Real Numbers

of 4ⁿ is 2. Also, we know from the fundamental theorem of arithmetic that the prime factorization of each number is unique.

∴ 4ⁿ can never end with 0.

9q² = 3m, 9q² + 6q + 1 = 3(3q²+ 2q) + 1 = 3m + 1,

9q² + 12q + 4 = 3(3q²+ 4q + 1) + 1 = 3m + 1.

Now, since 616 > 32, we apply division lemma to 616 and 32.

We have, 616 = 32 × 19 + 8

Here, remainder 8 ≠ O. So, we again apply division lemma to 32 and 8.

We have, 32 = 8 × 4 + O

Here, remainder is zero. So, HCF (616, 32) = 8

Hence, maximum number of columns is 8.

12 2² × 3, 15 = 3 × 5 and 21 = 3 × 7

Therefore, the HCF of these integers is 3.

2², 3¹, 5¹ and 7¹ and are the greatest powers involved in the prime factors of 12, 15 and 21.

So, LCM (12, 15, 21) = 2² × 3¹ × 5¹ × 7¹ = 420.

Thus, LCM (26, 91) = 2 × 7 × 13 = 182

HCF (26, 91) = 13

Now, LCM (26, 91) × HCF (26, 91) = 182 × 13 = 2366

and Product of the two numbers = 26 × 91 = 2366

Hence, LCM × HCF = Product of two numbers.

(ii) 144 = 24 × 32 and 198 = 2 × 32 × 11

∴ LCM(198,144)24 × 32 × 11 = 1584

HCF(198, 144) = 2 × 32 = 18

Now, LCM (198, 144) × HCF (198, 144) = 1584 × 18 = 28512

and product of 198 and 144 = 28512

Thus, product of LCM (198, 144) and HCF (198, 144)

= Product of 198 and 144.

Therefore, LCM of 18 and 12 = 2² × 3² = 36

So, they will meet again at the starting point after 36 minutes.

(ii)

then x = 0.3178178178 … … (i)

10x = 3.178178178 … …(ii)

10000x = 3178.178178… …(iii)

On subtracting (ii) from (iii), we get

Case – I When n = (4q + 1)

In this case n² – 1 = (4q + 1)²1 = 16q² + 8q = 8q(2q + 1)

which is clearly divisible by 8.

Case – II When n = (4q + 3)

In this case, we have

n2² = (4q + 3)² – 1 = 16q² + 24q + 8 = 8(2q²+ 3q + 1)

which is clearly divisible by 8.

Hence (n² – 1) is divisible by 8.

And x + 14 = 600 ⇒ 15x = 600 ⇒ x = 40

Then HCF = 40 and LCM = 14 × 40 = 560

∵ LCM × HCF = Product of the numbers

560 × 40 = 280 × Second number Second number $=\frac{560\times 40}{280}=80$

Then other number is 80.

Yes, value of x can be found without finding value of y or z as

x = 2 × 2 × 2 × 17 which arc prime 1tctors of x.

integer q ≥ O and O ≤ r < 6.

i.e., the possible remainders are 0, 1, 2, 3, 4, 5.

1’hus,a canbeoftheform6q,or6q + I,or6q + 2,orßq + 3,ör6q + 4,

or 6q + 5, where q is some quotient.

Since a. is odd integer, so a cannot be of the form , or + 2, or 6q + 4, (since they are even).

Thus, a is of the form 6q + 1, 6q + 3, or 6q + 5, where q is some integer.

Hence, any odd positive integer is of the form 6q + 1 or 6q -1- 3 or 6q + 5, where q is sorne integer.

(ii) We have, $\stackrel{-}{0.16}$ a non-terminating but repeating decimal expansion. So it is rational.

Let x = $\stackrel{-}{0.16}$

Then, x = 0.1616… (i)

100×16.1616… ..(ii)

On subtracting (i) from (ii), we get

100x – x = 16.1616 – 0.1616

⇒ 99x = 16 ⇒ x $=\frac{16}{99}=\frac{p}{q}$

The denominator (q) has factors other than 2 or 5.

