3. HCF and LCM
Step-by-step solutions to Practice Sets 10 to 14. Maharashtra State Board (Balbharati), Grade Seven Mathematics.
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Practice Set 10
1
A prime number has exactly two divisors and a composite number has more than two. The number 1 has only one divisor (1 itself), so it is neither prime nor composite.
- (i) 8, 14 — Not co-prime
2 is a common factor of both. - (ii) 4, 5 — Co-prime
Only common factor is 1. - (iii) 17, 19 — Co-prime
Only common factor is 1. - (iv) 27, 15 — Not co-prime
3 is a common factor of both.
∴ The pairs of co-primes are (ii) 4, 5 and (iii) 17, 19.
29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97
There are 16 prime numbers from 25 to 100.
(59, 61) ; (71, 73)
These are the only pairs of primes between 51 and 100 that differ by 2.
Note on the definition: twin primes are two prime numbers whose difference is 2. The textbook says “co-prime numbers”, which is a slip. For example, 9 and 11 are co-prime and differ by 2, but 9 is not prime, so (9, 11) is not a twin prime pair.
(3, 5) ; (5, 7) ; (11, 13) ; (17, 19) ; (29, 31)
(Another pair in this range is (41, 43).)
2 is the only even prime number.
Every other even number is divisible by 2 as well as by 1 and itself, so it has more than two divisors and is composite.
Practice Set 11
(i) 32
| 2 | 32 |
| 2 | 16 |
| 2 | 8 |
| 2 | 4 |
| 2 | 2 |
| 1 |
32 = 2 × 2 × 2 × 2 × 2
(ii) 57
| 3 | 57 |
| 19 | 19 |
| 1 |
57 = 3 × 19
(iii) 23
| 23 | 23 |
| 1 |
23 = 23
23 is itself a prime number, so it has no other prime factors.
(iv) 150
| 2 | 150 |
| 3 | 75 |
| 5 | 25 |
| 5 | 5 |
| 1 |
150 = 2 × 3 × 5 × 5
(v) 216
| 2 | 216 |
| 2 | 108 |
| 2 | 54 |
| 3 | 27 |
| 3 | 9 |
| 3 | 3 |
| 1 |
216 = 2 × 2 × 2 × 3 × 3 × 3
(vi) 208
| 2 | 208 |
| 2 | 104 |
| 2 | 52 |
| 2 | 26 |
| 13 | 13 |
| 1 |
208 = 2 × 2 × 2 × 2 × 13
(vii) 765
| 3 | 765 |
| 3 | 255 |
| 5 | 85 |
| 17 | 17 |
| 1 |
765 = 3 × 3 × 5 × 17
(viii) 342
| 2 | 342 |
| 3 | 171 |
| 3 | 57 |
| 19 | 19 |
| 1 |
342 = 2 × 3 × 3 × 19
(ix) 377
| 13 | 377 |
| 29 | 29 |
| 1 |
377 = 13 × 29
(x) 559
| 13 | 559 |
| 43 | 43 |
| 1 |
559 = 13 × 43
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Practice Set 12
(i) 25, 40
25 = 5 × 5
40 = 2 × 2 × 2 × 5
Common prime factors: 5. ∴ HCF = 5
(ii) 56, 32
56 = 2 × 2 × 2 × 7
32 = 2 × 2 × 2 × 2 × 2
Common prime factors: 2 × 2 × 2. ∴ HCF = 2 × 2 × 2 = 8
(iii) 40, 60, 75
40 = 2 × 2 × 2 × 5
60 = 2 × 2 × 3 × 5
75 = 3 × 5 × 5
Common prime factors: 5. ∴ HCF = 5
(iv) 16, 27
16 = 2 × 2 × 2 × 2
27 = 3 × 3 × 3
No prime factor is common to all the numbers. ∴ HCF = 1
(v) 18, 32, 48
18 = 2 × 3 × 3
32 = 2 × 2 × 2 × 2 × 2
48 = 2 × 2 × 2 × 2 × 3
Common prime factors: 2. ∴ HCF = 2
(vi) 105, 154
105 = 3 × 5 × 7
154 = 2 × 7 × 11
Common prime factors: 7. ∴ HCF = 7
(vii) 42, 45, 48
42 = 2 × 3 × 7
45 = 3 × 3 × 5
48 = 2 × 2 × 2 × 2 × 3
Common prime factors: 3. ∴ HCF = 3
(viii) 57, 75, 102
57 = 3 × 19
75 = 3 × 5 × 5
102 = 2 × 3 × 17
Common prime factors: 3. ∴ HCF = 3
(ix) 56, 57
56 = 2 × 2 × 2 × 7
57 = 3 × 19
No prime factor is common to all the numbers. ∴ HCF = 1
(x) 777, 315, 588
777 = 3 × 7 × 37
315 = 3 × 3 × 5 × 7
588 = 2 × 2 × 3 × 7 × 7
Common prime factors: 3 × 7. ∴ HCF = 3 × 7 = 21
(i) 275525
The remainder is 0 when the divisor is 25. ∴ HCF = 25
275525 = 275 ÷ 25525 ÷ 25 = 1121
(ii) 76133
The remainder is 0 when the divisor is 19. ∴ HCF = 19
76133 = 76 ÷ 19133 ÷ 19 = 47
(iii) 16169
The remainder is 0 when the divisor is 23. ∴ HCF = 23
16169 = 161 ÷ 2369 ÷ 23 = 73
Each line is one division written as dividend = divisor × quotient + remainder. The next division uses the previous divisor as the dividend and the remainder as the divisor.
