Class 8 Mathematics · Chapter 5

5. Number Play

Solutions to all Figure it Out and in-text questions. CBSE Class 8 Mathematics, NCERT Ganita Prakash Part 1.

5.1 Is This a Multiple Of?

1Can every natural number be written as a sum of consecutive numbers? Which numbers can be written in more than one way? Can even numbers and 0 be written so?

No. Numbers like 1, 2, 4, 8, 16 (powers of 2) cannot be written as a sum of two or more consecutive natural numbers.

Every odd number (except 1) is a sum of two consecutive numbers, like 9 = 4 + 5.

Some even numbers can be written so, like 6 = 1 + 2 + 3 and 10 = 1 + 2 + 3 + 4.

Numbers like 9 = 4 + 5 = 2 + 3 + 4 and 15 = 7 + 8 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5 can be written in more than one way.

Yes, 0 can be written using negative numbers: 0 = −1 + 0 + 1 or 0 = −2 + (−1) + 0 + 1 + 2.

2Take 4 consecutive numbers, place ‘+’ and ‘−’ signs between them in all 8 ways and evaluate. What do you observe?

3 + 4 + 5 + 6 = 18 3 + 4 + 5 − 6 = 6 3 + 4 − 5 + 6 = 8 3 + 4 − 5 − 6 = −4 3 − 4 + 5 + 6 = 10 3 − 4 + 5 − 6 = −2 3 − 4 − 5 + 6 = 0 3 − 4 − 5 − 6 = −12 5 − 6 − 7 − 8 = −16

5 + 6 + 7 + 8 = 26 5 + 6 + 7 − 8 = 10 5 + 6 − 7 + 8 = 12 5 + 6 − 7 − 8 = −4 5 − 6 + 7 + 8 = 14 5 − 6 + 7 − 8 = −2 5 − 6 − 7 + 8 = 0

All results are even. The values −4, −2 and 0 come every time.

Reason (algebra): Take n, n + 1, n + 2, n + 3. Then n + (n + 1) − (n + 2) − (n + 3) = −4, n − (n + 1) + (n + 2) − (n + 3) = −2 and n − (n + 1) − (n + 2) + (n + 3) = 0 for every n.

3Take any 4 numbers and place signs in the eight ways. What do you observe about their parities?

All eight results have the same parity (all even or all odd).

Changing +b to −b changes the value by 2b, which is even. So the parity does not change.

4Replace any negative sign in a + b − c − d with a positive sign and find the difference. What do you conclude?

(a + b + c − d) − (a + b − c − d) = 2c, which is even.

So switching a sign either way changes the value by an even number. All the expressions have the same parity.

5Is this phenomenon of all expressions having the same parity limited to taking 4 numbers?

No. It is true for any count of numbers. Switching any sign always changes the value by twice that number, which is even.

Breaking Even

6Without computing, find which of the following arithmetic expressions are even.

Even: 43 + 37 (odd + odd), 672 − 348 (even − even), 4 × 347 × 3 (has factor 4), 543 − 479 (odd − odd).

Odd: 708 − 477 (even − odd), 809 + 214 (odd + even), 119 × 303 (odd × odd), 513³ (odd × odd × odd).

7Which of the following algebraic expressions give an even number for any integer values? Give examples and non-examples.

2a + 2b — always even. 2a + 2b = 2(a + b).
3g + 5h — not always. g = 1, h = 1 gives 8 (even); g = 1, h = 2 gives 13 (odd).
4m + 2n — always even. 4m + 2n = 2(2m + n).
2u − 4v — always even. 2u − 4v = 2(u − 2v).
13k − 5k — always even. It equals 8k.
6m − 3n — not always. n = 2 gives even; n = 1 gives odd (6 − 3 = 3).
x² + 2 — not always. x = 6 gives 38; x = 3 gives 11.
b² + 1 — not always. b = 3 gives 10; b = 2 gives 5.
4k × 3j — always even. It equals 12kj = 2(6kj).

More expressions that are always even: 6p + 10q, 2x², 8a − 4, n(n + 1).

Pairs to Make Fours

8When is the sum of two even numbers a multiple of 4? What happens when we add a multiple of 4 to an even number that is not a multiple of 4?