Then by Euclid’s division algorithm, corresponding to the positive integers a and 3 there exist

non-negative integers q and r such that

a = 3q + r where 0 ≤ r <3

a² = 9q² + 6qr + r² ….(i) 0 ≤ r <3

Case – I: When r = 0 [putting in (i)]

a² = 9q² = 3(3q²) = 3m where m = 3q²

Case – II: r = 1

a² = 9q² + 6q + 1 = 3(3q² + 2q)+ 1 = 3m + 1 where m = 3q² + 2q

Case – III: r = 2

a² = 9q² + 12 + 4 = 3(3q² + 4q + 1) + 1 = 3m + 1 where m = (3q² + 4q + 1)

Hence, square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Therefore, n = 3q + r, where r = 0, 1, 2

n = 3q or n = 3q + 1 or n = 3q + 2

Case (i) if n = 3q, then n is divisible by 3, n + 2 and n +. 4 arc not divisible by 3.

Case (ii) if 71 = 3q + 1 then n + 2 = 3q + 3 = 3(q + 1), which is divisible by 3 and

n + 4 = 3q + 5, which is not divisible by 3.

So, only (n + 2) is divisible by 3.

Case (iii) If n = 3q + 2, then n + 2 = 3q + 4, which is not divisible by 3 and

(n + 4) = 3q + 6 = 3(q + 2), which is divisible by 3.

So, only (n + 4) is divisible by 3.

Hence one and only one out of n, (n + 2), (n + 4), is divisible by 3.

We have, 960 = 432 × 2 + 96

Since the remainder 96 ≠ 0, so we apply the division lemma to 432 and 96.

We have, 432 = 96 × 4 + 48

Again remainder 48 ≠ 0 so we again apply division lemma to 96 and 48.

We have, 96 = 48 × 2 + O

The remainder has now become zero. So our procedure stops.

Since the divisor at this stage is 48.

Hence, HOE of 960 and 432 is 48.

i.e., HCF (960, 432) = 4H

(ii) Since 12576 > 4052, we apply the division lemma to 12576 and 4052, to get

12576 = 4052 × 3 + 420

Since the remainder 420 ≠ 0, we apply the division lemma to 4052 and 420, to get

4052 = 420 × 9 + 272

We consider the new divisor 420 and the new remainder 272, and apply the division lemma to get

420 = 272 × 1 + 148

We consider the new divisor 272 and the new remainder 148, and apply the division lemma to get

272 = 148 × 1 + 124

We consider the new divisor 148 and the new remainder 124, and apply the division lemma to get

148 = 124 × 1 + 24

We consider the new divisor 124 and the new remainder 24, and apply the division lemma to get

124 = 24 × 5 + 4

We consider the new divisor 24 and the new remainder 4, and apply the division lemma to get

24 = 4 × 6 + O

The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 4, the HCF of 12576 and 4052 is 4.

30 = 2 × 3 × 5; 72 = 23 × 32 and 432 = 24 × 33

Here, 2’ and 31 are the smallest powers of the common factors 2 and 3 respectively.

So, HCF (30, 72, 432) = 21 × 31 = 2× 3 = 6

Again, 2, 33 and 51 arc the greatest powers of the prime factors 2, 3 and 5 respectively.

So, LCM (30, 72, 432) = 24 × 33 × 51 = 2160

HCF × LCM = 6 × 2160 = 12960

Product of numbers = 30 × 72 × 432 = 933120 .

Therefore, HCF × LCM ≠ Product of the numbers.

Then, there exist co-prime positive integers and such that

√7 = $\frac{a}{b}$ , b ≠ 0

So, a = √7 b

Squaring both sides, we have

a² = 7b² …… (i)

⇒ 7 divides a² ⇒ 7 divides a

So, we can write

a = 7c (where c is an integer)

Putting the value of a = 7c in (i), we have

49c² = 7b² 7² = b²

It means 7 divides b² and so 7 divides b.

So, 7 is a common factor of both a and b which is a contradiction.

