Common Ground of Numbers: Factors and Multiples — Questions and Answers
Welcome to HSLC Guru. In this lesson you will find the concepts of common factors, prime factorisation, factor trees, the Highest Common Factor (HCF), the Least Common Multiple (LCM) and co-prime numbers from ASSEB Class 7 New Mathematics Chapter 11 (Common Ground of Numbers: Factors and Multiples), together with fully worked answers to every “Work it Out” box and all of Exercise 11.
Summary
In this chapter we revise factors and multiples and learn to find the common factors, the Highest Common Factor (HCF) and the Least Common Multiple (LCM) of two or more numbers. A number that divides another exactly is its factor, and a number obtained by multiplying it by a whole number is its multiple.
The HCF is the greatest of the common factors, and the LCM is the smallest of the common multiples. Both can be found by prime factorisation, factor trees or the short-division method. For the HCF we take the lowest power of each common prime factor, and for the LCM we take the highest power of every prime factor involved.
HCF and LCM are used in many real-life problems — laying tiles, packing equal groups, ringing bells and blinking lights. For any two numbers there is a neat relationship: “LCM × HCF = product of the two numbers.” Two co-prime numbers have HCF 1 and an LCM equal to their product. We also meet two new ideas of mathematics — conjecture and generalisation.
Summary: This ASSEB Class 7 New Mathematics Chapter 11 (Common Ground of Numbers: Factors and Multiples) guide explains common factors, prime factorisation, factor trees, the short-division method, the HCF, the LCM, co-prime numbers, and the relation “LCM × HCF = product of the two numbers.” It gives fully worked solutions to every “Work it Out” box (11.1–11.6) and all of Exercise 11.
Key Concepts
Factor tree: A factor tree splits a number again and again into two factors until only prime numbers remain. The tree below shows the factor tree of 3825 (used in Work it Out 11.1, question 5).
Here 3825 = 3 × 1275, 1275 = 3 × 425, 425 = 5 × 85 and 85 = 5 × 17. So 3825 = 3 × 3 × 5 × 5 × 17, giving x = 3825, y = 5 and z = 17.
HCF and LCM with a Venn diagram: If the prime factors of two numbers are arranged in a Venn diagram, the product of the overlapping part gives the HCF and the product of everything gives the LCM. Below is the diagram for 18 = 2 × 3 × 3 and 24 = 2 × 2 × 2 × 3.
Overlapping part = 2, 3, so HCF = 2 × 3 = 6. Product of all factors = 3 × 2 × 3 × 2 × 2 = 72, so LCM = 72.
Textbook Questions and Answers
In-text Examples and Questions
1. How many prime factors are there in 720?
Answer: 720 = 2 × 2 × 2 × 2 × 3 × 3 × 5. So 720 has 7 prime factors (2 four times, 3 twice and 5 once).
2. What should be multiplied with 20 to get 180? (the teacher’s question to Nahid)
Answer: 180 = 2 × 2 × 3 × 3 × 5 = (2 × 2 × 5) × (3 × 3) = 20 × 9. So multiplying 20 by 9 gives 180; hence 20 is a factor of 180.
3. Determine the total number of factors in the prime factorisation of 96.
Answer: 96 = 2 × 2 × 2 × 2 × 2 × 3. Its factors are 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48 and 96 — so 96 has 12 factors.
4. Do the prime factorisation of 80, 120, 360 and 720 and state the total number of factors of each.
Answer: 80 = 2 × 2 × 2 × 2 × 5 (10 factors); 120 = 2 × 2 × 2 × 3 × 5 (16 factors); 360 = 2 × 2 × 2 × 3 × 3 × 5 (24 factors); 720 = 2 × 2 × 2 × 2 × 3 × 3 × 5 (30 factors).
5. Does a larger number always have more factors? (Nahid’s conjecture)
Answer: No. Since 943 = 23 × 41, the factors of 943 are only 1, 23, 41 and 943 — just 4 of them. So a large number does not necessarily have more factors; hence Nahid’s conjecture is wrong. Note that however large a prime number is, it has only two factors — 1 and the number itself.
6. Using the table method, find the HCF of 72 and 120.
Answer: Factors of 72: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72; factors of 120: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120. Common factors = 1, 2, 3, 4, 6, 8, 12, 24; so HCF = 24.
