Chapter 5 exercise answers: Number Play

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1. Let the four consecutive numbers be x, x+1, x+2 and x+3.
2. Their sum is x + (x+1) + (x+2) + (x+3) = 4x + 6, and this equals 34.
3. 4x = 34 − 6 = 28, so x = 7.
4. The numbers are 7, 8, 9 and 10.

Answer7, 8, 9 and 10

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Open NCERT p. 122Matches NCERT’s answer

1. Five consecutive numbers go up by 1 each time, and p is the last (greatest) one.
2. Counting backwards from p by 1 each time gives the other four: p−1, p−2, p−3, p−4.

Answerp−4, p−3, p−2 and p−1

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Open NCERT p. 122Matches NCERT’s answer

(i) The sum of two even numbers is a multiple of 3.

1. Write the two even numbers as 2a and 2b, where a and b are whole numbers. Their sum is 2a + 2b = 2(a + b).
2. 2(a + b) is a multiple of 3 only when a + b is a multiple of 3, and that depends on which even numbers are chosen.
3. Example: 2 + 4 = 6, which is a multiple of 3.
4. Non-example: 2 + 6 = 8, which is not a multiple of 3.

AnswerSometimes true: 2 + 4 = 6 is a multiple of 3, but 2 + 6 = 8 is not.

(ii) not divisible by 18, then it is also not divisible by 9.

1. Example where it holds: 10 is not divisible by 18, and it is not divisible by 9 either.
2. Non-example: 9 is not divisible by 18, yet it is divisible by 9. So are 27, 45, 63, …
3. In algebra: every number 18k + 9 = 9(2k + 1) leaves remainder 9 when divided by 18, so it is not divisible by 18, but it is 9 times a whole number, so it is divisible by 9.
4. The statement works for some numbers and fails for others.

AnswerSometimes true: it holds for 10, but fails for 9, which is not divisible by 18 yet is divisible by 9.

(iii) not divisible by 6, then their sum is not divisible by 6

1. Example where it holds: 1 and 2 are not divisible by 6, and 1 + 2 = 3 is not divisible by 6.
2. Non-example: 1 and 5 are not divisible by 6, but 1 + 5 = 6 is divisible by 6.
3. In algebra: write the numbers as 6a + r and 6b + s, where the remainders r and s are each between 1 and 5. The sum is 6(a + b) + (r + s).
4. r + s is between 2 and 10, so the sum is divisible by 6 exactly when r + s = 6: remainders 1 and 5, 2 and 4, or 3 and 3. For any other pair of remainders the statement holds.

AnswerSometimes true: it holds for 1 and 2 (sum 3), but fails for 1 and 5 (sum 6).

(iv) … of 6 and a multiple of 9 is a multiple of 3.

1. A multiple of 6 is 6a, and a multiple of 9 is 9b, for whole numbers a and b.
2. Their sum is 6a + 9b = 3(2a + 3b), which is 3 times a whole number.
3. So the sum is a multiple of 3 whatever a and b are. For example, 6 + 9 = 15 = 3 × 5.

AnswerAlways true, because 6a + 9b = 3(2a + 3b) is always 3 times a whole number.

(v) … of 6 and a multiple of 3 is a multiple of 9.

1. A multiple of 6 is 6a, and a multiple of 3 is 3b. Their sum is 6a + 3b = 3(2a + b).
2. This is always a multiple of 3, but it is a multiple of 9 only when 2a + b is a multiple of 3, which depends on a and b.
3. Example: 6 + 3 = 9, which is a multiple of 9.
4. Non-example: 6 + 6 = 12, which is not a multiple of 9.

AnswerSometimes true: 6 + 3 = 9 is a multiple of 9, but 6 + 6 = 12 is not.

