SOLUTION:

Soln. : (i) HCF of 135 and 225;
Applying the Euclid's lemma to 225 and 135, (where 225 > 135), we get
225 = (135 × 1) + 90, since 90 ≠ 0, therefore, applying the Euclid's lemma to 135 and 90, we have 135 = (90 × 1) + 45
But 45 ≠ 0
∴ Applying Euclid's lemma to 90 and 45, we get 90 = (45 × 2) + 0
Here, r = 0, so our procedure stops. Since, the divisor at the last step is 45,
∴ HCF of 225 and 135 is 45.
(ii) HCF of 196 and 38220 :
We start dividing the larger number 38220 by 196, we get 38220 = (196 × 195) + 0
Here, r = 0
∴ HCF of 38220 and 196 is 196.
(iii) HCF of 867 and 255
Here, 867 > 255
∴ Applying Euclid's Lemma to 867 and 255, we get
867 = (255 × 3) + 102, 102 ≠ 0
∴ Applying Euclid's Lemma to 255 and 102, we get
255 = (102 × 2) + 51, 51 ≠ 0
∴ Applying Euclid's Lemma to 102 and 51, we get
102 = (51× 2) + 0, r = 0
∴ HCF of 867 and 225 is 51.

SOLUTION:

Soln. : Let us consider a positive odd integer as ′a′.
On dividing ′a′ by 6, let q 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 i.e.,
a = 6q + 0 = 6q or a = 6q + 1
or a = 6q + 2 or a = 6q + 3
or a = 6q + 4 or a = 6q + 5
But, a = 6q, a = 6q + 2, a = 6q + 4 are even values of ′a′.
[∵ 6q = 2(3q) = 2m₁, 6q + 2 = 2(3q + 1) = 2m₂,
6q + 4 = 2(3q + 2) = 2m₃]
But ′a′ being an odd integer, we have :
a = 6q + 1, or a = 6q + 3, or a = 6q + 5

SOLUTION:

Soln. : Total number of members = 616
∴ The total number of members are to march behind an army band of 32 members is HCF of 616 and 32.
i.e., HCF of 616 and 32 is equal to the maximum number of columns such that the two groups can march in the same number of columns.
∴ Applying Euclid's lemma to 616 and 32, we get
616 = (32 × 19) + 8 ∵ 8 ≠ 0
∴ Again, applying Euclid's lemma to 32 and 8, we get
32 = (8 × 4) + 0, r = 0
∴ HCF of 616 and 32 is 8
Hence, the required number of maximum columns = 8.

SOLUTION:

Soln. : Let us consider an arbitrary positive integer as 'x' such that it is of the form
3q, (3q + 1) or (3q + 2).
For x = 3q, we have x² = (3q)²
⇒ x² = 9q² = 3(3q²) = 3m ... (1)
Putting 3q² = m where m is an integer.
x = 3q + 1,
x² = (3q + 1)² = 9q² + 6q + 1
= 3(3q² + 2q) + 1 = 3m + 1 ... (2)
Putting 3q² + 2q = m, where m is an integer.
For x = 3q + 2,
x² = (3q + 2)²
= 9q² + 12q + 4 = (9q² + 12q + 3) + 1
= 3(3q² + 4q + 1) + 1 = 3m + 1 ... (3)
Putting 3q² + 4q +1 = m, where m is an integer.
From (1), (2) and (3)
x² = 3m or 3m + 1
Thus, the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

SOLUTION:

Soln. : Let us consider an arbitrary positive integer x such that it is in the form of 3q, (3q + 1) or
(3q + 2).
For x = 3q
x³ = (3q)³= 27q³ = 9(3q³) = 9m ... (1)
Putting 3q³ = m, where m is an integer.
For x = 3q + 1
x³ = (3q + 1)³ = 27q³ + 27q² + 9q + 1
= 9(3q³ + 3q² + q) + 1 = 9m + 1 ... (2)
Putting (3q³ + 3q² + q) = m, where m is an integer.
For x = 3q + 2,
x³ = (3q + 2)³ = 27q³ + 54q² + 36q + 8
= 9(3q³ + 6q² + 4q) + 8 = 9m + 8 ... (3)
Putting (3q³ + 6q² + 4q) = m, where m is an integer.
From (1), (2), (3) we have:
x³ = 9m, (9m + 1) or (9m + 8)
Thus, cube of any positive integer can be in the form 9m, (9m + 1) or (9m + 8) for some integer m.