7. Find the sub-components of the factors of 63, and among 3 × 3 × 7 × 3 × 5 × 7, 3 × 3 × 5 × 7 and 3 × 7, which is a common factor of 63 and 105?
Answer: 63 = 3 × 3 × 7. Its sub-components — one at a time: 3, 7; two at a time: 3 × 3, 3 × 7; three at a time: 3 × 3 × 7. A common factor must be a sub-component of the factors of each number. Both 3 × 3 × 7 × 3 × 5 × 7 and 3 × 3 × 5 × 7 (= 315) are larger than 63, so they are not common factors. But 3 × 7 = 21 is a sub-component of both, so it is a common factor (and it is the HCF).
8. Is each of 210, 336 and 462 divisible by their HCF 42?
Answer: 210 ÷ 42 = 5, 336 ÷ 42 = 8, 462 ÷ 42 = 11 — all leave remainder 0. So all three are exactly divisible by 42; that is, two or more numbers are always completely divisible by their HCF.
9. What is the greatest common multiple of 4 and 6?
Answer: There is none. The multiples of any number continue endlessly (12, 24, 36, 48, …), so a greatest common multiple cannot be found. Only the least common multiple can be found — the LCM of 4 and 6 is 12.
10. Of 2 × 2 × 3 × 5 and 2 × 3 × 5, which is a common multiple of 12 and 20?
Answer: 2 × 2 × 3 × 5 = 60 contains all the prime factors of both 12 = 2 × 2 × 3 and 20 = 2 × 2 × 5, so it is a common multiple (and it is the LCM, 60). But 2 × 3 × 5 = 30 does not contain both 2’s of 12, so it is not divisible by 12 — not a common multiple.
11. Is the LCM 720 of 36, 48 and 90 divisible by all three numbers?
Answer: 720 ÷ 36 = 20, 720 ÷ 48 = 15, 720 ÷ 90 = 8 — all leave remainder 0. So the LCM of any set of numbers is always divisible by those numbers.
12. A box has some eggs. Dividing them into groups of 3 leaves 1 egg; dividing into groups of 4 also leaves 1 egg. What is the minimum number of eggs?
Answer: Numbers divisible by both 3 and 4 are their common multiples; adding 1 gives 13, 25, 37, … Each of these is 1 more than a multiple of the LCM of 3 and 4, which is 12. So the number of eggs = (LCM of 3 and 4) × n + 1 = 12n + 1. For the minimum, n = 1, so the minimum number of eggs is 13.
Work it Out 11.1
1. Find the common factors of the following: (a) 12, 16 (b) 24, 36 (c) 13, 19 (d) 1, 17 (e) 34, 51 (f) 6, 12, 24
Answer:
- (a) 12, 16: factors of 12 = 1, 2, 3, 4, 6, 12; factors of 16 = 1, 2, 4, 8, 16. Common = 1, 2, 4.
- (b) 24, 36: factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24; factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18, 36. Common = 1, 2, 3, 4, 6, 12.
- (c) 13, 19: both prime. Common = 1.
- (d) 1, 17: factors of 1 = 1; factors of 17 = 1, 17. Common = 1.
- (e) 34, 51: 34 = 2 × 17, 51 = 3 × 17. Common = 1, 17.
- (f) 6, 12, 24: factors of 6 = 1, 2, 3, 6, all of which divide 12 and 24. Common = 1, 2, 3, 6.
2. Factorise by the prime factorisation method: (a) 72 (b) 144 (c) 210 (d) 2005 (e) 480 (f) 1728
Answer: (a) 72 = 2 × 2 × 2 × 3 × 3; (b) 144 = 2 × 2 × 2 × 2 × 3 × 3; (c) 210 = 2 × 3 × 5 × 7; (d) 2005 = 5 × 401 (401 is prime); (e) 480 = 2 × 2 × 2 × 2 × 2 × 3 × 5; (f) 1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3.
3. Factorise by using the factor tree method: (a) 100 (b) 210 (c) 1024 (d) 1331
Answer: Split each number repeatedly into two factors until only primes remain. (a) 100 = 2 × 2 × 5 × 5; (b) 210 = 2 × 3 × 5 × 7; (c) 1024 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 (ten 2’s); (d) 1331 = 11 × 11 × 11.