Practice Set 13
(i) 12, 15
12 = 2 × 2 × 3
15 = 3 × 5
Taking each prime the greatest number of times it occurs: 2 at most twice, 3 at most once, 5 at most once.
∴ LCM = 2 × 2 × 3 × 5 = 60
(ii) 6, 8, 10
6 = 2 × 3
8 = 2 × 2 × 2
10 = 2 × 5
Taking each prime the greatest number of times it occurs: 2 at most 3 times, 3 at most once, 5 at most once.
∴ LCM = 2 × 2 × 2 × 3 × 5 = 120
(iii) 18, 32
18 = 2 × 3 × 3
32 = 2 × 2 × 2 × 2 × 2
Taking each prime the greatest number of times it occurs: 2 at most 5 times, 3 at most twice.
∴ LCM = 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288
(iv) 10, 15, 20
10 = 2 × 5
15 = 3 × 5
20 = 2 × 2 × 5
Taking each prime the greatest number of times it occurs: 2 at most twice, 3 at most once, 5 at most once.
∴ LCM = 2 × 2 × 3 × 5 = 60
(v) 45, 86
45 = 3 × 3 × 5
86 = 2 × 43
Taking each prime the greatest number of times it occurs: 2 at most once, 3 at most twice, 5 at most once, 43 at most once.
∴ LCM = 2 × 3 × 3 × 5 × 43 = 3870
(vi) 15, 30, 90
15 = 3 × 5
30 = 2 × 3 × 5
90 = 2 × 3 × 3 × 5
Taking each prime the greatest number of times it occurs: 2 at most once, 3 at most twice, 5 at most once.
∴ LCM = 2 × 3 × 3 × 5 = 90
(vii) 105, 195
105 = 3 × 5 × 7
195 = 3 × 5 × 13
Taking each prime the greatest number of times it occurs: 3 at most once, 5 at most once, 7 at most once, 13 at most once.
∴ LCM = 3 × 5 × 7 × 13 = 1365
(viii) 12, 15, 45
12 = 2 × 2 × 3
15 = 3 × 5
45 = 3 × 3 × 5
Taking each prime the greatest number of times it occurs: 2 at most twice, 3 at most twice, 5 at most once.
∴ LCM = 2 × 2 × 3 × 3 × 5 = 180
(ix) 63, 81
63 = 3 × 3 × 7
81 = 3 × 3 × 3 × 3
Taking each prime the greatest number of times it occurs: 3 at most 4 times, 7 at most once.
∴ LCM = 3 × 3 × 3 × 3 × 7 = 567
(x) 18, 36, 27
18 = 2 × 3 × 3
36 = 2 × 2 × 3 × 3
27 = 3 × 3 × 3
Taking each prime the greatest number of times it occurs: 2 at most twice, 3 at most 3 times.
∴ LCM = 2 × 2 × 3 × 3 × 3 = 108
(i) 32, 37
32 = 2 × 2 × 2 × 2 × 2
37 = 37 (prime)
HCF = 1, LCM = 1184
HCF × LCM = 1 × 1184 = 1184
Product of the numbers = 32 × 37 = 1184
∴ HCF × LCM = product of the numbers.
(ii) 46, 51
46 = 2 × 23
51 = 3 × 17
HCF = 1, LCM = 2346
HCF × LCM = 1 × 2346 = 2346
Product of the numbers = 46 × 51 = 2346
∴ HCF × LCM = product of the numbers.
(iii) 15, 60
15 = 3 × 5
60 = 2 × 2 × 3 × 5
HCF = 15, LCM = 60
HCF × LCM = 15 × 60 = 900
Product of the numbers = 15 × 60 = 900
∴ HCF × LCM = product of the numbers.
(iv) 18, 63
18 = 2 × 3 × 3
63 = 3 × 3 × 7
HCF = 9, LCM = 126
HCF × LCM = 9 × 126 = 1134
Product of the numbers = 18 × 63 = 1134
∴ HCF × LCM = product of the numbers.
(v) 78, 104
78 = 2 × 3 × 13
104 = 2 × 2 × 2 × 13
HCF = 26, LCM = 312
HCF × LCM = 26 × 312 = 8112
Product of the numbers = 78 × 104 = 8112
∴ HCF × LCM = product of the numbers.