Multiple of 4 + multiple of 4: 4p + 4q = 4(p + q). Always a multiple of 4.

(4p + 2) + (4q + 2) = 4(p + q + 1). Always a multiple of 4.

4p + (4q + 2) = 4(p + q) + 2. Never a multiple of 4; it leaves remainder 2.

Examples: 4 + 6 = 10, 8 + 10 = 18, 12 + 2 = 14. This is like even + odd = odd.

Always, Sometimes, or Never

9Statement 1 is always true. Determine if it is true with subtraction.

Yes. 8a − 8b = 8(a − b). So the difference of two multiples of 8 is also a multiple of 8. For example, 56 − 16 = 40 = 8 × 5.

10(6) If a number is divisible by both 9 and 4, it must be divisible by 36. (7) If a number is divisible by both 6 and 4, it must be divisible by 24.

(6) Always true. LCM of 9 and 4 is 36. Example: 72, 108.

(7) Sometimes true. LCM of 6 and 4 is 12, not 24. 48 is divisible by 24, but 12 and 36 are not.

What Remains?

11Find numbers that leave a remainder of 3 when divided by 5. Which expressions capture all such numbers? (i) 3k + 5 (ii) 3k − 5 (iii) 3k (iv) 5k + 3 (v) 5k − 2 (vi) 5k − 3

Numbers: 3, 8, 13, 18, 23, 28, …

Expressions: (iv) 5k + 3 (k ≥ 0) and (v) 5k − 2 (k ≥ 1).

Other expressions: 5k + 8 (k ≥ −1), 5k − 7 (k ≥ 2), and so on.

Figure it Out (Page 122)

1The sum of four consecutive numbers is 34. What are these numbers?

Let the numbers be n, n + 1, n + 2, n + 3.

4n + 6 = 34, so 4n = 28 and n = 7.

The numbers are 7, 8, 9 and 10

2Suppose p is the greatest of five consecutive numbers. Describe the other four numbers in terms of p.

The numbers are p − 4, p − 3, p − 2, p − 1 (and p).

3Is each statement always true, sometimes true or never true? Justify using algebra.

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

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

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

Sometimes true. 20 is divisible by neither. But 27 is not divisible by 18 and is divisible by 9.

(iii) If two numbers are not divisible by 6, then their sum is not divisible by 6.

Sometimes true. 7 + 8 = 15 is not divisible by 6. But 8 + 10 = 18 is divisible by 6.

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

Always true. 6a + 9b = 3(2a + 3b).

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

Sometimes true. 18 + 9 = 27 is a multiple of 9, but 12 + 9 = 21 is not. 6a + 3b = 3(2a + b), which is a multiple of 9 only when 2a + b is a multiple of 3.

4Find numbers that leave a remainder of 2 when divided by 3 and a remainder of 2 when divided by 4. Write an expression for all such numbers.

The number minus 2 must be a multiple of both 3 and 4, that is, a multiple of 12.

Numbers: 2, 14, 26, 38, 50, …

Expression: 12k + 2 (k = 0, 1, 2, …)

5The pebble riddle: groups of 3 leave 1, pairs leave 1, groups of 5 leave 1, groups of 7 leave 0, and fewer than 100 pebbles.

The number leaves remainder 1 when divided by 2, 3 and 5. So it is 1 more than a multiple of 30: 31, 61, 91.

Only 91 is divisible by 7 (91 = 7 × 13).

So the number of pebbles is 91

6Numbers leaving remainder 2 when divided by 6: Tathagat claims the sum of any three such numbers is a multiple of 6. Is it true?

(6a + 2) + (6b + 2) + (6c + 2) = 6a + 6b + 6c + 6 = 6(a + b + c + 1).

So the claim is true. Example: 2 + 8 + 14 = 24.

7661 leaves remainder 3 and 4779 leaves remainder 5 when divided by 7. Find the remainders of (i) 4779 + 661 (ii) 4779 − 661.

Let 4779 = 7p + 5 and 661 = 7q + 3.