4. Factorise by using the short division method: (a) 396 (b) 462 (c) 796 (d) 1016
Answer: Divide successively by the smallest primes. (a) 396 = 2 × 2 × 3 × 3 × 11; (b) 462 = 2 × 3 × 7 × 11; (c) 796 = 2 × 2 × 199 (199 is prime); (d) 1016 = 2 × 2 × 2 × 127 (127 is prime).
5. Answer the questions using the adjacent factor tree: (i) value of x? (ii) value of y? (iii) value of z? (iv) value of x + y + z? (See the factor tree in the Key Concepts section.)
Answer: In the tree, 3825 = 3 × 1275, 1275 = 3 × 425, 425 = 5 × 85, 85 = 5 × 17. So — (i) x = 3825 (option B); (ii) y = 5 (option A); (iii) z = 17 (option B); (iv) x + y + z = 3825 + 5 + 17 = 3847 (option B).
Work it Out 11.2
1. Find the HCF using prime factorisation: (a) 72, 96 (b) 210, 480 (c) 1728, 1296 (d) 56, 96, 144 (e) 720, 480, 2005
Answer:
- (a) 72, 96: 72 = 2 × 2 × 2 × 3 × 3, 96 = 2 × 2 × 2 × 2 × 2 × 3. Common = 2 × 2 × 2 × 3. HCF = 24.
- (b) 210, 480: 210 = 2 × 3 × 5 × 7, 480 = 2 × 2 × 2 × 2 × 2 × 3 × 5. Common = 2 × 3 × 5. HCF = 30.
- (c) 1728, 1296: 1728 = 2⁶ × 3³, 1296 = 2⁴ × 3⁴. Common = 2⁴ × 3³ = 16 × 27. HCF = 432.
- (d) 56, 96, 144: 56 = 2³ × 7, 96 = 2⁵ × 3, 144 = 2⁴ × 3². Only common prime is 2, lowest power 2³. HCF = 8.
- (e) 720, 480, 2005: 720 = 2⁴ × 3² × 5, 480 = 2⁵ × 3 × 5, 2005 = 5 × 401. Only common prime is 5. HCF = 5.
2. Find the HCF using the short division method: (a) 228, 560 (b) 132, 720 (c) 101, 288 (d) 72, 210, 462 (e) 700, 462, 1024
Answer:
- (a) 228, 560: common factors 2 × 2. HCF = 4.
- (b) 132, 720: common factors 2 × 2 × 3. HCF = 12.
- (c) 101, 288: 101 is prime and not a factor of 288, so no common factor other than 1. HCF = 1.
- (d) 72, 210, 462: common factors 2 × 3. HCF = 6.
- (e) 700, 462, 1024: only common factor is 2. HCF = 2.
Work it Out 11.3
1. Find the LCM by prime factorisation: (a) 34, 36 (b) 36, 102 (c) 45, 75 (d) 75, 300 (e) 91, 112 (f) 12, 16, 18 (g) 25, 95, 105 (h) 180, 264, 252 (i) 18, 45, 60, 100
Answer: For the LCM take the highest power of every prime factor.
- (a) 34, 36: 34 = 2 × 17, 36 = 2² × 3². LCM = 2² × 3² × 17 = 612.
- (b) 36, 102: 36 = 2² × 3², 102 = 2 × 3 × 17. LCM = 2² × 3² × 17 = 612.
- (c) 45, 75: 45 = 3² × 5, 75 = 3 × 5². LCM = 3² × 5² = 225.
- (d) 75, 300: 75 = 3 × 5², 300 = 2² × 3 × 5². LCM = 2² × 3 × 5² = 300.
- (e) 91, 112: 91 = 7 × 13, 112 = 2⁴ × 7. LCM = 2⁴ × 7 × 13 = 1456.
- (f) 12, 16, 18: 12 = 2² × 3, 16 = 2⁴, 18 = 2 × 3². LCM = 2⁴ × 3² = 144.
- (g) 25, 95, 105: 25 = 5², 95 = 5 × 19, 105 = 3 × 5 × 7. LCM = 3 × 5² × 7 × 19 = 9975.
- (h) 180, 264, 252: 180 = 2² × 3² × 5, 264 = 2³ × 3 × 11, 252 = 2² × 3² × 7. LCM = 2³ × 3² × 5 × 7 × 11 = 27720.
- (i) 18, 45, 60, 100: 18 = 2 × 3², 45 = 3² × 5, 60 = 2² × 3 × 5, 100 = 2² × 5². LCM = 2² × 3² × 5² = 900.
