Figure it Out — Divisibility, Conjectures and Cryptarithms — Number Play

1. 12 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 132

   If 31z5 is a multiple of 9, where z is a digit, what is the value of z? Explain why there are two answers to this problem.

   Hint. Apply the digit-sum test, then remember z can be any digit from 0 to 9.

   Step 1 — Apply the divisibility test for 9. The digit sum is 3 + 1 + z + 5 = 9 + z, and this must be a multiple of 9.

   Step 2 — Find which digits work. Since z is a single digit, 9 + z runs from 9 to 18. The multiples of 9 in that range are 9 and 18: • 9 + z = 9 → z = 0 • 9 + z = 18 → z = 9

   Step 3 — Explain why there are two answers. The digit sum already contains 9, so it is a multiple of 9 before z is added. Adding z keeps it a multiple of 9 only when z itself contributes a whole extra 9 — or nothing at all. Both 0 and 9 leave the remainder unchanged, since 9 ≡ 0 (mod 9). That is exactly why two digits work rather than one.

   More generally, whenever a digit-sum condition has one solution z, the value z + 9 also works if it is still a single digit. Here z = 0 gives the second solution z = 9.

   Check: 3105 → digit sum 9, and 3105 ÷ 9 = 345 ✓ Check: 3195 → digit sum 18, and 3195 ÷ 9 = 355 ✓

   ✦ z = 0 or z = 9. The digit sum is 9 + z, and both 0 and 9 leave it a multiple of 9 because 9 ≡ 0 (mod 9).

2. 23 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 132

   "I take a number that leaves a remainder of 8 when divided by 12. I take another number which is 4 short of a multiple of 12. Their sum will always be a multiple of 8", claims Snehal. Examine his claim and justify your conclusion.

   Hint. Write both numbers in terms of 12, add them, and test the result against 8 with actual values.

   Snehal's claim is false.

   Step 1 — Write both numbers algebraically. A number leaving remainder 8 on division by 12: a = 12n + 8. A number 4 short of a multiple of 12: b = 12m − 4.

   Step 2 — Add them. a + b = (12n + 8) + (12m − 4) = 12n + 12m + 4 = 12(n + m) + 4

   Writing k = n + m, the sum is always 12k + 4.

   Step 3 — Test that form against 8. The sum is certainly a multiple of 4, since 12k + 4 = 4(3k + 1). But being a multiple of 8 is a stronger demand, and it depends on k: • k = 1 → 16, which is a multiple of 8 ✓ • k = 2 → 28, which is not a multiple of 8 ✗ • k = 3 → 40 ✓ • k = 4 → 52 ✗

   So the sum is a multiple of 8 only when k is odd, and fails whenever k is even.

   Step 4 — Give a concrete counter-example. Take a = 20 (leaves remainder 8 on division by 12 ✓) and b = 8 (which is 4 short of 12 ✓). Sum = 28, and 28 ÷ 8 = 3.5 — not a whole number.

   So the claim fails, and the correct statement is that the sum is always a multiple of 4, but a multiple of 8 only half the time.

   ✦ False. The sum is always 12k + 4, which is a multiple of 4 but a multiple of 8 only when k is odd — for instance 20 + 8 = 28 is not divisible by 8.

3. 33 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 132

   When is the sum of two multiples of 3 a multiple of 6, and when is it not? Explain the different possible cases, and generalise the pattern.

   Hint. Write both as 3 times something and factorise the sum. What extra condition does 6 need?

   Step 1 — Write the two numbers and add. Let them be 3m and 3n. Then 3m + 3n = 3(m + n).

   So the sum is guaranteed to be a multiple of 3 — that part never fails.

   Step 2 — See what 6 additionally requires. Since 6 = 3 × 2, the sum is a multiple of 6 only if 3(m + n) also carries a factor of 2. The 3 supplies no factor of 2, so (m + n) must be even.

   Step 3 — Work through the cases.

   m	n	m + n	Sum a multiple of 6?
   even	even	even	Yes
   odd	odd	even	Yes
   even	odd	odd	No
   odd	even	odd	No

   Examples: • 6 + 12 = 18 = 6 × 3 ✓ (here m = 2, n = 4, both even) • 9 + 15 = 24 = 6 × 4 ✓ (m = 3, n = 5, both odd) • 6 + 9 = 15 ✗ (m = 2, n = 3, one of each)

   Step 4 — Generalise. A multiple of 3 is even exactly when its multiplier is even. So the rule can be restated in terms of the numbers themselves: The sum of two multiples of 3 is a multiple of 6 exactly when the two numbers are both even or both odd — that is, when they have the same parity.