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Open NCERT p. 122Matches NCERT’s answer

1. Numbers leaving remainder 2 when divided by 3 are 2, 5, 8, 11, 14, 17, 20, 23, 26, …
2. From this list, the ones that also leave remainder 2 when divided by 4 are 2, 14, 26, … — stepping by 12 each time.
3. 12 is the smallest number that both 3 and 4 divide into, so the matching numbers repeat every 12.
4. Any such number can be written as 12n + 2, for n = 0, 1, 2, …

Answer2, 14, 26, … ; algebraic expression 12n + 2

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1. Grouped in 3's with one left over, paired up (grouped in 2's) with one stubborn one left over, and grouped in 5's with one left over: these numbers are 1, 31, 61, 91, … (repeating every 30, since 3×2×5=30).
2. Grouped in 7's with none left over means the number must also be a multiple of 7.
3. Among 1, 31, 61, 91, only 91 is a multiple of 7 (91 = 7×13).
4. 91 is not more than a hundred, which fits the last line of the riddle.

Answer91 pebbles

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1. Each such number can be written as 6k + 2, for some whole number k.
2. Adding any three of them: (6a+2) + (6b+2) + (6c+2) = 6a + 6b + 6c + 6.
3. That is 6(a + b + c + 1), which is 6 times a whole number.
4. So the sum is always a multiple of 6 — Tathagat's claim is true.

AnswerYes, Tathagat's claim is true.

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Open NCERT p. 123Matches NCERT’s answer

(i) 4779 + 661

1. Write 661 = 7m + 3 and 4779 = 7n + 5, for some whole numbers m and n — that is what 'leaves remainder 3 (or 5)' means.
2. Adding: 661 + 4779 = 7m + 7n + 3 + 5 = 7(m+n) + 8.
3. 8 is one more group of 7 plus 1 left over, so the total is 7(m+n+1) + 1.
4. Picture it as stacks of 7: the leftover piles of 3 and 5 combine to make one more full stack of 7, with 1 counter left outside — so the remainder is 1.

Answer1

(ii) 4779 – 661

1. Subtracting: 4779 − 661 = (7n + 5) − (7m + 3) = 7(n−m) + 2.
2. 2 is already smaller than 7, so it is the remainder itself.
3. Picture it as taking 661's 3 leftover counters away from 4779's 5 leftover counters (both sit outside full stacks of 7): 5 − 3 = 2 counters remain — so the remainder is 2.

Answer2

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1. Each remainder is exactly 1 less than its divisor: 2 is 1 less than 3, 3 is 1 less than 4, and 4 is 1 less than 5.
2. So the number is 1 less than a number that 3, 4 and 5 all divide into.
3. The smallest number divisible by 3, 4 and 5 together is their LCM, which is 60.
4. So the smallest number fitting all three conditions is 60 − 1 = 59.

Answer59

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Open NCERT p. 126Matches NCERT’s answer

(i) 123

1. Add the digits of 123: 1+2+3 = 6.
2. 6 is not a multiple of 9, so 123 is not divisible by 9.

AnswerNo

(ii) 405

1. Add the digits of 405: 4+0+5 = 9.
2. 9 is a multiple of 9, so 405 is divisible by 9.

AnswerYes

(iii) 8888

1. Add the digits of 8888: 8+8+8+8 = 32.
2. 32 is not a multiple of 9, so 8888 is not divisible by 9.

AnswerNo

(iv) 93547

1. Add the digits of 93547: 9+3+5+4+7 = 28.
2. 28 is not a multiple of 9, so 93547 is not divisible by 9.

AnswerNo

(v) 358095

1. Add the digits of 358095: 3+5+8+0+9+5 = 30.
2. 30 is not a multiple of 9, so 358095 is not divisible by 9.

AnswerNo

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Open NCERT p. 126Matches NCERT’s answer

1. No odd digits means every digit is 0, 2, 4, 6 or 8, so the digit sum is even.
2. A multiple of 9 has a digit sum that is a multiple of 9. The digit sum cannot be 0 (the number is not 0) and cannot be 9 (it is even), so it is at least 18.
3. Two digits add up to at most 8 + 8 = 16, which is less than 18, so the number needs at least 3 digits.
4. To make a 3-digit number as small as possible, make the hundreds digit as small as possible. It must be even and not 0, so it is 2.
5. The other two digits must then add up to 18 − 2 = 16, and the only even digits that do so are 8 and 8. So the number is 288, and 288 = 9 × 32.