2. Decide whether each statement is true or false and choose the correct option: (P) In finding the LCM of 95 and 105, 19 appears once. (Q) 18 and 25 have no common factor other than 1. (A) P true, Q false (B) P false, Q true (C) Both P and Q true (D) Both P and Q false
Answer: 95 = 5 × 19, 105 = 3 × 5 × 7; LCM = 3 × 5 × 7 × 19 = 1995, in which 19 appears once — so P is true. 18 = 2 × 3 × 3 and 25 = 5 × 5 are co-prime, so their only common factor is 1 — Q is also true. Correct option: (C) Both P and Q true.
Work it Out 11.4
1. Find the LCM using the short division method: (a) 12, 60 (b) 78, 102 (c) 150, 205 (d) 133, 1197 (e) 12, 16, 32 (f) 60, 85, 35 (g) 91, 112, 49 (h) 131, 655, 393
Answer:
- (a) 12, 60: 60 is a multiple of 12, so LCM = 60.
- (b) 78, 102: 78 = 2 × 3 × 13, 102 = 2 × 3 × 17. LCM = 2 × 3 × 13 × 17 = 1326.
- (c) 150, 205: 150 = 2 × 3 × 5², 205 = 5 × 41. LCM = 2 × 3 × 5² × 41 = 6150.
- (d) 133, 1197: 1197 = 133 × 9, a multiple of 133. LCM = 1197.
- (e) 12, 16, 32: 12 = 2² × 3, 16 = 2⁴, 32 = 2⁵. LCM = 2⁵ × 3 = 96.
- (f) 60, 85, 35: 60 = 2² × 3 × 5, 85 = 5 × 17, 35 = 5 × 7. LCM = 2² × 3 × 5 × 7 × 17 = 7140.
- (g) 91, 112, 49: 91 = 7 × 13, 112 = 2⁴ × 7, 49 = 7². LCM = 2⁴ × 7² × 13 = 10192.
- (h) 131, 655, 393: 131 is prime, 655 = 5 × 131, 393 = 3 × 131. LCM = 3 × 5 × 131 = 1965.
2. Choose the correct statement: (i) LCM of 131 and 393 is 131. (ii) LCM of 20, 25 and 30 is 300. (iii) If the LCM of 12, 16 and 32 is 96, then 96 is divisible by 16. (A) (i), (ii) (B) (ii), (iii) (C) (i), (iii) (D) (i), (ii), (iii)
Answer: (i) is false — since 393 = 3 × 131, the LCM of 131 and 393 is 393, not 131. (ii) is true — LCM of 20, 25, 30 = 2² × 3 × 5² = 300. (iii) is true — 96 ÷ 16 = 6, so 96 is divisible by 16. Correct option: (B) (ii), (iii).
Work it Out 11.5
1. Complete the table using the formula “LCM × HCF = product of the two numbers.”
Answer: The missing value in each row = (LCM × HCF) ÷ the known number. The completed table is —
| Sl. No. | 1st No. | 2nd No. | LCM | HCF |
|---|---|---|---|---|
| 1 | 97 | 39 | 3783 | 1 |
| 2 | 6 | 33 | 66 | 3 |
| 3 | 68 | 8 | 136 | 4 |
| 4 | 8 | 28 | 56 | 4 |
| 5 | 189 | 54 | 378 | 27 |
| 6 | 208 | 32 | 416 | 16 |
| 7 | 510 | 60 | 1020 | 30 |
| 8 | 46 | 161 | 322 | 23 |
| 9 | 517 | 94 | 1034 | 47 |
| 10 | 77 | 539 | 539 | 77 |
2. Match the values from the columns and choose the correct answer. Column I (Numbers): (P) 6, 20 (Q) 12, 15, 21 (R) 17, 23, 29 (S) 8, 9, 25; Column II (HCF): (a) 1 (b) 1 (c) 2 (d) 3; Column III (LCM): (i) 420 (ii) 60 (iii) 1800 (iv) 11339.
Answer: P (6, 20): HCF = 2, LCM = 60 → c, (ii). Q (12, 15, 21): HCF = 3, LCM = 420 → d, (i). R (17, 23, 29): all prime, HCF = 1, LCM = 11339 → a, (iv). S (8, 9, 25): co-prime, HCF = 1, LCM = 1800 → b, (iii). Correct option: (D) P → c → (ii), Q → d → (i), R → a → (iv), S → b → (iii).