SOLUTION:

Soln. : (i) Using factor tree method, we have:

∴ 140 = 2 × 2 × 5 × 7 = 2² × 5 × 7
(ii) Using factor tree method, we have :

∴ 156 = 2 × 2 × 3 × 13 = 2² × 3 × 13
(iii) Using factor tree method, we have:

∴ 3825 = 3 × 3 × 5 × 5 × 17 = 3² × 5² × 17
(iv) Using factor tree method, we have:

∴ 5005 = 5 × 7 × 11 × 13
(v) Using the factor tree method, we have:

∴ 7429 = 17 × 19 × 23

SOLUTION:

Soln. : (i) Using factor-tree method, we have:

⇒ 26 = 2 × 13 and 91 = 7 × 13
∴ LCM of 26 and 91 = 2 × 7 × 13 = 182
HCF of 26 and 91 = 13
Now, LCM × HCF = 182 × 13 = 2366 and
26 × 91 = 2366
i.e., LCM × HCF = Product of two numbers.
(ii) Using the factor tree method, we have :

⇒ 510 = 2 × 3 × 5 × 17 and 92 = 2 × 2 × 23
∴ LCM of 510 and 92 = 2 × 2 × 3 × 5 × 17 × 23
= 23460
HCF of 510 and 92 = 2
⇒ LCM × HCF = 23460 × 2 = 46920
510 × 92 = 46920
i.e., LCM × HCF = Product of two numbers.
(iii) Using the factor tree method, we have:

∴ 336 = 2 × 2 × 2 × 2 × 3 × 7
∴ 54 = 2 × 3 × 3 × 3
LCM of 336 and 54 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7
= 3024
HCF of 336 and 54 = 2 × 3 = 6
LCM × HCF = 3024 × 6 = 18144
Also 336 × 54 = 18144
Thus, LCM × HCF = Product of two numbers.

SOLUTION:

Soln. : (i) we have

⇒ 12 = 2 × 2 × 3, 15 = 3 × 5 and 21 = 3 × 7
∴ HCF of 12, 15 and 21 is 3
LCM of 12, 15 and 21 is 2 × 2 × 3 × 5 × 7
i.e., 420
(ii) we have :

LCM of 17, 23 and 29 = 17 × 23 × 29 = 11339
(iii) we have :

⇒ 8 = 2 × 2 × 2 and 9 = 3 × 3 and 25 = 5 × 5
∴ HCF of 8, 9 and 25 is 1
LCM of 8, 9 and 25 is
2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800

SOLUTION:

Soln. : Here, HCF of 306 and 657 = 9 we have :
LCM × HCF = Product of the numbers.
∴ LCM × 9 = 306× 657
⇒ LCM = = 34 × 657 = 22338
Thus, LCM of 306 and 657 is 22338.

SOLUTION:

Soln. : Here, n is a natural number and let 6ⁿ end
in 0.
∴ 6ⁿ is divisible by 5.
But the prime factors of 6 are 2 and 3.
⇒ 6ⁿ = (2 × 3)ⁿ
i.e., In the prime factorisation of 6ⁿ, there is no factor 5.
So, by the fundamental theorem of Arithmetic, every composite number can be expressed as a product of primes and this factorisation is unique apart from the order in which the prime factorisation occurs.
∴ Our assumption that 6ⁿ ends in 0, is wrong.
Thus, there does not exist any natural number n for which 6ⁿ ends with zero.

SOLUTION:

Soln. : We have 7 × 11 × 13 + 13 = 13[(7 × 11) + 1]
= 13[78]
i.e., 13 × 78 cannot be a prime number because it has a factors 13 and 78.
∴ 13 × 78 is a composite number.
Also 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5[7 × 6 × 4 × 3 × 2 × 1 + 1] which is also not a prime number.
[∵ It has a factor 5]
Thus, '7 × 11 × 13 + 13' and '7 × 6 × 5 × 4 × 3 × 2 × 1 + 5' are composite numbers.

SOLUTION:

Soln. : Time taken by Sonia to drive one round of the field = 18 minutes
Time taken by Ravi to drive one round of the field = 12 minutes
The L.C.M. of 18 and 12 gives the exact number of minutes after which they meet at the starting point again.
We have

⇒ 18 = 2 × 3 × 3 and 12 = 2 × 2 × 3
∴ LCM of 18 and 12 = 2 × 2 × 3 × 3 = 36
Thus, they will meet again at the starting point after 36 minutes.

SOLUTION:

Soln. : Let be a rational number.
∴ We can find integers a and b (where, b ≠ 0 and a and b are co-prime) such that
⇒ ...(1)
Squaring both sides, we have a² = 5b²
∴ 5 divides a² ⇒ 5 divides a ...(2)
∴ a = 5c, where c is an integer.
∴ Putting a = 5c in (1), we have

⇒ 25c² = 5b² ⇒ 5c² = b²
⇒ 5 divides b² ⇒ 5 divides b ...(3)
From (2) and (3) a and b have at least 5 as a common factor. i.e., a and b are not co-prime.
∴ Our supposition that is rational is wrong.
Hence, is irrational.