   This matches the general fact that the sum of two numbers is even precisely when they share the same parity.

   ✦ The sum 3(m + n) is a multiple of 6 exactly when m + n is even — that is, when the two multiples of 3 are both even or both odd. If one is even and the other odd, the sum is a multiple of 3 but not of 6.

4. 43 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 132

   Sreelatha says, "I have a number that is divisible by 9. If I reverse its digits, it will still be divisible by 9." (i) Examine if her conjecture is true for any multiple of 9. (ii) Are any other digit shuffles possible such that the number formed is still a multiple of 9?

   Hint. The 9-test depends only on the digit sum. Does reversing change the digits themselves?

   (i) Yes — Sreelatha's conjecture is true for every multiple of 9.

   The test for divisibility by 9 depends only on the sum of the digits, not on their order. Reversing a number uses exactly the same digits, just written in the opposite sequence, so the digit sum is completely unchanged.

   If the original digit sum was a multiple of 9, the reversed number has the same digit sum and is therefore also a multiple of 9.

   Examples: • 189 → digit sum 18 ✓. Reversed: 981 → digit sum 18 ✓ (981 ÷ 9 = 109) • 4527 → digit sum 18 ✓. Reversed: 7254 → digit sum 18 ✓ (7254 ÷ 9 = 806)

   (ii) Yes — in fact any rearrangement of the digits works, not just reversal.

   The same reasoning applies with full force: shuffling digits into any order whatsoever leaves the multiset of digits untouched, so the digit sum is identical and divisibility by 9 survives.

   From 189 we can form 198, 819, 891, 918, 981 — every one of them is a multiple of 9.

   The one caution: a rearrangement must not put a 0 at the front, since that would not be a genuine number of the same length. And the quotient changes even though divisibility does not — 189 ÷ 9 = 21 while 981 ÷ 9 = 109.

   Same for 3. Because the test for 3 is also a digit-sum test, every shuffle of a multiple of 3 is again a multiple of 3. This does not extend to 2, 4, 5, 8 or 11, whose tests depend on which digits sit in which positions.

   ✦ (i) True — reversing leaves the digit sum unchanged, and the 9-test depends only on that sum. (ii) Yes, any shuffle of the digits works, for exactly the same reason (provided no leading zero).

5. 54 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 133

   If 48a23b is a multiple of 18, list all possible pairs of values for a and b.

   Hint. 18 = 2 × 9. Handle the two conditions separately, then combine.

   Step 1 — Split the condition. 18 = 2 × 9, and 2 and 9 share no common factor, so the number must be divisible by both 2 and 9.

   Step 2 — Use divisibility by 2. The last digit is b, so b must be even: b ∈ {0, 2, 4, 6, 8}.

   Step 3 — Use divisibility by 9. The digit sum is 4 + 8 + a + 2 + 3 + b = 17 + a + b, which must be a multiple of 9. Since a and b are digits, 17 + a + b runs from 17 to 35, so the only possible multiples of 9 are 18 and 27.

   Step 4 — Work through each even b.

   b	17 + a + b must be 18 or 27	a
   0	17 + a = 18 → a = 1 (27 → a = 10 ✗)	1
   2	19 + a = 27 → a = 8 (18 → a = −1 ✗)	8
   4	21 + a = 27 → a = 6	6
   6	23 + a = 27 → a = 4	4
   8	25 + a = 27 → a = 2	2

   Step 5 — Collect the answers. The pairs (a, b) are (1, 0), (8, 2), (6, 4), (4, 6) and (2, 8).

   Check two of them: • 481230 → digit sum 18 ✓, ends in 0 so even ✓, and 481230 ÷ 18 = 26735 ✓ • 482238 → digit sum 27 ✓, even ✓, and 482238 ÷ 18 = 26791 ✓

   Notice the pattern: apart from the first pair, a + b = 10 each time — a consequence of needing the digit sum to reach 27.