Answer288

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1. 6000 ÷ 9 = 666 remainder 6, so 9×666 = 5994 and 9×667 = 6003 are the two multiples of 9 nearest 6000.
2. 6000 − 5994 = 6, while 6003 − 6000 = 3, so 6003 is closer.

Answer6003

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Open NCERT p. 126Matches NCERT’s answer

1. 4300 ÷ 9 ≈ 477.8 and 4400 ÷ 9 ≈ 488.9, so the multiples of 9 in this range are 9×478 = 4302 up to 9×488 = 4392.
2. That is every whole number from 478 to 488, which is 488 − 478 + 1 = 11 multiples.

Answer11

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Open NCERT p. 131Matches NCERT’s answer

1. The digital root of a number is the same as its remainder when divided by 9 (writing 9 instead of 0).
2. 10 leaves remainder 1 when divided by 9, so adding 10 always shifts the digital root up by 1.
3. Starting digital root is 5, so the new digital root is 5 + 1 = 6.

Answer6

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Open NCERT p. 131One way to think about itAnswers can differ: one example

1. The starting number is your own choice, so every student's sequence is different, but the pattern below appears whatever number you start with. Here is one example.
2. Start with 14 (digital root 5) and keep adding 11: 14, 25, 36, 47, 58, 69, 80, 91, 102, 113, …
3. Their digital roots are 5, 7, 9, 2, 4, 6, 8, 1, 3, 5, …: each root is 2 more than the one before, counting on past 9 back round to 1 (so 8 is followed by 1, and 9 by 2).
4. This happens because 11 is 2 more than a multiple of 9 (11 = 9 + 2), so adding 11 always moves the digital root on by 2.
5. After 9 steps you have added 99 = 9 × 11, a multiple of 9, so the roots come back to where they started. Along the way they visit all nine values 1 to 9, once each.

In shortWhatever number you start with, the digital roots go up by 2 at each step (8 is followed by 1, and 9 by 2), pass through all nine values 1 to 9, and then repeat every 9 terms.

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Open NCERT p. 131Matches NCERT’s answer

1. 9a is a multiple of 9 for any whole number a, and 36b is also a multiple of 9 (since 36 = 9×4).
2. A multiple of 9 contributes nothing extra to a digital root once it is taken out.
3. So the digital root of 9a + 36b + 13 is the same as the digital root of just 13.
4. Digital root of 13 is 1 + 3 = 4.

Answer4

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Open NCERT p. 131One way to think about it

(i) the parity of a number and its digital root.

1. Even numbers: 10, 2, 12, 4, 14, 6, 16, 8, 18 have digital roots 1, 2, 3, 4, 5, 6, 7, 8, 9, so every root appears.
2. Odd numbers: 1, 11, 3, 13, 5, 15, 7, 17, 9 have digital roots 1, 2, 3, 4, 5, 6, 7, 8, 9, so again every root appears.
3. The reason: adding 9 to a number does not change its digital root, but it turns an odd number into an even one and an even number into an odd one. So every digital root belongs to odd and even numbers alike.
4. So the digital root does not tell you whether a number is odd or even, and being odd or even does not tell you the digital root.

In shortConjecture: there is no relation. Every digital root from 1 to 9 occurs for both odd and even numbers (for example, 1 is odd and 10 is even, and both have digital root 1).

(ii) the digital root of a number and … divided by 3 or 9.

1. Examples: 23 has digital root 5 and leaves remainder 5 when divided by 9. 47 has digital root 2 (4 + 7 = 11, 1 + 1 = 2) and 47 = 9 × 5 + 2. 36 has digital root 9 and leaves remainder 0.
2. Conjecture for 9: the digital root equals the remainder on division by 9, except that a multiple of 9 has digital root 9 where the remainder is 0.
3. Why: every place value (1, 10, 100, …) is 1 more than a multiple of 9, so a number and its digit sum leave the same remainder when divided by 9, and each round of adding digits keeps that remainder.
4. Conjecture for 3: since 9 is a multiple of 3, a number leaves the same remainder on division by 3 as its digital root does. Roots 3, 6 and 9 give remainder 0; roots 1, 4 and 7 give remainder 1; roots 2, 5 and 8 give remainder 2. For example, 47 has root 2 and 47 = 3 × 15 + 2.