SOLUTION:

Soln. : Let is rational.
∴ We can find two co-prime integers ′a′ and 'b' such that


... (1)
∵ a and b are integers,


⇒ from (1), is a rational
But this contradicts the fact that is irrational.
∴ Our supposition is wrong.
Hence, is an irrational.

SOLUTION:

Soln. : (i) We have
Let be rational, and
∴ is rational
Let such that a and b are co-prime integers and b ≠ 0.
∴ ... (1)
Since, the division of two integers is rational.
∴ is a rational.
From (1), is a rational number which contradicts the fact that is irrational.
∴ Our assumption is wrong.
Thus, is irrational.
(ii) Let us suppose that is rational.
Let there be two co-prime integers a and b.
such that where b ≠ 0
Now,

⇒ is a rational
This contradicts the fact that is irrational.
∴ We conclude that is irrational.
(iii) Let us suppose that is rational.
∴ We can find two coprime integers a and b (b ≠ 0), such that

... (1)
= Rational [ ∵ a and b are integers,]
is a rational number.
From (1), is a rational number, which contradicts the fact that is an irrational number.
∴ Our supposition is wrong.
is an irrational number.

SOLUTION:

Soln. : (i)
∵ 3125 = 5 × 5 × 5 × 5 × 5 = 5⁵, which is of the form (2^m × 5ⁿ)
∴ is a terminating decimal.
i.e., will have a terminating decimal expansion.
(ii)
∵ 8 = 2 × 2 × 2 = 2³ = 1 × 2³
= 5⁰ × 2³, which is of the form (2^m × 5ⁿ)
∴ will have a terminating decimal expansion.
(iii)
∵ 455 = 5 × 7 × 13, which is not of the form
2^m × 5ⁿ
∴ will have a non-terminating repeating decimal expansion.
(iv)
∵ 1600 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5
= 2⁶ × 5², which is of the form 2^m × 5ⁿ]
∴ will have a terminating decimal expansion.
(v)
∵ 343 = 7 × 7 × 7 = 7³, which is not of the form 2^m × 5ⁿ
∴ will have a non-terminating repeating decimal expansion.
(vi)
i.e., q = 2³ × 5², which is of the form 2^m × 5ⁿ.
∴ will have a terminating decimal expansion.
(vii)
i.e., q = 2² . 5⁷ . 7⁵, which is not of the form 2^m × 5ⁿ.
∴ will have a non-terminating repeating decimal expansion.
(viii)
which is of the form 2^m × 5ⁿ.
∴ will have a terminating decimal expansion.
(ix)
∵ 50 = 2 × 5 × 5 = 2¹ × 5², which is of the form 2^m × 5ⁿ.
∴ will have a terminating decimal expansion.
(x)
∵ 210 = 2 × 3 × 5 × 7 = 2¹.3¹.5¹.7¹, which is not of the form 2^m × 5ⁿ.
∴ will have a non-terminating repeating decimal expansion.

SOLUTION:

Soln. : (i)

Multiplying and dividing both sides by 2⁵, we have

(ii)
Multiplying and dividing by 5³, we have

(iii) , represents non-terminating repeating decimal expansion.
(iv)

Multiplying and dividing both sides, by 5⁴, we have

(v) , represents non-terminating repeating decimal expansion.
(vi)
Multiplying and dividing by 5, we have


(vii) represents non-terminating repeating decimal expansion.
(viii)

Multiplying and dividing by 2, we have


(ix)

(x) , represents non-terminating repeating decimal expansion.

SOLUTION:

Soln. : (i) 43.123456789
∵ The given decimal expansion terminates.
∴ It is rational of the form


Hence, p = 43123456789 and q = 2⁹ × 5⁹
Prime factors of q are 2⁹ and 5⁹.
(ii) 0.120120012000120000 ...
∵ The given decimal expansion is neither terminating nor non-terminating repeating,
∴ It is irrational number, hence cannot be written in p/q form.
(iii)
∵ The given decimal expansion is non-terminating repeating.
∴ It is rational number.
... (1)
Multiplying both sides by 1000000000, we have
1000000000x = 43123456789.123456789 ... (2)
Subtracting (1) from (2), we have
(1000000000x) - x = (43123456789.123456789.......)
- 43.123456789
⇒ 999999999x = 43123456746

Here, p = 4791495194 and q = 111111111, which is not of the form 2^m × 5ⁿ i.e., the prime factors of q are not of the form 2^m × 5ⁿ.