In-text — Digital Roots — Number Play

1. 12 marksGanita Prakash Cl-8 Part 1, in-text, page 130

   What property do you think the digital root will have? (Recall the divisibility shortcut for 9.)

   Hint. The digital root is what you get by adding digits repeatedly until one digit remains — and adding digits is exactly the 9-test.

   The digital root of a number is the remainder it leaves on division by 9 — except that a number exactly divisible by 9 has digital root 9 rather than 0.

   Why this is so: forming the digital root means adding the digits over and over until a single digit is left. But the divisibility shortcut for 9 works precisely because adding the digits does not change the remainder on division by 9 — every power of 10 leaves remainder 1, so each digit contributes just itself. Repeating a step that preserves the remainder must still preserve it at the end.

   So whatever single digit you finish on carries the same remainder mod 9 as the number you started with.

   Examples: • 3546 → 3+5+4+6 = 18 → 1+8 = 9. And 3546 ÷ 9 = 394 exactly, remainder 0, shown as digital root 9 ✓ • 3547 → 19 → 10 → 1. And 3547 ÷ 9 leaves remainder 1 ✓

   The only quirk to remember: the digital root is never 0, because the process always ends on a digit from 1 to 9. A remainder of 0 shows up as a digital root of 9.

   ✦ The digital root equals the remainder on division by 9, with a multiple of 9 having digital root 9 rather than 0 — because repeatedly adding digits never changes the remainder mod 9.

2. 23 marksGanita Prakash Cl-8 Part 1, in-text, page 130

   Between the numbers 600 and 700, which numbers have the digital root (i) 5, (ii) 7, (iii) 3?

   Hint. For a number 6ab, the digital root depends on 6 + a + b. Find which pairs (a, b) give the right total.

   Every number here has the form 6ab, so its digit sum is 6 + a + b and the digital root is that total reduced mod 9. Since a + b ranges from 0 to 18, the digit sum runs from 6 to 24, so only two totals can produce each target root.

   (i) Digital root 5 — need 6 + a + b ≡ 5 (mod 9), so the digit sum is 14 or 23, i.e. a + b = 8 or 17. a + b = 8: 608, 617, 626, 635, 644, 653, 662, 671, 680 a + b = 17: 689, 698 608, 617, 626, 635, 644, 653, 662, 671, 680, 689, 698

   (ii) Digital root 7 — digit sum 7 or 16, so a + b = 1 or 10. a + b = 1: 601, 610 a + b = 10: 619, 628, 637, 646, 655, 664, 673, 682, 691 601, 610, 619, 628, 637, 646, 655, 664, 673, 682, 691

   (iii) Digital root 3 — digit sum 12 or 21, so a + b = 6 or 15. a + b = 6: 606, 615, 624, 633, 642, 651, 660 a + b = 15: 669, 678, 687, 696 606, 615, 624, 633, 642, 651, 660, 669, 678, 687, 696

   The pattern to notice: within each list the numbers go up in steps of 9. That is no accident — adding 9 leaves the remainder mod 9 unchanged, so it cannot change the digital root. Each list therefore has 11 members, matching the 11 multiples-of-9 gaps across a span of 100.

   Spot check: 689 → 6+8+9 = 23 → 2+3 = 5 ✓ and 696 → 21 → 3 ✓

   ✦ (i) 608, 617, 626, 635, 644, 653, 662, 671, 680, 689, 698. (ii) 601, 610, 619, 628, 637, 646, 655, 664, 673, 682, 691. (iii) 606, 615, 624, 633, 642, 651, 660, 669, 678, 687, 696 — each list stepping up by 9.

3. 33 marksGanita Prakash Cl-8 Part 1, in-text, page 130

   Find the digital roots of some consecutive multiples of (i) 3, (ii) 4, and (iii) 6. What do you observe?

   Hint. List the multiples, take each digital root, and look for where the sequence starts repeating.

   (i) Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, … Digital roots: 3, 6, 9, 3, 6, 9, 3, 6, 9, … Cycle: 3, 6, 9 — length 3.

   (ii) Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, … Digital roots: 4, 8, 3, 7, 2, 6, 1, 5, 9, 4, … Cycle: 4, 8, 3, 7, 2, 6, 1, 5, 9 — length 9, and it uses every digit from 1 to 9.

   (iii) Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, … Digital roots: 6, 3, 9, 6, 3, 9, 6, 3, 9, … Cycle: 6, 3, 9 — length 3.

   Why the cycle lengths differ. The digital root is the remainder mod 9, so the digital roots of the multiples of k step forward by k each time, wrapping round at 9. The cycle closes as soon as those steps have covered a whole number of laps, which happens after 9 ÷ HCF(k, 9) steps. • k = 3: HCF(3, 9) = 3, so the cycle length is 9 ÷ 3 = 3 ✓ • k = 4: HCF(4, 9) = 1, so the cycle length is 9 ÷ 1 = 9 ✓ — and because 4 shares no factor with 9, it visits every root. • k = 6: HCF(6, 9) = 3, so the cycle length is 9 ÷ 3 = 3 ✓

   A useful consequence: since the roots of multiples of 3 are only ever 3, 6 or 9, seeing a digital root of 4 immediately tells you a number is not a multiple of 3.

   ✦ (i) 3, 6, 9 repeating. (ii) 4, 8, 3, 7, 2, 6, 1, 5, 9 repeating — all nine roots, since 4 shares no factor with 9. (iii) 6, 3, 9 repeating.

4. 43 marksGanita Prakash Cl-8 Part 1, in-text riddle, page 130

   "I'm made of digits, each tiniest and odd, No shared ground with root #1 — how odd! My digits count, their sum, my root — All point to one bold number's pursuit — The largest odd single-digit I proudly claim. What's my number? What's my name?"

   Hint. Work out the 'bold number' first from the last clue, then apply it to the digit count, the digit sum and the digital root at once.

   Step 1 — Identify the 'bold number'. "The largest odd single-digit" is 9. The verse says the digit count, the digit sum and the digital root all point to it.

   Step 2 — Identify the digit being used. "Each tiniest and odd" — the smallest odd digit is 1. So every digit of the number is 1.

   Step 3 — Put the two together. • The number has 9 digits, all of them 1. • Their sum is 9 × 1 = 9 ✓ matching the second clue. • The digital root of 9 is 9 ✓ matching the third.

   All three clues agree, so the number is 111111111

   Step 4 — Name it in the Indian system. Grouped as 11,11,11,111, this reads eleven crore eleven lakh eleven thousand one hundred eleven.

   Check: the digit sum is 9, so the number is divisible by 9. Indeed 111111111 ÷ 9 = 12345679 ✓ (a famous quotient — notice it contains every digit except 8).

   On the line "no shared ground with root #1": the digits are all 1, which shares no common factor beyond 1 with anything — a playful way of saying the number is built from the unit digit alone.

   ✦ The number is 111111111 — nine 1s, digit sum 9, digital root 9 — read as eleven crore eleven lakh eleven thousand one hundred eleven.

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.