In shortThe digital root equals the remainder on division by 9, with 9 standing for remainder 0. The remainder on division by 3 equals the remainder of the digital root divided by 3: roots 3, 6, 9 give 0; roots 1, 4, 7 give 1; roots 2, 5, 8 give 2.

Watch this explained “A report on division by nine”, 3:28 into Digital roots, and what survives repeated digit-summing · हिंदी में देखें

Open NCERT p. 132Matches NCERT’s answer

1. Add the digits: 3 + 1 + z + 5 = 9 + z.
2. For 31z5 to be a multiple of 9, this sum must also be a multiple of 9.
3. Since z is a single digit, 9 + z can only be 9 or 18 (the multiples of 9 it can reach).
4. 9 + z = 9 gives z = 0; 9 + z = 18 gives z = 9.

Answerz = 0 or z = 9.

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Open NCERT p. 132Matches NCERT’s answer

1. Write the first number as 12k + 8, because it leaves remainder 8 when divided by 12.
2. Write the second number as 12m − 4, because it is 4 short of a multiple of 12.
3. Their sum is (12k + 8) + (12m − 4) = 12k + 12m + 4 = 12(k + m) + 4.
4. Try 20 and 8: 20 = 12 × 1 + 8 and 8 = 12 × 1 − 4. Their sum is 28, and 28 is not a multiple of 8.
5. Sometimes it does work: 8 = 12 × 0 + 8 and 8 = 12 × 1 − 4, and 8 + 8 = 16 is a multiple of 8. In fact 12(k + m) + 4 is a multiple of 8 exactly when k + m is odd.
6. Snehal said the sum is ALWAYS a multiple of 8. One counterexample (20 + 8 = 28) is enough to show that is false.

AnswerNo — Snehal's claim is not true. The sum is a multiple of 8 only sometimes (8 + 8 = 16), not always (20 + 8 = 28).

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Open NCERT p. 132One way to think about it

1. Every multiple of 3 is 3 × (some whole number). That whole number is either even or odd, so every multiple of 3 is either an even multiple (like 6, 12, 18 — these are the multiples of 6) or an odd multiple (like 3, 9, 15).
2. Both even multiples: 3 × (even) + 3 × (even) = 3 × (even + even) = 3 × (even), which is a multiple of 3 × 2 = 6. Example: 6 + 12 = 18 = 6 × 3.
3. Both odd multiples: 3 × (odd) + 3 × (odd) = 3 × (odd + odd) = 3 × (even), again a multiple of 6. Example: 9 + 15 = 24 = 6 × 4.
4. One of each: 3 × (even) + 3 × (odd) = 3 × (even + odd) = 3 × (odd). This is an odd number, so it is a multiple of 3 but not of 6. Example: 6 + 9 = 15.
5. Generalising: the same argument works for any number n. n × a + n × b = n × (a + b), and this is a multiple of 2n exactly when a + b is even, that is, when a and b are both even or both odd.

In shortThe sum of two multiples of 3 is a multiple of 6 when both are even multiples of 3 or both are odd multiples of 3; it is not a multiple of 6 when one is even and the other is odd. In general, the sum of two multiples of n is a multiple of 2n exactly when both are even multiples of n or both are odd multiples of n.

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Open NCERT p. 132Matches NCERT’s answer

(i) Examine if her conjecture is true for any multiple of 9.

1. Divisibility by 9 depends only on the sum of the digits, never on the order they are written in.
2. Reversing a number uses exactly the same digits, just in the opposite order, so the digit sum does not change.
3. If the digit sum was a multiple of 9 before reversing, it is still the same multiple of 9 after.