   ✦ (a, b) = (1, 0), (8, 2), (6, 4), (4, 6) and (2, 8).

6. 64 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 133

   If 3p7q8 is divisible by 44, list all possible pairs of values for p and q.

   Hint. 44 = 4 × 11. Use the last-two-digits test for 4 and the alternating-sum test for 11.

   Step 1 — Factorise the divisor. 44 = 4 × 11, and 4 and 11 share no common factor, so the number must be divisible by both 4 and 11.

   Step 2 — Use divisibility by 4. A number is divisible by 4 when its last two digits are. Here they are q8, so we need q8 ÷ 4 to be whole: 08 ✓, 18 ✗, 28 ✓, 38 ✗, 48 ✓, 58 ✗, 68 ✓, 78 ✗, 88 ✓, 98 ✗ So q ∈ {0, 2, 4, 6, 8} — q must be even.

   Step 3 — Use divisibility by 11. Take the alternating sum from the units digit of 3 p 7 q 8: 8 − q + 7 − p + 3 = 18 − (p + q) This must be 0 or a multiple of 11.

   Since p and q are digits, p + q runs from 0 to 18, so 18 − (p + q) runs from 0 to 18. The candidates are 0 and 11.

   Case 18 − (p + q) = 0, i.e. p + q = 18. This needs p = q = 9, but q must be even — impossible.

   Case 18 − (p + q) = 11, i.e. p + q = 7. With q even: • q = 0 → p = 7 ✓ • q = 2 → p = 5 ✓ • q = 4 → p = 3 ✓ • q = 6 → p = 1 ✓ • q = 8 → p = −1 ✗

   Step 4 — Collect the answers. The pairs (p, q) are (7, 0), (5, 2), (3, 4) and (1, 6).

   Check two of them: • 37708 → last two 08 ÷ 4 ✓; alternating 8 − 0 + 7 − 7 + 3 = 11 ✓; and 37708 ÷ 44 = 857 ✓ • 31768 → last two 68 ÷ 4 ✓; alternating 8 − 6 + 7 − 1 + 3 = 11 ✓; and 31768 ÷ 44 = 722 ✓

   ✦ (p, q) = (7, 0), (5, 2), (3, 4) and (1, 6) — in every case p + q = 7.

7. 73 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 133

   Find three consecutive numbers such that the first is a multiple of 2, the second a multiple of 3, and the third a multiple of 4. Are there more such numbers? How often do they occur?

   Hint. Call the first n and write all three conditions in terms of n.

   Step 1 — Write the conditions. Let the three consecutive numbers be n, n + 1, n + 2. Then • n is a multiple of 2 • n + 1 is a multiple of 3 • n + 2 is a multiple of 4

   Step 2 — Tighten the first and third conditions. If n + 2 is a multiple of 4 then n is 2 less than a multiple of 4, so n leaves remainder 2 on division by 4. (This automatically makes n even, so the first condition is already covered.)

   Step 3 — Handle the middle condition. n + 1 divisible by 3 means n leaves remainder 2 on division by 3.

   Step 4 — Combine. n leaves remainder 2 on division by both 4 and 3. As in the earlier question, subtracting 2 clears both: n − 2 is a multiple of LCM(3, 4) = 12, so n = 12k + 2.

   Step 5 — List the solutions. k = 0 → n = 2, giving 2, 3, 4 ✓ (2 is a multiple of 2, 3 of 3, 4 of 4) k = 1 → n = 14, giving 14, 15, 16 ✓ k = 2 → n = 26, giving 26, 27, 28 ✓ k = 3 → n = 38, giving 38, 39, 40 ✓

   Step 6 — Answer the frequency question. Yes, there are infinitely many, and they occur once every 12 numbers — the pattern repeats with period 12, because 12 is the LCM of the divisors involved.

   ✦ 2, 3, 4 is the smallest set; in general n = 12k + 2, so such triples occur once every 12 numbers — 2/3/4, then 14/15/16, then 26/27/28, and so on.

8. 83 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 133

   Write five multiples of 36 between 45,000 and 47,000. Share your approach.

   Hint. 36 = 4 × 9, so combine the two tests — or simply find one multiple and add 36 repeatedly.

