Q1. Express each number as a product of its prime factors:
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429
Solutions:
(i) 140
Prime factors of 140 = 2, 2, 5, 7
= 2^{2} x 5 x 7
(ii) 156
Prime factors of 156 = 2 x 2 x 3 x 13
= 2^{2} x 3 x 13
(iii) 3825
Prime factors of 3825 = 3 x 3 x 5 x 5 x 17
= 3^{2 }x 5^{2} x 17
(iv) 5005
Prime factors of 5005 = 5 x 7 x 11 x 13
(v) 7429
Prime factors of 7429 = 17 x 19 x 23
Q2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.
(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54
Solutions:
(i) 26 and 91
Prime factors of 26 = 2 x 13
Prime factors of 91 = 7 x 13
HCF of 26 and 91 = 13
LCM of 26 and 91 = 2 x 7 x 13
= 14 x 17
= 182
Product of two numbers = 26 x 91
= 2366
LCM x HCF = 182 x13 = 2366
So, product of two numbers = LCM x HCF
(ii) 510 and 92
Prime factors of 510 = 2 x 3 x 5 x17
Prime factors of 92 = 2x 2 x 23
HCF of two numbers = 2
LCM of two numbers = 2 x 2 x 3 x 5 x 17 x 23
= 23460
Product of two numbers = 510x92
= 46920
LCM x HCF = 2 x 23460
=46920
Product of two numbers = LCM x HCF
(iii) 336 and 54
Prime factors of 336 = 2 x 2 x 2 x 2 x 3 x 7
Prime factors of 54 = 2 x 3 x 3 x 3
HCF of two numbers = 6
LCM of two numbers = 2 x 2 x 2 x 2 x 3 x 3 x 3 x 7
= 2^{4} x 3^{3} x 7 = 3024
Product of two numbers = 336 x 54
=18144
LCM x HCF = 3024 x 6 =18144
Product duct of two numbers = LCM x HCF
Q3. Find the LCM and HCF of the following integers by applying the prime factorisation method.
(i) 12, 15 and 21
(ii) 17, 23 and 29
(iii) 8, 9 and 25
Solutions:
(i) 12, 15 and 21
Prime factors of 12 = 2 x 2 x 3
= 2^{2} x 3
Prime factors of 15 = 3 x 5
Prime factors of 21 = 2 x 2 x 3
HCF of 12,15 and 21 = 3
LCM of 12,15 and 21 = 2^{2} x 3 x 5 x 7
= 420
(ii) 17, 23 and 29
Prime factors of 17 = 17 x 1
Prime factors of 23 = 23 x 1
Prime factors of 29 = 29 x 1
HCF of 17, 23 and 29 = 1
LCM of 17, 23 and 29 = 17 x 23 x 29
= 11339
(iii) 8, 9 and 25
Prime factors of 8 = 2 x 2 x 2 x1
= 2^{3} x 1
Prime factors of 9 = 3 x 3 x1
= 3^{2} x 1
Prime factors of 25 = 5 x 5 x 1
= 5^{2} x 1
HCF of 8, 9 and 25 = 1
LCM of 8,9 and 25 = 2 x 2 x 2 x 3 x 3 x 5 x 5
= 1800
Q4. Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution:
Given, HCF (306, 657) = 9.
We have to find,
LCM (306, 657) = ?
We know that
LCM x HCF = Product of two numbers
LCM x 9 = 306x 657
LCM = 34 x 657
LCM = 22338
Q5. Check whether 6^{n} can end with the digit 0 for any natural number n.
Solution:
Prime factors of 6n = (2 x 3)^{n} = (2)^{n} (3)^{n}
You can observe clearly, 5 is not in the prime factors of 6^{n}.
That means 6^{n} will not be divisible by 5.
Therefore, 6^{n} cannot end with the digit 0 for any natural number n.
Q6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Solution:
It can be observed that,
7 x 11 x 13 + 13 = 13 (7 x11 +1)
= 13( 77 + 1)
= 13 x 78
= 13 x13 x 6 x1
= 13 x13 x 2 x 3 x1
The given number has 2,3,13 and 1 as its factors. Therefore, it is a composite number.
7 x 6 x 5 x 4 x 3 x 2 x1 + 5 = 5 x( 7 x 6 x 4 x 3 x 2 x 1 + 1)
= 5 x(1008 + 1)
=5x1009x1
1009 cannot be factorised further. Therefore, the given expression has 5,1009 and 1 as its factors. Hence, it is a composite number.
Q7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?
Solution:
LCM of 18 and 12,
18 = 2 x 3 x 3
12 = 2 x 2 x 3
LCM of 12 and 18 = 2 x 2 x 3 x 3 = 36
Therefore, Ravi and Sonia will meet together at starting point after 36 minutes.
Q1. Use Euclid’s division algorithm to find the HCF of
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255
Ans:
(i) HCF of 135 and 225
Step 1: Since 225 > 135, we apply the division lemma to 225 and 135 to obtain
⇒ 225 = 135 × 1 + 90
Step 2: Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain
⇒ 135 = 90 × 1 + 45
Step 3: We consider the new divisor 90 and new remainder 45 and apply the division lemma to obtain: 90 = 2 × 45 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 45,
Therefore, the HCF of 135 and 225 is 45.
