Page No. 51
Figure it Out
Q1: List all the factors of the following numbers:
(a) 90
(b) 105
(c) 132
(d) 360 (this number has 24 factors)
(e) 840 (this number has 32 factors)
Ans:
(a) 90
90=2×3×3×5
Therefore, the factors of 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45 and 90.
Hence, 90 has 12 factors.
(b) 105
105=3×5×7
Therefore, the factors of 105 are 1, 3, 5, 7, 15, 21, 35 and 105.
Hence, 105 has 8 factors.
(c) 132
132=2×2×3×11
Therefore, the factors of 132 are 1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66 and 132.
Hence, 132 has 12 factors.
(d) 360
360=2×2×2×3×3×5
Therefore, the factors of 360 are 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180 and 360.
Hence, 360 has 24 factors.
(e) 840
840=2×2×2×3×5×7
Therefore, the factors of 840 are 1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 15, 20, 21, 24, 28, 30, 35, 40, 42, 56, 60, 70, 84, 105, 120, 140, 168, 210, 280, 420 and 840.
Hence, 840 has 32 factors.
Page No. 53
Figure it Out
Q1:Find the common factors and the HCF of the following numbers:
(a) 50, 60
(b) 140, 275
(c) 77, 725
(d) 370, 592
(e) 81, 243
Ans:
(a) Here, 50
50=2×5×5 and 60=2×2×3×5.
The common factors of 50 and 60 are 1, 2, 5 and 10.
Therefore, HCF(50,60)=2×5=10.
(b)
140=2×2×5×7 and 275=5×5×11.
The only common prime factor is 5. Thus, the common factor is 5 and HCF(140,275)=5.
(c) Here, 77 and 725
77=7×11 and 725=5×5×29.
There is no common prime factor. Therefore, the only common factor is 1 and HCF(77,725)=1.
(d) Here, 370 and 592
370=2×5×37 and 592=2×2×2×2×37.
The common factors are 1, 2, 37 and 74.
Therefore, HCF(370,592)=2×37=74.
(e) Here, 81 and 243
81=3×3×3×3 and 243=3×3×3×3×3.
The common factors include 1,3,9,27 and 81. Therefore, the common prime factors are 3×3×3×3, and HCF(81,243)=81.
Page No. 54
Figure it Out
Q1:Find the HCF of the following numbers:
(a) 24, 180
(b) 42, 75, 24
(c) 240, 378
(d) 400, 2500
(e) 300, 800
Ans:
(a) Given, 24, 180
The common prime factors are two 2s and one 3.
HCF(24,180)=2×2×3=4×3=12.
(b) Given, 42, 75, 24
The only prime factor common to 42, 75 and 24 is 3.
Therefore, HCF(42,75,24)=3.
(c) Given, 240, 378
The common prime factors are 2 and 3.
Therefore, HCF(240,378)=2×3=6.
(d) Here 400 and 2500
HCF(400,2500)=2×2×5×5=100.
(e) Here, 300, 800
HCF(300,800)=2×2×5×5=100.
Q2: Consider the numbers 72 and 144. Suppose they are factorised into composite numbers as: 72 = 6 × 12 and 144 = 8 × 18. Seeing this, can one say that these two numbers have no common factor other than 1? Why not?
Ans: No, one cannot say that 72 and 144 have no common factor other than 1 because these are factorisations into composite numbers, not prime factorisations.
Here, 72=6×12 and 144=8×18.
Both 6 and 12 are composite, and 8 and 18 are also composite.
The prime factorisations are 72=2×2×2×3×3 and 144=2×2×2×2×3×3.
Thus, HCF(72,144)=2×2×2×3×3=8×9=72.
Since the HCF is 72, which is greater than 1, the numbers have common factors other than 1.
Page No. 58
Figure it Out
Q1: Find the LCM of the following numbers:
(a) 30, 72
(b) 36, 54
(c) 105, 195, 65
(d) 222, 370
Ans:
(a) 30, 72
30=2×3×5
72=2×2×2×3×3
LCM(30,72)=2×2×2×3×3×5=8×9×5=360.
(b) 36, 54
36=2×2×3×3
54=2×3×3×3
LCM(36,54)=2×2×3×3×3=4×27=108.
(c) 105, 195, 65
105=3×5×7
195=3×5×13
65=5×13
LCM(105,195,65)=3×5×7×13=15×91=1365.
(d) 222, 370
222=2×3×37 and 370=2×5×37.
Therefore, LCM(222,370)=2×3×5×37=1110.
