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NCERT Solutions Ganita Prakash (Part 1) Chapter 5 .3 Digits in Disguise — Figure it Out

Book page 1325 Updated on2026-09-05

Q1.
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.
Answer

z = 0 or z = 9, giving 3105 and 3195.

Digit sum = 3 + 1 + z + 5 = 9 + z
z is a digit, so 9 + z runs from 9 to 18
Multiples of 9 in that range: 9 and 18
9 + z = 9 → z = 0   9 + z = 18 → z = 9

Check: 3105 = 9 × 345 ✓ and 3195 = 9 × 355 ✓

Why two answers: the digit sum without z is already 9 — a multiple of 9. Adding 0 keeps it a multiple of 9, and adding 9 lands on the next multiple. Because a digit can only stretch the sum by 9, exactly two multiples of 9 fall inside the range 9 to 18, so there are exactly two answers.
Q2.
“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.
Answer

Snehal's claim is false. The sum is sometimes a multiple of 8 and sometimes not.

First number: a = 12n + 8
Second number: b = 12m – 4
a + b = 12n + 12m + 4 = 12(n + m) + 4 = 4(3(n + m) + 1)

So the sum is always a multiple of 4, but for it to be a multiple of 8 the bracket 3(n + m) + 1 must be even — that is, n + m must be odd.

abSumMultiple of 8?
8 (n = 0)8 (m = 1)16Yes
20 (n = 1)8 (m = 1)28No
8 (n = 0)20 (m = 2)28No
20 (n = 1)20 (m = 2)40Yes
Conclusion: the counterexample 20 + 8 = 28 is enough to sink the claim. What Snehal could have said truthfully is that the sum is always a multiple of 4, and always leaves remainder 4 on division by 12.
Q3.
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.
Answer

The sum of two multiples of 3 is a multiple of 6 exactly when the two numbers have the same parity — both even or both odd.

3m + 3n = 3(m + n)
This is a multiple of 6 = 3 × 2 only when m + n is even
m + n is even ⇔ m and n have the same parity
CaseExampleSumMultiple of 6?
both even multiples of 3 (i.e. multiples of 6)6 + 1218Yes
both odd multiples of 39 + 1524Yes
one even, one odd6 + 915No
one even, one odd12 + 2133No
Generalising: this is the parity argument from the start of the chapter, wearing a coat of 3. Every multiple of 3 is either a multiple of 6 (remainder 0 on division by 6) or 3 more than one (remainder 3). Two of the same type add to a multiple of 6; one of each type leaves a remainder of 3 — exactly the structure we met with multiples of 4.
Q4.
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?
Answer

(i) Her conjecture is true for every multiple of 9.

Why it happens: reversing the digits moves them into different places but does not change the digits themselves — so the digit sum stays the same. Divisibility by 9 depends on nothing but the digit sum, so it survives the reversal.
189 → digit sum 18 → 189 = 9 × 21 ✓
981 → digit sum 18 → 981 = 9 × 109 ✓

5238 → digit sum 18 → 5238 = 9 × 582 ✓
8325 → digit sum 18 → 8325 = 9 × 925 ✓

(ii) Yes — any shuffle of the digits works, not just the reversal.

Take 189 and write down every arrangement:

189 = 9 × 21   198 = 9 × 22   819 = 9 × 91
891 = 9 × 99   918 = 9 × 102   981 = 9 × 109

All six are multiples of 9, for the same reason: rearranging digits never changes their sum.

Tip: the one thing to watch is a 0. Shuffling 405 could put the 0 in front, and 045 is not a 4-digit... not a 3-digit numeral at all — though its value 45 is still a multiple of 9. The same argument works for 3, since divisibility by 3 also depends only on the digit sum.
Q5.
If 48a23b is a multiple of 18, list all possible pairs of values for a and b.
Answer

The pairs are (a, b) = (1, 0), (8, 2), (6, 4), (4, 6) and (2, 8).

18 = 2 × 9, and 2 and 9 are coprime
So 48a23b must be divisible by 2 and by 9

Divisible by 2: the units digit b must be even → b = 0, 2, 4, 6 or 8.

Divisible by 9: the digit sum must be a multiple of 9.

4 + 8 + a + 2 + 3 + b = 17 + a + b
a + b ≤ 18, so 17 + a + b lies between 17 and 35
The only multiple of 9 in that range is 27 → a + b = 10
(a + b = 1 would give 18, but that needs a + b = 1 — check it too)

Both possibilities have to be tried: 17 + a + b = 18 gives a + b = 1, and 17 + a + b = 27 gives a + b = 10.

ba from a + b = 1a from a + b = 10Valid pair
0a = 1 ✓a = 10 ✗(1, 0) → 481230
2a = – 1 ✗a = 8 ✓(8, 2) → 488232
4✗a = 6 ✓(6, 4) → 486234
6✗a = 4 ✓(4, 6) → 484236
8✗a = 2 ✓(2, 8) → 482238

Check one: 481230 → digit sum 18 ✓, ends in 0 ✓, and 481230 = 18 × 26735 ✓

Q6.
If 3p7q8 is divisible by 44, list all possible pairs of values for p and q.
Answer

The pairs are (p, q) = (7, 0), (5, 2), (3, 4) and (1, 6).