(i) 4779 + 661 = 7p + 7q + 8 = 7(p + q + 1) + 1. Remainder = 1

(ii) 4779 − 661 = 7p − 7q + 2 = 7(p − q) + 2. Remainder = 2

Visually: Arrange each number in rows of 7 dots. The extra dots are 5 and 3. Together they are 8 = one more row of 7 and 1 extra dot. On removing, 5 − 3 = 2 extra dots remain.

8Find a number that leaves remainder 2 when divided by 3, remainder 3 when divided by 4 and remainder 4 when divided by 5. What is the smallest such number? Why?

Each remainder is 1 less than the divisor. So the number + 1 is a multiple of 3, 4 and 5, that is, of 60.

The smallest such number is 60 − 1 = 59. It is the smallest because 60 is the smallest common multiple of 3, 4 and 5.

5.2 Checking Divisibility Quickly

12Explain using algebra why the divisibility shortcuts for 5, 2, 4 and 8 work.

A number … dcba = … + 1000d + 100c + 10b + a.

5 and 2: 10b, 100c, … are multiples of 10, so of 5 and 2. So the number is divisible by 5 (or 2) if the units digit a is divisible by 5 (or 2).

4: 100c, 1000d, … are multiples of 100, so of 4. So we only check the last two digits (10b + a).

8: 1000d, 10000e, … are multiples of 1000, so of 8. So we only check the last three digits (100c + 10b + a).

13Is 10 divisible by 9? Find the remainders when multiples of 10 and of 100 are divided by 9.

No. 10 = 9 + 1, so the remainder is 1.

20 leaves 2, 30 leaves 3, …; 100 = 99 + 1 leaves 1, 200 leaves 2, …

So the remainder equals the number of tens (or hundreds).

14Which statements are correct? (i) If a number is divisible by 9, the sum of its digits is divisible by 9. (ii) If the sum of digits is divisible by 9, the number is divisible by 9. (iii) If a number is not divisible by 9, the sum of its digits is not divisible by 9. (iv) If the sum of the digits is not divisible by 9, the number is not divisible by 9.

All four are correct. A number and the sum of its digits always leave the same remainder when divided by 9. So both are divisible by 9 together, or both are not.

Figure it Out (Page 126)

1Find, without dividing, whether the following numbers are divisible by 9.

(i) 123: 1 + 2 + 3 = 6 → No

(ii) 405: 4 + 0 + 5 = 9 → Yes

(iii) 8888: sum = 32 → No

(iv) 93547: sum = 28 → No

(v) 358095: sum = 30 → No

2Find the smallest multiple of 9 with no odd digits.

The digit sum must be a multiple of 9. With even digits, the sum is even, so it must be 18.

Two even digits give at most 8 + 8 = 16. So we need 3 digits.

Smallest: hundreds digit 2, then 8 + 8 = 16. The number is 288

3Find the multiple of 9 that is closest to the number 6000.

6 + 0 + 0 + 0 = 6. So 6000 is 6 more than a multiple of 9 (5994) and 3 less than the next (6003).

The closest multiple is 6003

4How many multiples of 9 are there between 4300 and 4400?

First multiple after 4300 is 4302 (digit sum 9). Last before 4400 is 4392.

Number of multiples = (4392 − 4302) ÷ 9 + 1 = 10 + 1 = 11

15Explore the remainders when powers of 10 are divided by 3. Explain the divisibility rule for 3.

10 = 9 + 1, 100 = 99 + 1, 1000 = 999 + 1. Each power of 10 leaves remainder 1 when divided by 3.

So a number leaves the same remainder as the sum of its digits. It is divisible by 3 if the sum of its digits is divisible by 3.

16Is 462 divisible by 11?

400 is 4 more than a multiple of 11. 60 is 6 less. 2 is 2 more.

4 − 6 + 2 = 0. So 462 is divisible by 11 (462 = 11 × 42).

17If the difference (excess − short) is 11 or a multiple of 11, what does it say about the remainder?

The remainder is 0. The number is divisible by 11.