AnswerYes — her conjecture is true for every multiple of 9, not just the one she picked.

(ii) Are any other digit shuffles possible such that the number formed …

1. The digit-sum test only cares which digits are present, not the positions they sit in.
2. So any rearrangement of the digits — not only the exact reversal — keeps the digit sum the same.
3. That means every possible shuffle of the digits is also divisible by 9.

AnswerYes — any shuffle of the digits works, not only reversing them.

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Open NCERT p. 132Matches NCERT’s answer

1. 18 = 2 × 9, so the number must be divisible by both 2 and 9.
2. Divisible by 2 needs the last digit b to be even: b could be 0, 2, 4, 6 or 8.
3. Divisible by 9 needs the digit sum 4+8+a+2+3+b = 17+a+b to be a multiple of 9.
4. Since a and b are digits (0–9), 17+a+b can only reach 18 or 27, so a+b = 1 or a+b = 10.
5. Matching a+b = 1 or 10 with each even value of b gives the possible pairs.

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

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Open NCERT p. 133Matches NCERT’s answer

1. 44 = 4 × 11, and 4 and 11 share no factor, so 3p7q8 must be divisible by both 4 and 11.
2. Divisible by 4: the last two digits, q8, must make a multiple of 4. 08, 28, 48, 68 and 88 are multiples of 4, but 18, 38, 58, 78 and 98 are not. So q is even: q = 0, 2, 4, 6 or 8.
3. Divisible by 11: add and subtract the digits alternately, starting from the units digit: 8 − q + 7 − p + 3 = 18 − (p + q). This must be a multiple of 11.
4. p + q is between 0 and 18, so 18 − (p + q) is between 0 and 18. The only multiples of 11 in that range are 0 and 11, so p + q = 18 or p + q = 7.
5. p + q = 18 needs p = 9 and q = 9, but q must be even. So p + q = 7.
6. q = 0 gives p = 7; q = 2 gives p = 5; q = 4 gives p = 3; q = 6 gives p = 1; q = 8 would need p = −1, which is not a digit.
7. Check: 37708 = 44 × 857, 35728 = 44 × 812, 33748 = 44 × 767, 31768 = 44 × 722.

Answer(p, q) = (7, 0), (5, 2), (3, 4), (1, 6).

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Open NCERT p. 133Matches NCERT’s answer

1. Call the first number n; then n+1 must be a multiple of 3 and n+2 a multiple of 4.
2. n+2 a multiple of 4 means n leaves remainder 2 on division by 4.
3. n+1 a multiple of 3 means n leaves remainder 2 on division by 3.
4. A number leaving remainder 2 on division by 4 is automatically even, so the first condition is already satisfied.
5. Combining 'remainder 2 mod 4' and 'remainder 2 mod 3' (which share no common factor) gives 'remainder 2 mod 12'.
6. The smallest such n is 2 (triple 2, 3, 4); the next is 14 (triple 14, 15, 16), and so on every 12 numbers.

AnswerExample: 2, 3, 4. Yes, there are infinitely many more — a fresh triple starts every 12 numbers (n = 2, 14, 26, 38, …).

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Open NCERT p. 133Checked by computerAnswers can differ: one example

1. 45,000 ÷ 36 = 1250 exactly, so 45,000 is itself a multiple of 36.
2. Keep adding 36 to get the next multiples: 45,036, then 45,072, 45,108, 45,144, 45,180.
3. All five are less than 47,000, so they fit inside the given range.

Answer45,036; 45,072; 45,108; 45,144; 45,180 (found by adding 36 repeatedly, starting just above 45,000).

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Open NCERT p. 133Matches NCERT’s answer

1. Consecutive even numbers are 2 apart, and 5p sits in the middle of the five.
2. The two numbers just before 5p are 5p − 2 and 5p − 4.
3. The two numbers just after 5p are 5p + 2 and 5p + 4.

Answer5p − 4, 5p − 2, 5p, 5p + 2, 5p + 4 — so the other four numbers are 5p−4, 5p−2, 5p+2, 5p+4.