44 = 4 × 11, and 4 and 11 are coprime
So 3p7q8 must be divisible by 4 and by 11

Divisible by 4: the last two digits q8 must form a multiple of 4 → 08, 28, 48, 68, 88, so q = 0, 2, 4, 6 or 8.

Divisible by 11: take the alternating sum from the units digit.

8 – q + 7 – p + 3 = 18 – (p + q)
This must be 0 or a multiple of 11
18 – (p + q) = 0 → p + q = 18 → p = q = 9, but q must be even ✗
18 – (p + q) = 11 → p + q = 7 ✓
18 – (p + q) = – 11 → p + q = 29, impossible ✗
qp = 7 – qNumber÷ 44
0737708857 ✓
2535728812 ✓
4333748767 ✓
6131768722 ✓
8– 1 ✗——
Q7.
Find three consecutive numbers such that the first number is a multiple of 2, the second number is a multiple of 3, and the third number is a multiple of 4. Are there more such numbers? How often do they occur?
Answer

One such set is 2, 3, 4. There are infinitely many, and they occur once in every 12 numbers.

Let the numbers be n, n + 1, n + 2
n + 2 is a multiple of 4 → n leaves remainder 2 on ÷ 4
(this already makes n even, so the first condition is free)
n + 1 is a multiple of 3 → n leaves remainder 2 on ÷ 3

n – 2 is a multiple of 4 and of 3 → a multiple of LCM (3, 4) = 12
n = 12k + 2
nThe three numbersCheck
22, 3, 42 = 2 × 1, 3 = 3 × 1, 4 = 4 × 1 ✓
1414, 15, 1614 = 2 × 7, 15 = 3 × 5, 16 = 4 × 4 ✓
2626, 27, 2826 = 2 × 13, 27 = 3 × 9, 28 = 4 × 7 ✓
3838, 39, 4038 = 2 × 19, 39 = 3 × 13, 40 = 4 × 10 ✓
Why every 12: the two conditions fix the remainder of n on division by 4 and on division by 3. Those two remainders together repeat with period LCM (3, 4) = 12, so the whole pattern shifts along by 12 each time.
Q8.
Write five multiples of 36 between 45,000 and 47,000. Share your approach with the class.
Answer

Five of them: 45036, 45072, 45108, 45144, 45180.

The approach. Find one multiple of 36 in the range, then keep adding 36 — every step lands on the next one.

45000 ÷ 36 = 1250 exactly, so 45000 = 36 × 1250
Next: 36 × 1251 = 45036
Then add 36 each time: 45072, 45108, 45144, 45180

Checking without dividing. 36 = 4 × 9 and 4 and 9 are coprime, so a multiple of 36 must pass both tests:

NumberLast two digits ÷ 4Digit sum ÷ 9÷ 36
4503636 = 4 × 9 ✓18 ✓1251
4507272 = 4 × 18 ✓18 ✓1252
4510808 = 4 × 2 ✓18 ✓1253
4514444 = 4 × 11 ✓18 ✓1254
4518080 = 4 × 20 ✓18 ✓1255
Tip: 45000 itself is a multiple of 36, but the question asks for multiples between 45,000 and 47,000, so the list starts at 45036.
Q9.
The middle number in the sequence of 5 consecutive even numbers is 5p. Express the other four numbers in sequence in terms of p.
Answer

The other four are 5p – 4, 5p – 2, 5p + 2 and 5p + 4.

The full sequence in order is
5p – 4, 5p – 2, 5p, 5p + 2, 5p + 4

Consecutive even numbers differ by 2, not 1 — that is the only thing to be careful about here.

Check with numbers: if 5p = 20 (so p = 4), the five numbers are 16, 18, 20, 22, 24 ✓. Their sum is 25p = 100, five times the middle number — the same tidy fact that holds for any five consecutive terms with a constant gap.
Q10.
Write a 6-digit number that it is divisible by 15, such that when the digits are reversed, it is divisible by 6.
Answer

One answer is 210045. Here is how to build such a number rather than hunt for one.

Step 1 — divisible by 15. 15 = 3 × 5, so the number must end in 0 or 5 and have a digit sum that is a multiple of 3.

Step 2 — the last digit must be 5, not 0. If it ended in 0, the reversed number would begin with 0 and would not be a 6-digit number at all.