   Approach 1 — find the first one, then step by 36. 45000 ÷ 36 = 1250 exactly, so 45000 is itself a multiple of 36. The next ones come by adding 36 each time: 36 × 1251 = 45036 36 × 1252 = 45072 36 × 1253 = 45108 36 × 1254 = 45144 36 × 1255 = 45180

   All five lie between 45,000 and 47,000 ✓

   Approach 2 — use the divisibility tests (the method the question is really after). Since 36 = 4 × 9 and 4 and 9 share no common factor, a number is a multiple of 36 exactly when it is a multiple of both 4 and 9. • For 4: the last two digits must be divisible by 4 — so they must be 00, 04, 08, 12, 16, 20, … • For 9: the digit sum must be a multiple of 9.

   Checking 45036: last two digits 36 ÷ 4 = 9 ✓, digit sum 4+5+0+3+6 = 18 ✓ — a multiple of 36. Checking 45072: last two 72 ÷ 4 = 18 ✓, digit sum 18 ✓ Checking 45108: last two 08 ÷ 4 = 2 ✓, digit sum 18 ✓

   A useful check: the digit sums all come to 18 or 27 here, which is a good quick filter — but remember divisibility by 9 alone is not enough, since 45018 has digit sum 18 yet 18 ÷ 4 fails the test for 4.

   ✦ 45036, 45072, 45108, 45144, 45180 — found by noting 45000 = 36 × 1250 and stepping up by 36, or by testing for divisibility by both 4 and 9.

9. 92 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 134

   The middle number in a sequence of 5 consecutive even numbers is 5p. Express the other four numbers in terms of p.

   Hint. Consecutive even numbers differ by 2, not 1.

   Step 1 — Note the step size. Consecutive even numbers go up in steps of 2, not 1. This is the whole point of the question.

   Step 2 — Build outwards from the middle. The middle number is given as 5p. Two numbers sit below it and two above: • two below: 5p − 4 and 5p − 2 • the middle: 5p • two above: 5p + 2 and 5p + 4

   So the other four numbers are 5p − 4, 5p − 2, 5p + 2 and 5p + 4.

   Step 3 — Check with a value. Take p = 4, so the middle number is 20. The five numbers become 16, 18, 20, 22, 24 ✓ — five consecutive even numbers with 20 in the middle.

   A bonus worth noticing. Their sum is (5p − 4) + (5p − 2) + 5p + (5p + 2) + (5p + 4) = 25p, since the −4 and +4 cancel, as do the −2 and +2. So the sum is always 5 times the middle number — the same rule that holds for any odd count of equally spaced numbers.

   The common error is writing 5p − 2, 5p − 1, 5p + 1, 5p + 2, which would be right for consecutive whole numbers but skips the fact that these are even.

   ✦ The other four numbers are 5p − 4, 5p − 2, 5p + 2 and 5p + 4.

10. 104 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 134

    Write a 6-digit number that is divisible by 15, such that when the digits are reversed, it is divisible by 6.

    Hint. 15 = 3 × 5 and 6 = 2 × 3. Note that the digit sum survives reversal but the end digits swap.

    Step 1 — Break both conditions down. Divisible by 15 means divisible by 3 and 5. Divisible by 6 means divisible by 2 and 3.

    Step 2 — Notice what reversal does and does not change. The digit sum is unchanged by reversal, so if the original passes the test for 3, so does the reversed number. That handles the shared factor of 3 in one stroke. What does change is which digit is at each end — and both remaining conditions are about end digits.

    Step 3 — Fix the last digit of the original. Divisibility by 5 requires the original to end in 0 or 5. If it ended in 0, the reversed number would start with 0, which is not a genuine 6-digit number. So the original must end in 5.

    Step 4 — Fix the first digit of the original. The reversed number must be even, and its last digit is the original's first digit. So the original must start with an even digit: 2, 4, 6 or 8 (not 0, since a number cannot start with 0).

    Step 5 — Satisfy the digit sum. The digit sum must be a multiple of 3.

    Step 6 — Build one. Take 200025: • ends in 5 ✓ so divisible by 5 • digit sum 2 + 0 + 0 + 0 + 2 + 5 = 9, a multiple of 3 ✓ • so 200025 ÷ 15 = 13335 ✓ Reversed: 520002 • even ✓ so divisible by 2 • digit sum still 9 ✓ so divisible by 3 • so 520002 ÷ 6 = 86667 ✓

    Other valid answers include 200055, 200085, 202005, 405015 and 800001 — any 6-digit number starting with an even digit, ending in 5, with digit sum a multiple of 3.