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Open NCERT p. 133Checked by computerAnswers can differ: one example

1. Divisible by 15 means divisible by 3 (digit sum is a multiple of 3) and by 5 (ends in 0 or 5).
2. Reversing keeps the same digits, so the digit sum — and hence divisibility by 3 — does not change; only the reversed number's last digit (the original's FIRST digit) decides its divisibility by 2.
3. So pick a first digit that is even, a last digit that is 0 or 5, and make sure the digit sum is a multiple of 3: 200025 works.
4. Check: 200025 ÷ 15 = 13,335 exactly, and its reverse 520002 ÷ 6 = 86,667 exactly.

Answer200025 (its reverse, 520002, is divisible by 6).

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Open NCERT p. 133Matches NCERT’s answer

1. Any multiple of 11 can be written as 11k for some whole number k.
2. Doubling it gives 2 × 11k = 11 × (2k), which is still 11 times a whole number.
3. So every multiple of 11, without exception, stays a multiple of 11 when doubled.
4. Deepak's claim that only SOME multiples behave this way (while others don't) is therefore wrong.

AnswerNo — Deepak's conjecture is false. Every multiple of 11 remains a multiple of 11 when doubled, not just some of them.

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Open NCERT p. 133Matches NCERT’s answer

(i) The product of a multiple of 6 and a multiple of 3 …

1. A multiple of 6 is 6a and a multiple of 3 is 3b, for whole numbers a and b.
2. Their product is 6a × 3b = 18ab.
3. 18ab = 9 × (2ab), which is always a multiple of 9.

AnswerAlways True.

(ii) The sum of three consecutive even numbers will be divisible by 6.

1. Three consecutive even numbers can be written as n, n+2, n+4 (n even).
2. Their sum is 3n + 6 = 3(n + 2).
3. Since n is even, n+2 is also even, so 3(n+2) is 3 × an even number, which is always a multiple of 6.

AnswerAlways True.

(iii) If abcdef is a multiple of 6, then badcef …

1. badcef has exactly the same six digits as abcdef, only with the first two swapped and the middle two swapped; the last digit f is unchanged.
2. Divisibility by 6 needs an even last digit and a digit sum that is a multiple of 3.
3. Since the last digit and the digit sum are unchanged by the swap, badcef is a multiple of 6 whenever abcdef is.

AnswerAlways True.

(iv) 8 (7b – 3) – 4 (11b + 1) is a multiple of 12.

1. Expand: 8(7b−3) − 4(11b+1) = 56b − 24 − 44b − 4 = 12b − 28.
2. 12b is always a multiple of 12, so 12b − 28 leaves the same remainder as −28 does when divided by 12.
3. −28 = 12×(−3) + 8, so that remainder is always 8, never 0.
4. So the expression can never be a multiple of 12, whatever whole number b is.

AnswerNever True.

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Open NCERT p. 133One way to think about it

1. Every whole number leaves remainder 0, 1 or 2 when divided by 3, so each number is (a multiple of 3) + its remainder.
2. When the three numbers are added, the multiples of 3 add up to a multiple of 3. So the sum is divisible by 3 exactly when the three remainders add up to a multiple of 3.
3. All three remainders the same: 0 + 0 + 0 = 0, 1 + 1 + 1 = 3, 2 + 2 + 2 = 6. All are multiples of 3, so the sum is divisible by 3. Example: 4 + 7 + 10 = 21.
4. All three remainders different: 0 + 1 + 2 = 3, so the sum is divisible by 3. Example: 3 + 4 + 5 = 12. (This is why any three consecutive numbers add up to a multiple of 3.)
5. Exactly two remainders the same: 0 + 0 + 1 = 1, 0 + 0 + 2 = 2, 1 + 1 + 0 = 2, 1 + 1 + 2 = 4, 2 + 2 + 0 = 4, 2 + 2 + 1 = 5. None of these is a multiple of 3, so the sum is not divisible by 3. Example: 3 + 6 + 4 = 13.
6. These three cases cover every possible choice of three numbers.