Page No. 59
Figure it Out
Q1:Make a general statement about the HCF for the following pairs of numbers. You could consider examples before coming up with general statements. Look for possible explanations of why they hold.
(a) Two consecutive even numbers
(b) Two consecutive odd numbers
(c) Two even numbers
(d) Two consecutive numbers
(e) Two co-prime numbers
Share your observations with the class.
Ans: (a) Two Consecutive Even Numbers
Examples: (2,4), (6,8) and (10,12).
The HCF in each case is 2.
Therefore, the HCF of any two consecutive even numbers is 2. Every even number is divisible by 2, and two consecutive even numbers differ by 2, so they cannot have another common factor.
(b) Two Consecutive Odd Numbers
Examples: (3,5), (7,9) and (11,13).
The HCF in each case is 1.
Therefore, any two consecutive odd numbers are co-prime and have HCF 1.
(c) Two Even Numbers
Examples: (4,10) has HCF 2, (8,12) has HCF 4, and (14,20) has HCF 2.
Therefore, the HCF of two even numbers is always even because 2 is a factor of both numbers. It may be 2 or a higher multiple of 2.
(d) Two Consecutive Numbers
Examples: (7,8), (14,15) and (20,21).
The HCF in each case is 1.
Therefore, the HCF of any two consecutive numbers is 1. Consecutive numbers differ by 1 and cannot have a common factor greater than 1.
(e) Two Co-prime Numbers
Examples: (4,9), (5,8) and (7,10).
The HCF in each case is 1.
Therefore, the HCF of two co-prime numbers is always 1 because co-prime numbers have no common factor other than 1.
Q2: The LCM of 3 and 24 is 24 (it is one of the two given numbers).
(a) Find more such number pairs where the LCM is one of the two numbers.
(b) Make a general statement about such numbers. Describe such number pairs using algebra.
Ans: (a) Some such pairs are given below.
(i) For 2 and 4, LCM(2,4)=4, because 4 is a multiple of 2.
(ii) For 5 and 10, LCM(5,10)=10, because 10 is a multiple of 5.
(iii) For 6 and 12, LCM(6,12)=12, because 12 is a multiple of 6.
(iv) For 7 and 49, LCM(7,49)=49, because 49 is a multiple of 7.
(v) For 10 and 100, LCM(10,100)=100, because 100 is a multiple of 10.
(b) Let the two positive integers be a and b, where a<b. The LCM is one of the given numbers, namely the larger number b, if and only if the smaller number a is a factor of b.
For example, LCM(3,24)=24 because 3 is a factor of 24 and 24÷3=8.
Algebraically, LCM(a,b)=b if and only if b=k×a, where k is a positive integer.
Q3: Make a general statement about the LCM for the following pairs of numbers. You could consider examples before coming up with these general statements. Look for possible explanations of why they hold.
(a) Two multiples of 3
(b) Two consecutive even numbers
(c) Two consecutive numbers
(d) Two co-prime numbers
Ans: (a) Two Multiples of 3
Examples: (6,9), (9,12) and (12,18). Their LCMs are 18, 36 and 36 respectively.
Therefore, the LCM of two multiples of 3 is always a multiple of 3 because both numbers are divisible by 3.
(b) Two Consecutive Even Numbers
Examples: (2,4), (6,8) and (10,12). Their LCMs are 4, 24 and 60 respectively.
Two consecutive even numbers have 2 as their only common factor. Therefore, their LCM is half their product.
LCM(2n,2n+2)=(2n)(2n+2)2=n(2n+2)=2n2+2n
(c) Two Consecutive Numbers
Examples: (7,8), (9,10) and (10,11). Their LCMs are 56, 90 and 110 respectively.
Consecutive numbers have no common factor other than 1. Therefore, the LCM of two consecutive numbers is their product.
(d) Two Co-prime Numbers
Examples: (4,9), (5,8) and (7,10). Their LCMs are 36, 40 and 70 respectively.
Co-prime numbers have no common factor other than 1. Therefore, the LCM of two co-prime numbers is equal to their product.
Page No. 62
Q: Explore whether the LCM is a factor of the product in the following cases. If yes, identify the number that the LCM should be multiplied by to get the product. Do you see any pattern? Use these numbers:
(a) 45, 105
(b) 275, 352
(c) 222, 370
Ans: For two numbers, the product is related to the HCF and LCM by
Product=HCF×LCM.
Therefore, the number by which the LCM is multiplied is the HCF.