    ✦ 200025 works: it is divisible by 15, and reversed to 520002 it is divisible by 6. The recipe is start with an even digit, end in 5, and make the digit sum a multiple of 3.

11. 113 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 134

    Deepak claims, "There are some multiples of 11 which, when doubled, are still multiples of 11. But other multiples of 11 don't remain multiples of 11 when doubled." Examine his conjecture and explain your conclusion.

    Hint. Write a general multiple of 11 and double it. Does anything depend on which multiple you picked?

    Deepak's conjecture is false. Every multiple of 11, without exception, stays a multiple of 11 when doubled.

    Step 1 — Write a general multiple of 11. Let n be any multiple of 11, so n = 11k for some whole number k.

    Step 2 — Double it. 2n = 2 × 11k = 22k = 11 × (2k)

    Step 3 — Read off the conclusion. Since 2n is 11 times the whole number 2k, it is a multiple of 11 — and this holds for every value of k. There is no room for exceptions, because k never appears in a position where it could break the factor of 11.

    Examples: 33 × 2 = 66 = 11 × 6 ✓, 121 × 2 = 242 = 11 × 22 ✓, 55 × 2 = 110 = 11 × 10 ✓

    Where the intuition goes wrong. Multiplying by 2 can destroy other properties — doubling an odd number makes it even, doubling a multiple of 9 need not give a multiple of 18. But multiplying can never remove a factor that is already present. Once 11 divides n, it divides any multiple of n.

    The general rule: if a number is a multiple of d, then so is every multiple of that number. Deepak has confused this with the reverse situation, where dividing can lose a factor.

    ✦ False. Any multiple of 11 is 11k, and doubling gives 22k = 11(2k), which is a multiple of 11 for every k — multiplying can never remove a factor that is already there.

12. 124 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 134

    Determine whether the statements below are 'Always True', 'Sometimes True' or 'Never True', explaining your reasoning. (i) The product of a multiple of 6 and a multiple of 3 is a multiple of 9. (ii) The sum of three consecutive even numbers will be divisible by 6. (iii) If abcdef is a multiple of 6, then badcef will be a multiple of 6. (iv) 8(7b − 3) − 4(11b + 1) is a multiple of 12.

    Hint. For each one, do the algebra rather than testing examples — three of these are decided outright.

    (i) Product of a multiple of 6 and a multiple of 3 is a multiple of 9 — ALWAYS TRUE. Let the numbers be 6a and 3b. 6a × 3b = 18ab = 9 × (2ab) Since the product is 9 times a whole number, it is a multiple of 9 for every a and b. The reason it always works is that 6 contributes one factor of 3 and 3 contributes another, and two factors of 3 make 9. Example: 12 × 15 = 180 = 9 × 20 ✓

    (ii) The sum of three consecutive even numbers is divisible by 6 — ALWAYS TRUE. Let them be 2n, 2n + 2, 2n + 4. Sum = 6n + 6 = 6(n + 1) This is 6 times a whole number every time. Example: 8 + 10 + 12 = 30 = 6 × 5 ✓

    (iii) If abcdef is a multiple of 6, then badcef is a multiple of 6 — ALWAYS TRUE. Divisibility by 6 needs divisibility by both 2 and 3. • By 3: the test is the digit sum, and swapping a↔b and c↔d only rearranges the digits, leaving the sum unchanged ✓ • By 2: the test is the last digit, and f is untouched by these swaps ✓ Both conditions survive, so the new number is still a multiple of 6. Example: 123456 ÷ 6 = 20576 ✓, and 213546 ÷ 6 = 35591 ✓

    (iv) 8(7b − 3) − 4(11b + 1) is a multiple of 12 — NEVER TRUE. Expand carefully: 8(7b − 3) = 56b − 24 4(11b + 1) = 44b + 4 Difference = 56b − 24 − 44b − 4 = 12b − 28

    Now test against 12: 12b − 28 = 12(b − 2) − 4, so it is always 4 short of a multiple of 12. Since a leftover of −4 (equivalently 8) can never vanish whatever b is, the expression is never a multiple of 12. Check: b = 1 → −16 ✗, b = 3 → 8 ✗, b = 5 → 32 ✗ — none divisible by 12.