Work it Out 11.6
1. In which of the following cases can a general statement be formed regarding HCF? Explain. (a) two consecutive odd numbers (b) two consecutive numbers (c) two co-prime numbers (d) two prime numbers (e) two even numbers (f) two multiples of 3
Answer: A definite general statement can be formed for (a), (b), (c) and (d), because in each case the HCF = 1 — (a) two consecutive odd numbers differ by 2 and both are odd, so 2 cannot be a common factor; (b) two consecutive numbers differ by 1; (c) co-prime numbers are defined to have HCF 1; (d) two different primes share only 1. But for (e) two even numbers and (f) two multiples of 3, the exact HCF cannot be stated (it varies) — we can only say the HCF is at least 2 in (e) and at least 3 in (f).
2. In which of the following cases can a general statement be formed regarding LCM? Explain. (a) two consecutive numbers (b) two co-prime numbers (c) two prime numbers (d) two even numbers (e) two multiples of 3
Answer: A definite general statement can be formed for (a), (b) and (c) — in all three the HCF is 1, so LCM = product of the two numbers. But for (d) two even numbers and (e) two multiples of 3, the HCF is not 1 and it varies, so no fixed general formula for the LCM can be formed.
Exercise 11
1. Find the smallest number that is divisible by both 12 and 18.
Answer: Required number = LCM of 12 and 18. 12 = 2² × 3, 18 = 2 × 3²; LCM = 2² × 3² = 36.
2. Among how many maximum people can 24 sweets and 4 chocolates be distributed equally?
Answer: Maximum number of people = HCF of 24 and 4 = 4. Each person gets 6 sweets and 1 chocolate.
3. A number of mangoes in a basket can be divided equally into groups of 5, 6 and 9. What is the minimum number of mangoes?
Answer: Minimum number = LCM of 5, 6 and 9. 5, 6 = 2 × 3, 9 = 3²; LCM = 2 × 3² × 5 = 90 mangoes.
4. Find these values and arrange them in ascending order: (P) LCM of 10 and 9 (Q) HCF of 150 and 185 (R) LCM of 10 and 15 (S) HCF of 128 and 144. (A) S, R, Q, P (B) P, Q, R, S (C) Q, S, R, P (D) R, P, Q, S
Answer: P = 10 × 9 = 90 (co-prime); Q = HCF(150, 185) = 5; R = LCM(10, 15) = 30; S = HCF(128, 144) = 16. Ascending order: 5 < 16 < 30 < 90, i.e. Q, S, R, P. Correct option: (C) Q, S, R, P.
5. Three bells ring at intervals of 30, 45 and 60 minutes. If they ring together once, after how much time will they ring together again?
Answer: Time = LCM of 30, 45 and 60 = 2² × 3² × 5 = 180 minutes = 3 hours.
6. Haradhan wants to pack 210 oranges, 252 apples and 294 guavas into small boxes with an equal number of each type in every box. What is the maximum number of fruits he can put in each box?
Answer: Maximum = HCF of 210, 252 and 294. 210 = 2 × 3 × 5 × 7, 252 = 2² × 3² × 7, 294 = 2 × 3 × 7²; HCF = 2 × 3 × 7 = 42 fruits per box.
7. Match the statements in the columns. (P) HCF of two co-prime numbers — 1. 29 (Q) LCM of 20, 30 and 40 — 2. 1 (R) HCF of 638 and 783 — 3. 18 (S) 18 is a multiple of 6, so LCM of 6 and 18 — 4. 120. Options: (A) P-4, Q-3, R-2, S-1 (B) P-3, Q-4, R-1, S-2 (C) P-2, Q-4, R-1, S-3 (D) P-1, Q-2, R-3, S-4
Answer: P = 1 (HCF of co-primes) → 2; Q = LCM(20, 30, 40) = 120 → 4; R = HCF(638, 783) = 29 → 1 (638 = 2 × 11 × 29, 783 = 3³ × 29); S = LCM(6, 18) = 18 → 3. Correct option: (C) P-2, Q-4, R-1, S-3.
8. Mukul plants 45 brinjal, 81 chili and 63 tomato seedlings so that each row has an equal number of only one type. What is the maximum number of seedlings in one row?
Answer: Maximum = HCF of 45, 81 and 63. 45 = 3² × 5, 81 = 3⁴, 63 = 3² × 7; HCF = 3² = 9 seedlings per row.