18Find whether the following numbers are divisible by 11, and find the remainder if not: (i) 158 (ii) 841 (iii) 481 (iv) 5529 (v) 90904 (vi) 857076

(i) 158: 8 − 5 + 1 = 4 → remainder 4

(ii) 841: 1 − 4 + 8 = 5 → remainder 5

(iii) 481: 1 − 8 + 4 = −3 → 3 short, so remainder 8

(iv) 5529: 9 − 2 + 5 − 5 = 7 → remainder 7

(v) 90904: 4 − 0 + 9 − 0 + 9 = 22 → divisible

(vi) 857076: 6 − 7 + 0 − 7 + 5 − 8 = −11 → divisible

19Is the alternating-sign method similar to or different from the excess/short method?

It is the same method. The digits with ‘+’ give the excess and the digits with ‘−’ give the shortage. We just do both steps in one expression.

20Fill in the divisibility table. Find a quick way to do this.

Number	2	3	4	5	6	8	9	10	11
128	Yes	No	Yes	No	No	Yes	No	No	No
990	Yes	Yes	No	Yes	Yes	No	Yes	Yes	Yes
1586	Yes	No	No	No	No	No	No	No	No
275	No	No	No	Yes	No	No	No	No	Yes
6686	Yes	No	No	No	No	No	No	No	No
639210	Yes	Yes	No	Yes	Yes	No	No	Yes	Yes
429714	Yes	Yes	No	No	Yes	No	Yes	No	No
2856	Yes	Yes	Yes	No	Yes	Yes	No	No	No
3060	Yes	Yes	Yes	Yes	Yes	No	Yes	Yes	No
406839	No	Yes	No	No	No	No	No	No	No

Quick way: Use the shortcuts. Divisible by 6 means Yes for both 2 and 3. Divisible by 10 means Yes for both 2 and 5. Divisible by 9 means Yes for 3. Divisible by 8 means Yes for 4.

The textbook’s first row shows “No” for 4 under 128. But 128 = 4 × 32, so 128 is divisible by 4. The correct entry is Yes.

21Check divisibility by 6 using the shortcuts for 2 and 3: 38, 225, 186, 64.

38: even, digit sum 11 → not divisible by 6.

225: odd → not divisible by 6.

186: even, digit sum 15 → divisible by 6 (186 = 6 × 31).

64: even, digit sum 10 → not divisible by 6.

22Explain using prime factorisation why checking 3 and 8 works for 24, but checking 4 and 6 is not sufficient.

24 = 2 × 2 × 2 × 3. 8 = 2 × 2 × 2 and 3 have no common factor. A number divisible by both has all the factors 2 × 2 × 2 × 3, so it is divisible by 24.

4 = 2 × 2 and 6 = 2 × 3 share the factor 2. A number divisible by both need only have 2 × 2 × 3 = 12. For example, 12 is divisible by 4 and 6 but not by 24.

Digital Roots

23What property do you think the digital root has?

The digital root is the remainder when the number is divided by 9. If the number is a multiple of 9, the digital root is 9.

24Between 600 and 700, which numbers have the digital root (i) 5 (ii) 7 (iii) 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

25Write the digital roots of any 12 consecutive numbers. What do you observe?

For 20 to 31: 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4.

The digital roots go 1 to 9 and then repeat.

26Find the digital roots of consecutive multiples of (i) 3 (ii) 4 (iii) 6. What are the digital roots of numbers 1 more than a multiple of 6?

(i) 3, 6, 9, 3, 6, 9, …

(ii) 4, 8, 3, 7, 2, 6, 1, 5, 9, and then repeat.

(iii) 6, 3, 9, 6, 3, 9, …

1 more than multiples of 6 (1, 7, 13, 19, 25, 31, …): digital roots 1, 7, 4, 1, 7, 4, … They repeat 1, 7, 4.

Reason: Adding 6 adds 6 to the digital root (going round after 9).

27Riddle: I’m made of digits, each tiniest and odd … What’s my number?

The tiniest odd digit is 1. The number of digits, their sum and the digital root are all 9 (the largest odd single digit).

So the number is 11,11,11,111 (eleven crore eleven lakh eleven thousand one hundred eleven).

Figure it Out (Page 131)

1The digital root of an 8-digit number is 5. What will be the digital root of 10 more than that number?

10 has digital root 1. Adding 10 adds 1 to the digital root.

So the digital root becomes 5 + 1 = 6

2Write any number. Repeatedly add 11. What are the digital roots of this sequence?