    ✦ (i) Always true, since 6a × 3b = 9(2ab). (ii) Always true, since the sum is 6(n + 1). (iii) Always true, because the swaps preserve both the digit sum and the last digit. (iv) Never true — the expression simplifies to 12b − 28, which is always 4 short of a multiple of 12.

13. 133 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 134

    Choose any 3 numbers. When is their sum divisible by 3? Explore all possible cases and generalise.

    Hint. Work with the remainders each number leaves on division by 3, rather than the numbers themselves.

    Step 1 — Reduce to remainders. On division by 3 the only possible remainders are 0, 1 and 2. Write the three numbers as n₁ = 3a + r₁, n₂ = 3b + r₂, n₃ = 3c + r₃. Adding: S = 3(a + b + c) + (r₁ + r₂ + r₃)

    Since the first part is already a multiple of 3, the sum is divisible by 3 exactly when r₁ + r₂ + r₃ is.

    Step 2 — Enumerate the cases. The remainders total anywhere from 0 to 6, so the workable totals are 0, 3 and 6.

    Case	Remainders	Total	Divisible?
    All the same	0,0,0 / 1,1,1 / 2,2,2	0, 3, 6	Yes
    All different	0, 1, 2 in any order	3	Yes
    Exactly two the same	e.g. 0,0,1 / 1,1,0 / 2,2,1	1, 2, 4, 5	No

    Step 3 — Generalise. The sum of three numbers is divisible by 3 exactly when their remainders mod 3 are either all equal or all different.

    Examples: • 4, 7, 10 — all leave remainder 1. Sum 21 = 3 × 7 ✓ • 5, 6, 7 — remainders 2, 0, 1, all different. Sum 18 ✓ • 4, 5, 9 — remainders 1, 2, 0, all different. Sum 18 ✓ • 4, 7, 9 — remainders 1, 1, 0, two the same. Sum 20 ✗

    Why 'all different' works: the three distinct remainders must be 0, 1 and 2, which total 3. Why 'all equal' works: the total is 3r, a multiple of 3 whatever r is.

    ✦ The sum is divisible by 3 exactly when the sum of the three remainders is 0, 3 or 6 — which happens when the three remainders mod 3 are either all the same or all different.

14. 144 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 135

    Is the product of two consecutive integers always a multiple of 2? Why? What about the product of three consecutive integers — is it always a multiple of 6? What can you say about the product of 4 consecutive integers, and of five?

    Hint. Among any k consecutive integers, how many are guaranteed to be multiples of 2? Of 3? Of 4?

    The governing principle: among any k consecutive integers, at least one is a multiple of k, and more generally at least one is a multiple of each number up to k. This is because the multiples of m occur every m steps, so a run of m consecutive integers cannot avoid one.

    Two consecutive integers — always a multiple of 2. ✓ Of any two consecutive integers, one is odd and the other even, so the product always contains a factor of 2. Example: 7 × 8 = 56 = 2 × 28 ✓

    Three consecutive integers — always a multiple of 6. ✓ Among any three consecutive integers there is at least one multiple of 2 and at least one multiple of 3. Since 2 and 3 share no common factor, the product carries both, so it is a multiple of 2 × 3 = 6. Example: 4 × 5 × 6 = 120 = 6 × 20 ✓

    Four consecutive integers — always a multiple of 24. ✓ Among any four consecutive integers there is a multiple of 4, another even number (giving an extra factor of 2), and a multiple of 3. Together these give 4 × 2 × 3 = 24. Example: 3 × 4 × 5 × 6 = 360 = 24 × 15 ✓ Example: 5 × 6 × 7 × 8 = 1680 = 24 × 70 ✓

    Five consecutive integers — always a multiple of 120. ✓ Adding a multiple of 5 to the previous argument gives 24 × 5 = 120. Example: 2 × 3 × 4 × 5 × 6 = 720 = 120 × 6 ✓ Example: 4 × 5 × 6 × 7 × 8 = 6720 = 120 × 56 ✓

    The general pattern: the product of k consecutive integers is always a multiple of k factorial (1 × 2 × 3 × … × k) — here 2, 6, 24 and 120 respectively.