In shortThe sum of three numbers is divisible by 3 exactly when their remainders on division by 3 are all the same, or all different (one each of 0, 1 and 2). When exactly two of the remainders are the same, the sum is not divisible by 3.

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Open NCERT p. 133Matches NCERT’s answer

1. Two consecutive integers: one of them is even, so their product is always a multiple of 2. Example: 7 × 8 = 56.
2. Three consecutive integers: at least one is even, and one is a multiple of 3 (every third integer is). So the product is a multiple of 2 and of 3. Since 2 and 3 share no factor, it is a multiple of 6. Example: 4 × 5 × 6 = 120 = 6 × 20.
3. Four consecutive integers: two of them are even, and one of those two is a multiple of 4. So the product is a multiple of 4 × 2 = 8. One of the four is also a multiple of 3. Since 8 and 3 share no factor, the product is a multiple of 8 × 3 = 24.
4. Five consecutive integers: the first four already make the product a multiple of 24, and one of the five is a multiple of 5. Since 24 and 5 share no factor, the product is a multiple of 24 × 5 = 120.
5. No larger number works: 1 × 2 × 3 × 4 = 24 and 1 × 2 × 3 × 4 × 5 = 120 are themselves such products.

AnswerYes — the product of two consecutive integers is always a multiple of 2. Yes — the product of three consecutive integers is always a multiple of 6. The product of four consecutive integers is always a multiple of 24, and the product of five consecutive integers is always a multiple of 120.

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Open NCERT p. 133Matches NCERT’s answer

(i) EF × E = GGG

1. EF is a two-digit number and GGG is a three-digit number with all three digits equal, so GGG = G × 111 = G × 3 × 37.
2. So EF × E = 3 × 37 × G. 37 is a prime number, so it must divide EF or E. E is a single digit, so 37 divides EF.
3. The two-digit multiples of 37 are 37 and 74.
4. EF = 74 means E = 7, and 74 × 7 = 518, whose digits are not all equal. EF = 37 means E = 3, and 37 × 3 = 111.
5. So E = 3, F = 7 and G = 1, three different digits.

AnswerE = 3, F = 7, G = 1 (37 × 3 = 111).

(ii) WOW × 5 = MEOW

1. Units column: the units digit of WOW × 5 is the units digit of W × 5, and it must be W. Any digit times 5 ends in 0 or 5, so W = 0 or W = 5. W is the first digit of WOW, so it cannot be 0. So W = 5.
2. Then WOW = 5O5, and 5O5 × 5 lies between 505 × 5 = 2525 and 595 × 5 = 2975, so M = 2.
3. 5O5 × 5 = 2525 + O × 50. If O is even, O × 50 is a whole number of hundreds and the tens digit stays 2. If O is odd, the tens digit becomes 7.
4. The tens digit of MEOW is O itself, so either O = 2 (even) or O = 7 (odd).
5. O = 2 gives 525 × 5 = 2625, but then M = 2 and O = 2: two different letters with the same digit, which is not allowed.
6. O = 7 gives 575 × 5 = 2875, so M = 2, E = 8, O = 7 and W = 5, all different.

AnswerW = 5, O = 7, M = 2, E = 8 (575 × 5 = 2875).

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Open NCERT p. 133Matches NCERT’s answer

1. Since 4 divides 8, and 8 divides 32, every multiple of 32 is also a multiple of 8, and every multiple of 8 is also a multiple of 4.
2. So the three sets sit one fully inside the other, not just partly overlapping: multiples of 32 is inside multiples of 8, which is inside multiples of 4.
3. Diagrams (i) and (ii) draw only partly-overlapping circles, which is wrong for a full containment.
4. Diagram (iii) draws three nested circles but puts 'multiples of 32' as the OUTER ring and 'multiples of 4' as the INNERMOST — backwards, since multiples of 4 is the biggest set.
5. Diagram (iv) correctly shows 'multiples of 4' as the outer circle, 'multiples of 8' in the middle, and 'multiples of 32' as the innermost circle.

AnswerDiagram (iv).

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