Practice Set 14
- (i) The HCF of 120 and 150 is — (1) 30
120 = 2 × 2 × 2 × 3 × 5, 150 = 2 × 3 × 5 × 5; HCF = 2 × 3 × 5 = 30. - (ii) The HCF of this pair of numbers is not 1. — (3) 40, 20
20 divides 40, so the HCF is 20. The other pairs (13, 17), (29, 20) and (14, 15) are co-prime.
(i) 14, 28
14 = 2 × 7
28 = 2 × 2 × 7
HCF = 14, LCM = 28
(ii) 32, 16
32 = 2 × 2 × 2 × 2 × 2
16 = 2 × 2 × 2 × 2
HCF = 16, LCM = 32
(iii) 17, 102, 170
17 = 17 (prime)
102 = 2 × 3 × 17
170 = 2 × 5 × 17
HCF = 17, LCM = 510
(iv) 23, 69
23 = 23 (prime)
69 = 3 × 23
HCF = 23, LCM = 69
(v) 21, 49, 84
21 = 3 × 7
49 = 7 × 7
84 = 2 × 2 × 3 × 7
HCF = 7, LCM = 588
In (i), (ii) and (iv) the smaller number divides the larger one, so the smaller number is the HCF and the larger number is the LCM.
(i) 36, 42
36 = 2 × 2 × 3 × 3
42 = 2 × 3 × 7
Taking each prime the greatest number of times it occurs: 2 at most twice, 3 at most twice, 7 at most once.
∴ LCM = 2 × 2 × 3 × 3 × 7 = 252
(ii) 15, 25, 30
15 = 3 × 5
25 = 5 × 5
30 = 2 × 3 × 5
Taking each prime the greatest number of times it occurs: 2 at most once, 3 at most once, 5 at most twice.
∴ LCM = 2 × 3 × 5 × 5 = 150
(iii) 18, 42, 48
18 = 2 × 3 × 3
42 = 2 × 3 × 7
48 = 2 × 2 × 2 × 2 × 3
Taking each prime the greatest number of times it occurs: 2 at most 4 times, 3 at most twice, 7 at most once.
∴ LCM = 2 × 2 × 2 × 2 × 3 × 3 × 7 = 1008
(iv) 4, 12, 20
4 = 2 × 2
12 = 2 × 2 × 3
20 = 2 × 2 × 5
Taking each prime the greatest number of times it occurs: 2 at most twice, 3 at most once, 5 at most once.
∴ LCM = 2 × 2 × 3 × 5 = 60
(v) 24, 40, 80, 120
24 = 2 × 2 × 2 × 3
40 = 2 × 2 × 2 × 5
80 = 2 × 2 × 2 × 2 × 5
120 = 2 × 2 × 2 × 3 × 5
Taking each prime the greatest number of times it occurs: 2 at most 4 times, 3 at most once, 5 at most once.
∴ LCM = 2 × 2 × 2 × 2 × 3 × 5 = 240
8 = 2 × 2 × 2
9 = 3 × 3
10 = 2 × 5
15 = 3 × 5
20 = 2 × 2 × 5
Taking each prime the greatest number of times it occurs: 2 at most 3 times, 3 at most twice, 5 at most once.
∴ LCM = 2 × 2 × 2 × 3 × 3 × 5 = 360
A number that leaves remainder 5 on division by each of these is 5 more than a common multiple of them. The smallest such number is
LCM + remainder = 360 + 5 = 365
(i) 348319
∴ HCF = 29
348319 = 348 ÷ 29319 ÷ 29 = 1211
(ii) 221247
∴ HCF = 13
221247 = 221 ÷ 13247 ÷ 13 = 1719
(iii) 437551
∴ HCF = 19
437551 = 437 ÷ 19551 ÷ 19 = 2329
Product of the two numbers = HCF × LCM
216 × other number = 72 × 432
∴ other number = 72 × 432216 = 31104216 = 144
Check: HCF of 216 and 144 is 72 and their LCM is 432. ✓
HCF × LCM = product of the numbers
3 × LCM = 765
∴ LCM = 7653 = 255
(The two numbers are 15 and 51: 15 × 51 = 765 and their HCF is 3.)
The length of each piece must divide 392, 308 and 490 exactly, and it must be the greatest such length. So we find the HCF.
392 = 2 × 2 × 2 × 7 × 7
308 = 2 × 2 × 7 × 11
490 = 2 × 5 × 7 × 7
Common prime factors: 2 × 7. ∴ HCF = 2 × 7 = 14
The greatest length of each piece is 14 m.
(The bundles give 28, 22 and 35 pieces respectively.)
Two consecutive even numbers can be written as 2n and 2n + 2. Their HCF is always 2.
HCF × LCM = product of the numbers
2 × 180 = 360, so the product of the two numbers is 360.
Consecutive even numbers whose product is 360: 18 × 20 = 360.
∴ The numbers are 18 and 20.
Check: 18 = 2 × 3 × 3, 20 = 2 × 2 × 5; LCM = 2 × 2 × 3 × 3 × 5 = 180. ✓