So, our assumption that √7 is rational is wrong.

Hence, we conclude that √7 is an irrational number.

So, 5 – √3 may be written as

5 – √3 = $\frac{p}{q}$, where p and q are integers, having no common factor except 1 and q ≠ 0.

⇒ 5 – $\frac{p}{q}$ = √3 ⇒ √3 = $\frac{5q-p}{q}$

Since $\frac{5q-p}{q}$ is a rational number as p and q are integers.

∴ √3 is also a rational number which is a contradiction.

Thus, our assumption is wrong.

Hence, 5 – √3 is an irrational number.

we have, 2160 847 × 2 + 466

Since remainder 466 ≠ 0. So, we apply the division lemma to 847 and 466

847 = 466 × 1+ 381

Again remainder 381 ≠ 0. So we again apply the division lemma to 466 and 381.

466 = 381 × 1 + 85

Again remainder 85 ≠0. So, we again apply the division lemma to 381 and 85

381 = 85 × 4 + 41

Again remainder 41 ≠ 0. So, we again apply the division lemma to 85 and 41.

85 = 41 × 2 + 3

Again remainder 3 ≠ 0. So, we again apply the division lemma to 4 1 and 3.

41 = 3 × 13 + 2

Again remainder 2 ≠ 0. So, we again apply the division lemma to 3 and 2.

3 = 2 × 1 + 1

Again remainder 1 ≠ 0. So, we apply division lemma to 2 and 1

2 = 1 × 2 + 0

The remainder now becomes O. So, our procedure stops.

Since the divisor at this stage is 1.

Hence, HCF of 847 and 2160 is 1 and numbers are co-prime.

⇒ $\sqrt{n-1}$is also perfect. square of positive integer From (A) and (B)

$\sqrt{n+1}$and $\sqrt{n-1}$ are perfect squares of positive integer. It contradict the fact that two perfect squares differ at least by 3.

Hence, there is no positive integer n for which $\sqrt{n-1}$ + $\sqrt{n+1}$ is rational.

Similarly, required positive integer is a Íctor of 436 – 11 = 425 and 542 – 15 = 527

Clearly, the required number is the HCF of 391, 425 and 527.

Using the factor tree, we get the prime factorisations of 391, 425 and 527 as follows:

391 = 17 × 23, 425 52 × 17 and 527 17 × 31

∴ HCF of 391, 425, and 527 is 17.

Hence, the required number = 17.

Solution:

For a number to end in zero it must be divisible by 5, but 4ⁿ = 2²ⁿ is never divisible by 5. So, 4ⁿ never ends in zero for any value of n.

Solution:

We know that product of two rational numbers is also a rational number.

So, a² = a × a = rational number.

a³ = a² × a = rational number.

a⁴ = a³ × a = rational number.

………………………………..

………………………………..

aⁿ = a^n-1 × a = rational number.

Solution:

Let x = 2m + 1 and y = 2k + 1

Then x² +y² = (2m + 1)² + (2k + 1)²

= 4m² + 4m + 1 + 4k² + 4k + 1

= 4(m² + k² + m + k) + 2

So, it is even but not divisible by 4.

Solution:

Let three consecutive positive integers be n, n + 1 and n + 2.

We know that when a number is divided by 3, the remainder obtained is either 0 or 1 or 2.

So, n = 3p or 3p + 1 or 3p + 2, where pis some integer.

If n = 3p, then 2 is divisible by 3.

If n = 3p + 1, then n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3.

If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.

So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 3.

Solution:

Any odd number is of the form of (2k + 1), where k is any integer.

So, n² – 1 = (2k + 1)² – 1 = 4k² + 4k

For k = 1, 4k² + 4k = 8, which is divisible by 8.

Similarly, for k = 2, 4k² + 4k = 24, which is divisible by 8.

And fork= 3, 4k² + 4k = 48, which is also divisible by 8.

So, 4k² + 4k is divisible by 8 for all integers k, i.e., n² – 1 is divisible by 8 for all odd values of n.