(ii) HCF of 196 and 38220
Step 1: Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain
⇒ 38220 = 196 × 195 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 196,
Therefore, HCF of 196 and 38220 is 196.
(iii) HCF of 867 and 255
Step 1: Since 867 > 255, we apply the division lemma to 867 and 255 to obtain
⇒ 867 = 255 × 3 + 102
Step 2: Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102 to obtain
⇒ 255 = 102 × 2 + 51
Step 3: We consider the new divisor 102 and new remainder 51 and apply the division lemma to obtain
⇒ 102 = 51 × 2 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 51, Therefore, HCF of 867 and 255 is 51.
Q2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Ans: Let us consider a positive odd integer as ‘a’.
On dividing ‘a’ by 6, Let ‘a’ be the quotient and ‘r’ be the remainder.
∴ Using Euclid’s lemma, we have:
a = 6q + r where 0 ≤ r < 6 i.e., r = 0, 1, 2, 3, 4, or 5
Now substituting the value of r, we get:
If r = 0, then a = 6q
Similarly, for r= 1, 2, 3, 4 and 5, the value of a is 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5, respectively.
If a = 6q, 6q+2, 6q+4, then a is an even number and divisible by 2.
A positive integer can be either even or odd.
Therefore, any positive odd integer is of the form of 6q+1, 6q+3 and 6q+5, where q is some integer.
Q3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Ans: Given,
Number of army contingent members = 616
Number of army band members = 32
If the two groups have to march in the same column, we have to find out the highest common factor between the two groups.
HCF(616, 32) gives the maximum number of columns in which they can march.
By Using Euclid’s algorithm to find their HCF, we get,
Since, 616 > 32
∴ 616 = 32 × 19 + 8
Since, 8 ≠ 0, therefore, taking 32 as the new divisor, we have,
32 = 8 × 4 + 0
Now we have got remainder as 0, therefore, HCF (616, 32) = 8.
Hence, the maximum number of columns in which they can march is 8.
Q4. Use Euclid’s division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.
[Hint: Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in form 3m or 3m + 1.]
Ans: Let us consider an arbitrary positive integer as ‘x’ such that it is of the form 3q, (3q + 1) or (3q + 2).
Note:
For any positive integer x and b = 3,
x = 3q + r, where q is quotient and r is remainder such that 0 < r < 3.
If r = 0 then x = 3q
If r = 1 then x = 3q + 1
If r = 2 then x = 3q + 2
For x = 3q, we have
x^{2} = (3q)^{2}
⇒ x^{2} = 9q^{2}
= 3 (3q^{2})
= 3m ...(1) (Putting 3q^{2} = m where m is an integer)
For x = 3q + 1
x^{2} = (3q + 1)^{2}
= 9q^{2} + 6q + 1
= 3 (3q^{2} + 2q) + 1
= 3m + 1 ...(2) (Puting 3q^{2} + 2q = m, where m is an integer)
For x = 3q + 2
x^{2} = (3r + 2)^{2}
= 9q^{2} + 12q + 4
= (9q^{2} + 12q + 3) + 1
= 3 (3q^{2} + 4q + 1) + 1
= 3m + 1 ...(3) (Putting 3q^{2} + 4q + 1 = m, where m is an integer)
From (1), (2) and (3)
x^{2} = 3m or 3m + 1
Thus, the square of any positive integer is either of form 3m or 3m + 1 for some integer m.
Q5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Ans: Let us consider an arbitrary positive integer x such that it is in the form of 3q, (3q + 1) or (3q + 2).
Note:
For any positive integer x and b = 3,
x = 3q + r, where q is quotient and r is remainder such that 0 < r < 3.
If r = 0 then x = 3q
If r = 1 then x = 3q + 1
If r = 2 then x = 3q + 2
∴ x is of the form 3q, (3q + 1) or (3q + 2).
For x = 3q
x^{3} = (3q)^{3}
= 27q^{3}
= 9 (3q^{3})
= 9m ...(1) (Putting 3q^{3} = m, where m is an integer)
For x = 3q + 1
x^{3} = (3q + 1)^{3}
= 27q^{3} + 27q^{2} + 9q + 1
= 9 (3q^{3} + 3q^{2} + q) + 1
= 9m + 1 ...(2) (Putting (3q^{3} + 3q^{2} + 1) = m, where m is an integer)
For x = 3q + 2
x^{3} = (3q + 2)^{3}
= 27q^{3} + 54q^{2} + 36q + 8
= 9 (3q^{3} + 6q^{2} + 4q) + 8
= 9m + 8 ...(3) (Putting (3q^{3} + 6q^{2} + 4q) = m, where m is an integer)
From (1), (2), (3) we have:
x^{3} = 9m, (9m + 1) or (9m + 8)
Thus, x^{3} of any positive integer can be in the form 9m, (9m + 1) or (9m + 8).
Check out the NCERT Solutions of all the exercises of Real Numbers:
Exercise 1.2 NCERT Solutions: Real Numbers
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