    ✦ Yes to all. Two consecutive integers give a multiple of 2, three give a multiple of 6, four give a multiple of 24 and five give a multiple of 120 — in general, k consecutive integers give a multiple of k!

15. 153 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 135

    Solve the cryptarithms: (i) EF × E = GGG (ii) WOW × 5 = MEOW

    Hint. For (i), note that GGG is a repdigit — factorise 111. For (ii), a number ending in W times 5 must end in W.

    (i) EF × E = GGG

    Step 1 — Factorise the repdigit. GGG = 111 × G, and 111 = 3 × 37. So EF × E = 3 × 37 × G, meaning 37 must divide the left side.

    Step 2 — Locate the 37. E is a single digit, so 37 cannot divide E. Therefore 37 divides EF, and since EF is a two-digit number the only options are 37 and 74.

    Step 3 — Test each. • EF = 37 gives E = 3, and 37 × 3 = 111 = GGG with G = 1 ✓ • EF = 74 would need E = 7, but 74 × 7 = 518, which is not a repdigit ✗

    So E = 3, F = 7, G = 1. Check: 37 × 3 = 111 ✓

    (ii) WOW × 5 = MEOW

    Step 1 — Pin down W from the units column. The product ends in W, and it equals 5 × W in the units place. A multiple of 5 ends in 0 or 5, so W = 0 or W = 5. But W is the leading digit of WOW, so it cannot be 0. Hence W = 5.

    Step 2 — Bound the number. WOW = 5O5, which lies between 505 and 595. Multiplying by 5 gives a product between 2525 and 2975, so MEOW is a four-digit number starting with 2 — that is, M = 2.

    Step 3 — Solve for O. 5O5 × 5 = MEOW = 2 E O 5. Trying O = 7: 575 × 5 = 2875. Reading off MEOW = 2875 gives M = 2 ✓, E = 8, O = 7 ✓ (matching the O in WOW), W = 5 ✓

    So W = 5, O = 7, M = 2, E = 8. Check: 575 × 5 = 2875 ✓ — and all four letters stand for different digits ✓

    ✦ (i) E = 3, F = 7, G = 1 (since 111 = 3 × 37). (ii) W = 5, O = 7, M = 2, E = 8 (since 575 × 5 = 2875).

16. 162 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 135

    Which of the given Venn diagrams captures the relationship between the multiples of 4, 8 and 32?

    Hint. Is every multiple of 8 also a multiple of 4? Is every multiple of 4 a multiple of 8?

    The correct diagram is (iv) — three fully nested circles, with multiples of 32 inside multiples of 8, inside multiples of 4.

    Step 1 — Check whether multiples of 8 sit inside multiples of 4. Since 8 = 4 × 2, any multiple of 8 can be written as 8k = 4(2k), which is a multiple of 4. So every multiple of 8 is a multiple of 4. The reverse fails: 4, 12 and 20 are multiples of 4 but not of 8. So the containment is strict — the multiples of 8 form a proper subset.

    Step 2 — Check whether multiples of 32 sit inside multiples of 8. Since 32 = 8 × 4, any multiple of 32 is 32k = 8(4k), a multiple of 8. So every multiple of 32 is a multiple of 8 (and therefore of 4 too). Again strict: 8, 16 and 24 are multiples of 8 but not of 32.

    Step 3 — Conclude the shape. Multiples of 32 ⊂ multiples of 8 ⊂ multiples of 4 — three concentric regions, each wholly inside the next. Diagrams showing partial overlap or separate circles would wrongly suggest some multiple of 8 is not a multiple of 4, which is impossible.

    A concrete illustration: • 32, 64, 96 — in all three regions • 8, 16, 24 — in the multiples of 8 and 4, but not 32 • 4, 12, 20 — only in the multiples of 4

    The general rule: whenever one number divides another, the multiples of the larger form a subset of the multiples of the smaller — so the Venn diagram is always nested, never overlapping.

    ✦ Diagram (iv) — fully nested, since 4 divides 8 and 8 divides 32, so multiples of 32 ⊂ multiples of 8 ⊂ multiples of 4.

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.