Step 3 — the reverse must be divisible by 6. 6 = 2 × 3. Divisibility by 3 is automatic, since reversing does not change the digit sum. Divisibility by 2 needs the reverse to end in an even digit — that is, the first digit of the original number must be even.

Recipe: first digit even, last digit 5, digit sum a multiple of 3

210045 → digit sum 12 ✓, ends in 5 ✓ → 210045 = 15 × 14003 ✓
reversed: 540012 → even ✓, digit sum 12 ✓ → 540012 = 6 × 90002 ✓
More answers: 200025 (reverse 520002), 402015 (reverse 510204), 810345 (reverse 543018) — all built by the same recipe.
Q11.
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 if his conjecture is true; explain your conclusion.
Answer

Deepak's conjecture is false. Every multiple of 11 stays a multiple of 11 when doubled — there are no exceptions.

Let the number be N = 11k
2N = 2 × 11k = 22k = 11 × (2k)

11 is still a factor, so 2N is a multiple of 11 no matter what k is.

Multiple of 11Doubled÷ 11
11222 ✓
5511010 ✓
14328626 ✓
9021804164 ✓
Why: this is simply the rule from page 119 — if A is divisible by k, then all multiples of A are divisible by k. Doubling produces a multiple of A, so the divisor 11 cannot be lost. Nothing about 11 or about doubling is special; tripling, or multiplying by 100, works the same way.
Q12.
Determine whether the statements below are ‘Always True’, ‘Sometimes True’, or ‘Never True’. Explain 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.
Answer

(i) Always True.

6m × 3n = 18mn = 9 × (2mn)

9 is a factor of the product for every choice of m and n. Examples: 6 × 3 = 18 = 9 × 2 ✓; 12 × 9 = 108 = 9 × 12 ✓; 30 × 21 = 630 = 9 × 70 ✓.

(ii) Always True.

2n + (2n + 2) + (2n + 4) = 6n + 6 = 6(n + 1)

Examples: 4 + 6 + 8 = 18 = 6 × 3 ✓; 10 + 12 + 14 = 36 = 6 × 6 ✓; – 2 + 0 + 2 = 0 = 6 × 0 ✓.

Tip: the sum is 3 times the middle number, and the middle number is even — so it is a multiple of 3 × 2 = 6.

(iii) Always True.

Going from abcdef to badcef swaps a with b and c with d, leaving e and f where they are. Test the two halves of divisibility by 6 separately:

  • By 2: the units digit is still f, so nothing changes.
  • By 3: the digits are the same digits, just moved — the digit sum is unchanged.

Both tests survive, so the new number is a multiple of 6.

Example: 123456 = 6 × 20576
badcef = 214356 = 6 × 35726 ✓
One caution: if b = 0 the rearranged string starts with 0, so it is no longer a 6-digit numeral — but its value is still a multiple of 6.

(iv) Never True.

8(7b – 3) – 4(11b + 1)
= 56b – 24 – 44b – 4
= 12b – 28
= 12(b – 2) – 4

Whatever b is, the expression is 4 short of a multiple of 12, so it can never be a multiple of 12.

b = 1 → – 16   b = 2 → – 4   b = 3 → 8   b = 5 → 32
Another way to see it: 12b – 28 = 4(3b – 7). The factor 4 is always there, but 3b – 7 always leaves remainder 2 on division by 3, so the third factor of 3 that 12 requires is never present.
Q13.
Choose any 3 numbers. When is their sum divisible by 3? Explore all possible cases and generalise.
Answer

Divide each number by 3 and keep only its remainder — 0, 1 or 2. The sum is divisible by 3 exactly when the three remainders add to 0, 3 or 6.

n₁ = 3a + r₁, n₂ = 3b + r₂, n₃ = 3c + r₃
n₁ + n₂ + n₃ = 3(a + b + c) + (r₁ + r₂ + r₃)
The bracket 3(a + b + c) never affects divisibility by 3

Since each remainder is 0, 1 or 2, their total lies between 0 and 6. Only 0, 3 and 6 work, and that leaves exactly two shapes of answer:

CaseRemaindersTotalExample
all three remainders equal0, 0, 00 ✓3 + 9 + 12 = 24
1, 1, 13 ✓4 + 7 + 10 = 21
2, 2, 26 ✓2 + 5 + 8 = 15
all three remainders different0, 1, 23 ✓3 + 7 + 11 = 21
exactly two the same1, 2, 25 ✗4 + 5 + 8 = 17
0, 0, 11 ✗3 + 6 + 7 = 16

Generalisation: the sum of three numbers is divisible by 3 when their remainders on division by 3 are all equal or all different — and in no other case.