(a) Numbers: 45 and 105
45=3×3×5
105=3×5×7
HCF=3×5=15
LCM=3×3×5×7=315
45×105=4725
4725÷315=15
Thus, Product=LCM×15, and 15 is the HCF.
(b) Numbers: 275 and 352
275=5×5×11
352=2×2×2×2×2×11
HCF=11
LCM=25×52×11=8800
275×352=96800
96800÷8800=11
Thus, Product=LCM×11, and 11 is the HCF.
(c) Numbers: 222 and 370
222=2×3×37
370=2×5×37
HCF=2×37=74
LCM=2×3×5×37=1110
222×370=82140
82140÷1110=74
Thus, Product=LCM×74, and 74 is the HCF.
Q: Do you see that, in each case, the number by which the LCM is multiplied to get the product is actually the HCF?
Ans: Yes. In every case, the number by which the LCM is multiplied to obtain the product is the HCF.
Therefore, for any two natural numbers,
HCF×LCM=Product of the two numbers
Q: Why does this happen? Can you give an explanation or proof?
Ans: Consider the prime factorisations of the two numbers.
Some prime factors may be common to both numbers, while some may occur in only one number.
HCF: It contains only the common prime factors, each taken with the smaller power.
LCM: It contains every prime factor appearing in either number, each taken with the greater power.
When the HCF and LCM are multiplied, each common prime factor occurs once from each number, and every non-common factor is also included. Thus, the prime factors on both sides are the same.
Therefore, HCF×LCM is always equal to the product of the two numbers.
Q: Explore whether this property holds when 3 numbers are considered.
Ans: No, this property generally does not hold for three numbers.
Example: Take 4, 6 and 8.
Product=4×6×8=192
HCF(4,6,8)=2
LCM(4,6,8)=24
HCF×LCM=2×24=48
Since 48 is not equal to 192, the relation does not hold for these three numbers.
For two numbers, the common and non-common prime factors fit together exactly through the HCF and LCM. For three or more numbers, prime factors may occur in different numbers and with different powers. Therefore, the product of the HCF and LCM usually does not contain all the factors in the required amounts.
Page No. 63
Figure it Out
Q1: In the two rows below, colours repeat as shown. When will the blue stars meet next?
Ans: In the first row, the blue stars are at positions 4 and 10. They repeat every 10−4=6 positions, so their positions are 4,10,16,22,28,….
In the second row, the blue stars are at positions 4 and 8. They repeat every 8−4=4 positions, so their positions are 4,8,12,16,20,….
The first common position after the starting position is 16. Hence, the blue stars meet again at the 16th position.
Q2: (a) Is 5 × 7 × 11 × 11 a multiple of 5 × 7 × 7 × 11 × 2?
(b) Is 5 × 7 × 11 × 11 a factor of 5 × 7 × 7 × 11 × 2?
Ans: (a) Let a=5×7×11×11 and b=5×7×7×11×2.
For a to be a multiple of b, every prime factor of b must occur in a with at least the same power. However, a has no factor 2 and has only one factor 7, whereas b has two factors 7. Therefore, a is not a multiple of b.
(b) For a to be a factor of b, every prime factor of a must occur in b with at least the same power. The number a has two factors 11, but b has only one factor 11. Therefore, 5×7×11×11 is not a factor of 5×7×7×11×2.
Q3: Find the HCF and LCM of the following (state your answers in the form of prime factorisations):
(a) 3 × 3 × 5 × 7 × 7 and 12 × 7 × 11
(b) 45 and 36
Ans:
(a) The first number is 3×3×5×7×7.
The second number is 12×7×11=2×2×3×7×11.
HCF=3×7=21.
LCM=2×2×3×3×5×7×7×11=97020.
(b) 45=3×3×5
36=2×2×3×3
HCF=3×3=9.
LCM=2×2×3×3×5=180.
Q4: Find two numbers whose HCF is 1 and LCM is 66.
Ans: For two numbers a and b,
a×b=HCF(a,b)×LCM(a,b)
Here, HCF=1 and LCM=66.
Therefore, a×b=1×66=66.
The factor pairs of 66 are (1,66), (2,33), (3,22) and (6,11).
Each of these pairs has HCF 1 and LCM 66.
Q5: A cowherd took all his cows to graze in the fields. The cows came to a crossing with 3 gates. An equal number of cows passed through each gate. Later, at another crossing with 5 gates again an equal number of cows passed through each gate. The same happened at the third crossing with 7 gates. If the cowherd had less than 200 cows, how many cows did he have? (Based on the folklore mathematics from Karnataka.)