Take 10: 10, 21, 32, 43, 54, 65, 76, 87, 98, 109, …

Digital roots: 1, 3, 5, 7, 9, 2, 4, 6, 8, 1, …

Each time the digital root increases by 2 (since 11 has digital root 2). After 9 steps it comes back.

3What will be the digital root of the number 9a + 36b + 13?

9a + 36b + 13 = 9(a + 4b + 1) + 4.

So the number is 4 more than a multiple of 9. Its digital root is 4

4Make conjectures about (i) the parity of a number and its digital root (ii) the digital root and the remainder on division by 3 or 9.

(i) There is no relation. 13 is odd with root 4; 22 is even with root 4.

(ii) The remainder on division by 9 is the digital root (0 if the root is 9).

For division by 3: root 1, 4, 7 → remainder 1; root 2, 5, 8 → remainder 2; root 3, 6, 9 → remainder 0.

5.3 Digits in Disguise

28Solve the cryptarithms: (i) A1 + 1B = B0 (ii) AB + 37 = 6A (iii) ON + ON + ON = PO (iv) QR + QR + QR = PRR

(i) A = 7, B = 9 (71 + 19 = 90)

(ii) A = 2, B = 5 (25 + 37 = 62)

(iii) 3 × ON = PO. Solutions: O = 3, N = 1, P = 9 (31 × 3 = 93); also 17 × 3 = 51 and 24 × 3 = 72.

(iv) Q = 8, R = 5, P = 2 (85 + 85 + 85 = 255)

29(vi) GH × H = 9K. Pick the solution from the options. (vii) BYE × 6 = RAY.

(vi) The units digit of GH is H, and the product’s units digit K must be different from H.

24 × 4 = 96 fits (G = 2, H = 4, K = 6). 11 × 9 and 16 × 6 repeat digits; 12 × 8 and 46 × 2 do not use H as multiplier.

Answer: 24 × 4 = 96

(vii) B = 1. Y × 6 + carry must end in Y, and E × 6 must end in Y. Trying gives 105 × 6 = 630 (B = 1, Y = 0, E = 5, R = 6, A = 3).

30Solve: (i) UT × 3 = PUT (ii) AB × 5 = BC (iii) L2N × 2 = 2NP (iv) XY × 4 = ZX (v) PP × QQ = PRP (vi) JK × 6 = KKK

(i) 50 × 3 = 150 (U = 5, T = 0, P = 1)

(ii) 19 × 5 = 95 (A = 1, B = 9, C = 5)

(iii) 125 × 2 = 250 (L = 1, N = 5, P = 0). Another answer: 124 × 2 = 248.

(iv) 23 × 4 = 92 (X = 2, Y = 3, Z = 9)

(v) 22 × 11 = 242 (P = 2, Q = 1, R = 4). Others: 33 × 11 = 363, 44 × 11 = 484.

(vi) 74 × 6 = 444 (J = 7, K = 4)

Figure it Out (Page 132)

1If 31z5 is a multiple of 9, find z. Why are there two answers?

3 + 1 + z + 5 = 9 + z must be a multiple of 9. So z = 0 or 9.

There are two answers because both 9 + 0 = 9 and 9 + 9 = 18 are multiples of 9, and z can be any digit from 0 to 9.

2Snehal: a number with remainder 8 (÷ 12) plus a number 4 short of a multiple of 12 is always a multiple of 8. Examine.

(12m + 8) + (12n − 4) = 12(m + n) + 4.

This is not always a multiple of 8. For m + n = 1, we get 16 (multiple of 8). For m + n = 2, we get 28 (not a multiple of 8).

So the claim is false. The sum is always a multiple of 4.

3When is the sum of two multiples of 3 a multiple of 6?

3m + 3n = 3(m + n). It is a multiple of 6 when m + n is even.

So the sum is a multiple of 6 when both numbers are even multiples of 3 (like 6 + 12) or both are odd multiples of 3 (like 3 + 9). It is not when one is odd and one is even (like 3 + 6 = 9).

4Sreelatha: “If I reverse the digits of a multiple of 9, it is still divisible by 9.” (i) Is this true for any multiple of 9? (ii) Are other digit shuffles possible?

(i) Yes. Reversing does not change the sum of digits, so it stays a multiple of 9.