9. A tailor cuts equal-length pieces from ribbons of 72 cm, 90 cm and 108 cm with none left over. What is the maximum length of each piece?
Answer: Maximum length = HCF of 72, 90 and 108. 72 = 2³ × 3², 90 = 2 × 3² × 5, 108 = 2² × 3³; HCF = 2 × 3² = 18 cm.
10. Assertion (A): If the product of the LCM and HCF of 30 and another number is 720, then the other number is 24. Reason (R): For two numbers, LCM × HCF = product of the two numbers.
Answer: LCM × HCF = 30 × (other number) = 720, so the other number = 720 ÷ 30 = 24 — Assertion is true. Reason is also true and correctly explains it. So (A) Both are true and R is the correct explanation of A.
11. Find the smallest 4-digit number divisible by 18, 24 and 32.
Answer: First, LCM(18, 24, 32) = 2⁵ × 3² = 288. 288 × 3 = 864 (3-digit), 288 × 4 = 1152 (4-digit). So the smallest 4-digit number is 1152.
12. Three friends’ steps measure 63 cm, 70 cm and 77 cm. What is the minimum distance they must cover to meet again at the same spot?
Answer: Minimum distance = LCM of 63, 70 and 77. 63 = 3² × 7, 70 = 2 × 5 × 7, 77 = 7 × 11; LCM = 2 × 3² × 5 × 7 × 11 = 6930 cm (= 69.3 m).
13. What is the largest number that divides 35, 45 and 50 without leaving a remainder?
Answer: Required number = HCF of 35, 45 and 50. 35 = 5 × 7, 45 = 3² × 5, 50 = 2 × 5²; HCF = 5.
14. The capacities of three milk containers are 180 L, 264 L and 252 L. What is the maximum capacity of a measuring vessel that can measure all three exactly?
Answer: Maximum capacity = HCF of 180, 264 and 252. 180 = 2² × 3² × 5, 264 = 2³ × 3 × 11, 252 = 2² × 3² × 7; HCF = 2² × 3 = 12 L.
15. Biradatta, Farhan and Ratul complete one circular track in 12, 15 and 18 minutes respectively. For how many hours must they cycle to reach the start together?
Answer: Time = LCM of 12, 15 and 18 = 2² × 3² × 5 = 180 minutes = 3 hours.
16. Bithika has 18 pencils and 36 erasers. Each set must have an equal number of pencils and an equal number of erasers. What is the maximum number of sets she can prepare?
Answer: Maximum number of sets = HCF of 18 and 36 = 18 sets (each with 1 pencil and 2 erasers).
17. Assertion (A): If the LCM of two numbers is 60 and the HCF is 6, then the numbers are 24 and 15. Reason (R): The LCM of two numbers is always divisible by both numbers.
Answer: For 24 and 15, HCF = 3 (not 6) and LCM = 120 (not 60), so Assertion is false. But the Reason is true (the LCM is always divisible by both numbers). So (D) Assertion (A) is false but Reason (R) is true.
18. Red lights blink every 4 seconds and green lights every 6 seconds, starting together. In 60 seconds, how many times will both blink at the same time?
Answer: Both blink together every LCM(4, 6) = 12 seconds, i.e. at 12, 24, 36, 48 and 60 seconds = 5 times.
19. Ankur leads exercise every 3 days and Ritu every 2 days. Today both led together. In the next 30 days, on how many days will they lead together?
Answer: Both lead together every LCM(2, 3) = 6 days, i.e. on days 6, 12, 18, 24 and 30 = 5 days.
Appendix: Finding the Number of Factors
The total number of factors of a number can be found from its prime factorisation. If a number’s prime factorisation is $N = a^{p} \times b^{q} \times c^{r}$ (where a, b, c are primes), then the total number of factors of N = $(p+1)(q+1)(r+1)$. For example, 72 = 2³ × 3², so its total number of factors = (3 + 1)(2 + 1) = 12.
Additional Questions and Answers
Multiple Choice Questions (MCQ)
1. The prime factorisation of 24 is — (a) 2 × 2 × 6 (b) 2 × 2 × 2 × 3 (c) 4 × 6 (d) 3 × 8
Answer: (b) 2 × 2 × 2 × 3.
2. The HCF of two co-prime numbers is — (a) 0 (b) 1 (c) their product (d) 2
Answer: (b) 1.