Ans: The number of cows must be divisible by 3, 5 and 7. Therefore, it must be a multiple of LCM(3,5,7).
LCM(3,5,7)=3×5×7=105.
The multiples of 105 are 105, 210, 315 and so on.
Only 105 is less than 200. Hence, the cowherd had 105 cows.
Q6: The length, width, and height of a box are 12 cm, 18 cm, and 36 cm, respectively. Which of the following-sized cubes can be packed in this box without leaving gaps?
(a) 9 cm (b) 6 cm (c) 4 cm (d) 3 cm (e) 2 cm
Ans: (b)
Explanation: The side of a cube that fits exactly must divide all three dimensions, 12, 18 and 36. Their HCF is 6, since the common prime factors are 2×3. Although 3 cm and 2 cm also divide all three dimensions, the largest cube that fits without gaps has side 6 cm. Therefore, option (b) is correct.
Q7: Among the numbers below, which is the largest number that perfectly divides both 306 and 36?
(a) 36
(b) 612
(c) 18
(d) 3
(e) 2
(f) 360
Ans: (c)
Explanation: 306=2×3×3×17 and 36=2×2×3×3. Their common prime factors give HCF(306,36)=2×3×3=18. Therefore, the largest number that divides both numbers is 18, so option (c) is correct.
Q8: Find the smallest number that is divisible by 3, 4, 5 and 7, but leaves a remainder of 10 when divided by 11.
Ans: LCM(3,4,5,7)=3×4×5×7=420.
Let the required number be N=420k, where k is a positive integer.
Since 420=11×38+2, we have 420≡2(mod11).
The required remainder is 10, so 2k≡10(mod11).
This gives k≡5(mod11). The smallest positive value of k is 5.
Therefore, N=420×5=2100.
Q9: Children are playing ‘Fire in the Mountain’. When the number 6 was called out, no one got out. When the number 9 was called out, no one got out. But when the number 10 was called out, some people got out. How many children could have been playing initially?
Options:
(a) 72
(b) 90
(c) 45
(d) 36
(e) 3
(f) None of these
Ans: (a) and (d)
Explanation: No one got out when 6 and 9 were called, so the total number N must be divisible by both 6 and 9. Hence, N must be divisible by LCM(6,9)=18.
Some people got out when 10 was called, so N must not be divisible by 10.
72 is divisible by 18 but not by 10, so it is possible.
90 is divisible by both 18 and 10, so it is not possible.
45 and 3 are not divisible by 18, so they are not possible.
36 is divisible by 18 but not by 10, so it is also possible.
Therefore, 36 and 72 could have been the initial numbers of children.
Q10: Tick the correct statement(s). The LCM of two different prime numbers (m, n) can be:
(a) Less than both numbers
(b) In between the two numbers
(c) Greater than both numbers
(d) Less than m × n
(e) Greater than m × n
Ans: (c)
Explanation: Different prime numbers m and n have no common factor other than 1. Therefore, LCM(m,n)=m×n, which is greater than both m and n. Hence, option (c) is correct. It is not less than m×n, so option (d) is also false.
Q11. A dog is chasing a rabbit that has a head start of 150 feet. It jumps 9 feet every time the rabbit jumps 7 feet. In how many leaps does the dog catch up with the rabbit?
Ans: The dog gains 9−7=2 feet in each leap.
Head start =150 feet.
Number of leaps =150÷2=75.
Therefore, the dog catches the rabbit in 75 leaps.
Q12: What is the smallest number that is a multiple of 1, 2, 3, 4, 5, 6, 8, 9, 10? Do you remember the answer from Grade 6, Chapter 5?
Ans: 1=1
2=2
3=3
4=2×2
5=5
6=2×3
8=2×2×2
9=3×3
10=2×5
LCM(1,2,3,4,5,6,8,9,10)=2×2×2×3×3×5
=8×9×5
=360
Thus, the smallest number is 360.
Q13: Here is a problem posed by the ancient Indian Mathematician Mahaviracharya (850 C.E.). Add together
815+120+736+1163+121
What do you get? How can we find this sum efficiently?
Ans: We need to add 815,120,736,1163 and 121.
The denominators have the following prime factors:
15=3×5
20=2×2×5
36=2×2×3×3
63=3×3×7
21=3×7
The LCM of the denominators is 2×2×3×3×5×7=1260.
Using 1260 as the common denominator,
815+120+736+1163+121=8×84+1×63+7×35+11×20+1×601260
=672+63+245+220+601260
=12601260=1
Therefore, the sum is 1.