(ii) Yes. Any shuffle of the digits gives a multiple of 9, for the same reason.

5If 48a23b is a multiple of 18, list all possible pairs of values for a and b.

It must be divisible by 2 and 9. So b is even, and 4 + 8 + a + 2 + 3 + b = 17 + a + b is a multiple of 9.

So a + b = 1 or 10.

Pairs (a, b): (1, 0), (8, 2), (6, 4), (4, 6), (2, 8)

6If 3p7q8 is divisible by 44, list all possible pairs of values for p and q.

Divisible by 4: q8 must be divisible by 4, so q is even (08, 28, 48, 68, 88).

Divisible by 11: (8 + 7 + 3) − (q + p) = 18 − (p + q) must be 0 or a multiple of 11. So p + q = 7.

Pairs (p, q): (7, 0), (5, 2), (3, 4), (1, 6)

7Find three consecutive numbers: the first a multiple of 2, the second a multiple of 3, the third a multiple of 4. Are there more? How often?

2, 3, 4 is one set. The next sets are 14, 15, 16 and 26, 27, 28.

They occur every 12 numbers (LCM of 2, 3, 4 is 12). The first number is of the form 12k + 2.

8Write five multiples of 36 between 45,000 and 47,000.

A multiple of 36 must be divisible by 4 and 9. 45000 is divisible by 36, so keep adding 36.

45036, 45072, 45108, 45144, 45180

9The middle number of 5 consecutive even numbers is 5p. Express the other four numbers.

5p − 4, 5p − 2, 5p + 2, 5p + 4

10Write a 6-digit number divisible by 15 such that when the digits are reversed, it is divisible by 6.

Divisible by 15: ends in 5 (not 0, or the reversed number would start with 0) and digit sum divisible by 3.

Reversed divisible by 6: the first digit must be even, and the digit sum divisible by 3.

Example: 200025. 200025 = 15 × 13335, and 520002 = 6 × 86667.

11Deepak: “Some multiples of 11 remain multiples of 11 when doubled, but others don’t.” Examine.

If n = 11k, then 2n = 22k = 11 × 2k. So every multiple of 11 stays a multiple of 11 when doubled.

Deepak’s conjecture is false.

12Always True, Sometimes True or Never True?

(i) The product of a multiple of 6 and a multiple of 3 is a multiple of 9.

Always true. 6a × 3b = 18ab = 9 × 2ab.

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

Always true. 2n + (2n + 2) + (2n + 4) = 6n + 6 = 6(n + 1).

(iii) If abcdef is a multiple of 6, then badcef will be a multiple of 6.

Always true. The units digit f stays the same (so it is still even), and the digit sum stays the same (so it is still divisible by 3).

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

Never true. It equals 56b − 24 − 44b − 4 = 12b − 28 = 12(b − 3) + 8. It always leaves remainder 8.

13Choose any 3 numbers. When is their sum divisible by 3? Explore all cases.

Look at the remainders when each number is divided by 3 (0, 1 or 2).

The sum is divisible by 3 when the sum of the remainders is 0, 3 or 6. This happens when:

all three numbers leave the same remainder (0, 0, 0 or 1, 1, 1 or 2, 2, 2), or
all three leave different remainders (0, 1, 2).

In all other cases, the sum is not divisible by 3.

14Is the product of two consecutive integers always a multiple of 2? Of three consecutive integers a multiple of 6? What about 4 and 5 consecutive integers?

Two: Yes. One of them is even.

Three: Yes, a multiple of 6. One of them is even and one is a multiple of 3.

Four: Always a multiple of 24. One is a multiple of 4, another is even, and one is a multiple of 3. So it has 2 × 4 × 3 = 24.

Five: Always a multiple of 120 (24 × 5), since one of them is a multiple of 5 as well.

15Solve the cryptarithms: (i) EF × E = GGG (ii) WOW × 5 = MEOW

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

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

16Which Venn diagram captures the relationship between the multiples of 4, 8 and 32?

Every multiple of 32 is a multiple of 8, and every multiple of 8 is a multiple of 4.

So multiples of 32 lie inside multiples of 8, which lie inside multiples of 4. The answer is (iv)

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