Did you notice? Three consecutive numbers always have remainders 0, 1 and 2 in some order, so their sum is always a multiple of 3. That matches n + (n + 1) + (n + 2) = 3(n + 1).
Q14.
Is the product of two consecutive integers always multiple of 2? Why? What about the product of these consecutive integers? Is it always a multiple of 6? Why or why not? What can you say about the product of 4 consecutive integers? What about the product of five consecutive integers?
Answer

Two consecutive integers — always a multiple of 2.

n(n + 1)
One of any two consecutive integers is even, so 2 is a factor

Examples: 3 × 4 = 12, 8 × 9 = 72, 15 × 16 = 240 — all even.

Three consecutive integers — always a multiple of 6. (The book's phrase “these consecutive integers” here means three, as the mention of 6 shows.)

n(n + 1)(n + 2)
At least one of the three is even → a factor of 2
Exactly one of any three consecutive integers is a multiple of 3 → a factor of 3
2 and 3 are coprime → a factor of 6

Examples: 1 × 2 × 3 = 6, 4 × 5 × 6 = 120 = 6 × 20, 9 × 10 × 11 = 990 = 6 × 165.

Four consecutive integers — always a multiple of 24.

Among four consecutive integers, two are even,
and one of those two is a multiple of 4
→ factors 4 × 2 = 8
At least one is a multiple of 3 → factor 3
8 × 3 = 24

Examples: 1 × 2 × 3 × 4 = 24, 2 × 3 × 4 × 5 = 120 = 24 × 5, 5 × 6 × 7 × 8 = 1680 = 24 × 70.

Five consecutive integers — always a multiple of 120.

The four-in-a-row argument already gives 24
Exactly one of any five consecutive integers is a multiple of 5
24 × 5 = 120

Examples: 1 × 2 × 3 × 4 × 5 = 120, 2 × 3 × 4 × 5 × 6 = 720 = 120 × 6, 6 × 7 × 8 × 9 × 10 = 30240 = 120 × 252.

The pattern: 2 = 2!, 6 = 3!, 24 = 4!, 120 = 5!. The product of k consecutive integers is always a multiple of k! — because among any k consecutive integers you are guaranteed a multiple of 1, of 2, of 3, …, of k, arranged so that all the needed prime factors are present.
Q15.
Solve the cryptarithms — (i) EF × E = GGG (ii) WOW × 5 = MEOW
Answer

(i) E = 3, F = 7, G = 1 → 37 × 3 = 111

GGG = 111 × G = 3 × 37 × G
So EF × E must be one of 111, 222, …, 999

The two-digit number EF has tens digit E, and E is also the multiplier. Check the factorisations:

GGGAs (2-digit) × (1-digit)Tens digit = multiplier?
11137 × 33 = 3 ✓
22237 × 63 ≠ 6 ✗
33337 × 93 ≠ 9 ✗
44474 × 67 ≠ 6 ✗
66674 × 97 ≠ 9 ✗

Only 111 = 37 × 3 fits, so E = 3, F = 7, G = 1 — three different digits ✓

(ii) W = 5, O = 7, M = 2, E = 8 → 575 × 5 = 2875

Units: 5 × W must end in W → W = 0 or 5
W leads WOW, so W ≠ 0 → W = 5

WOW = 505 + 10O
(505 + 10O) × 5 = 2525 + 50O
MEOW = 1000M + 100E + 10O + 5
2525 + 50O = 1000M + 100E + 10O + 5
2520 + 40O = 1000M + 100E

The right side is a multiple of 100, so 2520 + 40O must end in 00 → O = 2 or 7.

O2520 + 40OM, EValid?
22600M = 2, E = 6No — M = 2 clashes with O = 2
72800M = 2, E = 8Yes
Check: 575 × 5 = 2875 = MEOW with M = 2, E = 8, O = 7, W = 5 ✓
Q16.
Which of the following Venn diagrams captures the relationship between the multiples of 4, 8, and 32?
Answer

Diagram (iv) — the one that nests multiples of 32 inside multiples of 8 inside multiples of 4.

32 = 8 × 4, so every multiple of 32 is a multiple of 8
8 = 4 × 2, so every multiple of 8 is a multiple of 4
multiples of 32 ⊂ multiples of 8 ⊂ multiples of 4
Multiples of 4 Multiples of 8 Multiples of 32 4, 12, 20 8, 24, 40 32, 64
Each circle sits wholly inside the next, and none of the three is empty of its own.

Each ring must be non-empty, and it is:

Number÷ 4÷ 8÷ 32Where it sits
4YesNoNoouter ring only
8YesYesNomiddle ring
32YesYesYesinnermost circle
Why the other three are wrong: they either reverse the nesting (multiples of 4 drawn inside multiples of 32, which would mean 4 is a multiple of 32) or place the circles side by side as if some multiples of 8 were not multiples of 4. Both readings contradict 8 = 4 × 2 and 32 = 8 × 4.
Note: the four Venn circles are missing from the printed page in some copies of this reprint — only the labels and their pointer arrows appear. The arrows still show the intended nesting, and the mathematics fixes the answer as (iv).
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