, since 5 − 3 = 2.

8. 83 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 123

   Find a number that leaves a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5. What is the smallest such number? Can you give a simple explanation of why it is the smallest?

   Hint. Compare each remainder with its divisor. What do you notice?

   Step 1 — Spot the pattern in the remainders. 2 is one less than 3. 3 is one less than 4. 4 is one less than 5. Every remainder is exactly one short of its divisor.

   Step 2 — Use that to simplify the problem. If a number n is one short of a multiple of 3, of 4 and of 5, then n + 1 is divisible by 3, 4 and 5 all at once. Adding 1 turns all three awkward remainders into zero simultaneously.

   Step 3 — Find the smallest such n + 1. n + 1 must be a common multiple of 3, 4 and 5, so it is a multiple of LCM(3, 4, 5) = 60. The smallest positive option is n + 1 = 60.

   Step 4 — Solve. n = 60 − 1 = 59.

   Check: 59 ÷ 3 = 19 remainder 2 ✓, 59 ÷ 4 = 14 remainder 3 ✓, 59 ÷ 5 = 11 remainder 4 ✓

   Step 5 — Why it is the smallest. Any valid n must have n + 1 divisible by 3, 4 and 5, and 60 is the smallest number with that property — that is precisely what 'least common multiple' means. So no smaller positive n can work.

   All such numbers are 60k − 1: 59, 119, 179, 239, …

   ✦ 59. Each remainder is one less than its divisor, so n + 1 must be divisible by 3, 4 and 5; the smallest such value is LCM(3,4,5) = 60, giving n = 59.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 8 Mathematics textbook Ganita Prakash Part 1, Reprint 2026-27 (hegp105.pdf). The chapter is about divisibility reasoning, digital roots and cryptarithms — questions sit in four 'Figure it Out' blocks (pages 122, 126, 131 and 132-135) plus several in-text drills. Every numeric answer here was independently recomputed before being compared with the book's printed answer key: the full 10-row divisibility table was re-derived rule by rule, every cryptarithm was re-solved from scratch, and the divisibility-by-44 and multiple-of-18 digit pairs were found by exhaustive search over the digits.. Questions are referenced from the NCERT textbook for identification.