3. The LCM of 12 and 18 is — (a) 6 (b) 36 (c) 72 (d) 216
Answer: (b) 36.
4. For any number n and its multiple 5n, the HCF is — (a) 5n (b) 5 (c) n (d) 1
Answer: (c) n.
5. If the LCM of two numbers is 60 and the HCF is 5, then the product of the two numbers is — (a) 12 (b) 65 (c) 300 (d) 55
Answer: (c) 300 (LCM × HCF = 60 × 5 = 300).
6. The LCM of 1 and any number n is — (a) 1 (b) n (c) n + 1 (d) 0
Answer: (b) n.
7. Which of the following is a prime number? (a) 91 (b) 51 (c) 101 (d) 87
Answer: (c) 101 (91 = 7 × 13, 51 = 3 × 17, 87 = 3 × 29).
8. The HCF of two consecutive numbers is always — (a) 2 (b) 1 (c) their sum (d) cannot be determined
Answer: (b) 1.
9. From the prime factorisation of 60, its total number of factors is — (a) 6 (b) 8 (c) 10 (d) 12
Answer: (d) 12 [60 = 2² × 3 × 5, factors = (2+1)(1+1)(1+1) = 12].
10. A mathematical statement made only by observation or intuition, without proof, is called a — (a) generalisation (b) conjecture (c) theorem (d) formula
Answer: (b) conjecture.
Fill in the Blanks
1. The smallest common multiple divisible by each number is called the ______.
Answer: Least Common Multiple (LCM).
2. The greatest of the common factors of two or more numbers is called the ______.
Answer: Highest Common Factor (HCF).
3. Two numbers whose HCF is 1 are called ______ numbers.
Answer: co-prime.
4. LCM × HCF of two numbers = ______ of the two numbers.
Answer: product.
5. Any prime number has exactly ______ factors.
Answer: two (1 and the number itself).
True or False
1. 1 is a factor of every number.
Answer: True.
2. The HCF of two numbers is always greater than their LCM.
Answer: False (the HCF is always less than or equal to the LCM).
3. Two consecutive numbers are always co-prime.
Answer: True.
4. The LCM of any set of numbers is always divisible by each of those numbers.
Answer: True.
5. For three or more numbers, “LCM × HCF = product” is always true.
Answer: False (it is not always true for three or more numbers).
Short Answer Questions
1. What is the difference between a factor and a multiple?
Answer: A number that divides another exactly is its factor (e.g. 3 is a factor of 12), while a number obtained by multiplying it by a whole number is its multiple (e.g. 12, 24, 36 are multiples of 12). A factor is less than or equal to the number, whereas a multiple is greater than or equal to it.
2. Find the HCF and LCM of 36 and 48 and verify that LCM × HCF = product.
Answer: 36 = 2² × 3², 48 = 2⁴ × 3. HCF = 2² × 3 = 12; LCM = 2⁴ × 3² = 144. LCM × HCF = 144 × 12 = 1728, and 36 × 48 = 1728 — so the relationship holds.
3. What are co-prime numbers? Give one example.
Answer: Two numbers that have no common factor other than 1 — that is, whose HCF is 1 — are called co-prime numbers. For example, 8 and 15 are co-prime (HCF = 1) even though neither is a prime number.
4. Find the LCM of 24 and 36 by the short division method.
Answer: Dividing by common primes: 2 | 24, 36 → 2 | 12, 18 → 3 | 6, 9 → 2, 3. LCM = 2 × 2 × 3 × 2 × 3 = 72.
Key Terms
| Term | Meaning |
|---|---|
| Factor | A number that divides another number exactly |
| Multiple | A number obtained by multiplying a number by a whole number |
| Prime factor | A factor that is a prime number |
| Prime factorisation | Expressing a number as a product of its prime factors |
| Common factor | A factor shared by two or more numbers |
| Highest Common Factor (HCF/GCD) | The greatest of the common factors of two or more numbers |
| Least Common Multiple (LCM) | The smallest of the common multiples of two or more numbers |
| Factor tree | A tree diagram used to find the prime factors of a number |
| Short division method | Finding HCF/LCM by dividing by common primes |
| Co-prime numbers | Two numbers whose HCF is 1 |
| Conjecture | A mathematical statement not yet proved true or false |
| Generalisation | Turning a particular result into a general rule |