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NCERT Solutions Ganita Prakash (Part 1) Chapter 5 Divisibility Shortcuts for Other Numbers / Digital Roots — In-text Questions

Book page 130 Updated on2026-09-05

Q1.
How about checking divisibility by 24? Will checking the divisibility by its factors, 4 and 6, work? Why or why not?
Answer

No, it does not work. The smallest counterexample is 12.

12 ÷ 4 = 3 ✓   12 ÷ 6 = 2 ✓   but 12 ÷ 24 ✗
Why not: 4 = 2 × 2 and 6 = 2 × 3 share the factor 2. Being divisible by both only guarantees divisibility by LCM (4, 6) = 12, not by 24. Other numbers that pass 4 and 6 but fail 24: 36, 60, 84.

The chapter's replacement test is to check divisibility by 3 and by 8 instead.

Q2.
Explain using prime factorisation why checking divisibility by 3 and 8 works for checking divisibility by 24, but checking divisibility by 4 and 6 is not sufficient for checking divisibility by 24.
Answer

Write everything in prime factors:

24 = 2 × 2 × 2 × 3 = 2³ × 3

Using 3 and 8. Divisibility by 8 puts 2³ into the number's prime factorisation; divisibility by 3 puts a 3 there. These use different primes, so both blocks sit side by side:

the number contains 2³ × 3 = 24 ✓
LCM (3, 8) = 3 × 8 = 24

Using 4 and 6. Divisibility by 4 puts 2² in; divisibility by 6 puts 2 × 3 in. But the 2 inside 6 can be the same 2 that is already inside 4 — the two demands overlap.

4 = 2², 6 = 2 × 3
LCM (4, 6) = 2² × 3 = 12, one factor of 2 short of 24
The general point: two tests k and m together prove divisibility by LCM (k, m), never by more. So a pair of tests works for a number N only when LCM (k, m) = N — which happens when k and m are coprime and multiply to N.
Q3.
What property do you think this digital root will have? Recall that we did this while finding the divisibility shortcut for 9.
Answer

The digital root of a number is its remainder on division by 9 — except that a multiple of 9 has digital root 9 rather than 0.

NumberDigital rootRemainder ÷ 9
48971029 → 11 → 22
42713 → 44
730919 → 10 → 11
40590
Why it happens: replacing a number by its digit sum never changes the remainder on division by 9, because the number equals (a multiple of 9) + (its digit sum). Repeating the step keeps the remainder fixed all the way down to a single digit — and the only single digits are 1 to 9, so remainder 0 shows up as 9.
Q4.
Between the numbers 600 and 700, which numbers have the digital root: (i) 5, (ii) 7, (iii) 3?
Answer

Every number here is 6ab, so the digit sum is 6 + a + b. Work out what a + b must be in each case.

(i) Digital root 5 → 6 + a + b must be 5, 14 or 23 → a + b = 8 or 17

608, 617, 626, 635, 644, 653, 662, 671, 680, 689, 698

(ii) Digital root 7 → 6 + a + b must be 7 or 16 → a + b = 1 or 10

601, 610, 619, 628, 637, 646, 655, 664, 673, 682, 691

(iii) Digital root 3 → 6 + a + b must be 12 or 21 → a + b = 6 or 15

606, 615, 624, 633, 642, 651, 660, 669, 678, 687, 696
Did you notice? Each list has 11 numbers and they step by 9. That is no accident — adding 9 leaves the remainder on division by 9 unchanged, so it keeps the digital root the same.
Q5.
Write the digital roots of any 12 consecutive numbers. What do you observe?
Answer

Take 25 to 36:

Number252627282930313233343536
Digital root789123456789

Observation: the digital roots go up by 1 each time, and after 9 they start again at 1. The pattern repeats every 9 numbers — that is why the 12 roots show the first three of them twice.

Why it happens: adding 1 to the number adds 1 to its remainder on division by 9, until the remainder returns to 0 — which the digital root records as 9.
Q6.
We saw that the digital root of multiples of 9 is always 9. Now, find the digital roots of some consecutive multiples of (i) 3, (ii) 4, and (iii) 6.
Answer
Multiples of 3369121518212427
Digital root369369369

(i) The roots cycle 3, 6, 9 — only three values, repeating every 3 multiples.

Multiples of 44812162024283236
Digital root483726159

(ii) The roots cycle 4, 8, 3, 7, 2, 6, 1, 5, 9 — all nine values appear before repeating.

Multiples of 661218243036
Digital root639639

(iii) The roots cycle 6, 3, 9 — again only three values.

Why the cycle lengths differ: each step adds the multiplier to the root, working modulo 9. For 3 and 6 the step shares the factor 3 with 9, so only the multiples of 3 among the roots are ever reached — three of them. For 4, which shares no factor with 9, the steps eventually land on every root, so the cycle has length 9.
Q7.
What are the digital roots of numbers that are 1 more than a multiple of 6? What do you notice? Try to explain the patterns noticed.
Answer

The roots cycle 7, 4, 1.

6k + 171319253137434955
Digital root741741741

What you notice: only three of the nine possible roots ever appear, and they are exactly the roots that leave remainder 1 on division by 3.

Explanation: moving from one such number to the next adds 6, so the digital root moves on by 6 and then drops back by 9 when it overshoots: 7 → 13 gives 4, 4 → 10 gives 1, 1 → 7. Since 6 and 9 share the factor 3, the steps can only reach roots that differ from 7 by a multiple of 3 — namely 7, 4 and 1. The cycle length is 9 ÷ 3 = 3, exactly as it was for the multiples of 3 and of 6.
Try This: repeat with numbers that are 2 more than a multiple of 6 — 8, 14, 20, 26, … The roots cycle 8, 5, 2. Same length, shifted by one.
Q8.
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?
Answer

The number is 11,11,11,111 — nine 1s in a row.

Its name in the Indian system: eleven crore eleven lakh eleven thousand one hundred eleven.

Read the clues one by one:

ClueWhat it fixes
“made of digits, each tiniest and odd”every digit is 1 — the smallest odd digit
“the largest odd single-digit I proudly claim”the target value is 9
“my digits count … all point to” 9there are 9 digits
“their sum … all point to” 91 + 1 + … + 1 (nine times) = 9 ✓
“my root … all point to” 9digit sum 9, so the digital root is 9 ✓
11,11,11,111 = 111111111
digits: 9   digit sum: 9   digital root: 9
and 111111111 = 9 × 12345679 ✓
Did you know? That last division is a small curiosity in itself — 111111111 ÷ 9 = 12345679, the digits 1 to 9 with the 8 missing.
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