Euclids Algorithm And Real Numbers

Given here, n is an odd integer for some k Z

where Z is the set of all integers.

Since, we know that every odd integer is of the form 4k + 1 and 4k – 1.

Consider two cases:

Case 1: For n = 4k + 1

)

Therefore, it is divisible by 16.

Case 2: For n = 4k – 1

)

Therefore, it is divisible by 16.

Thus, n⁴ + 4n² + 11 is divisible by 16.

Given here, n is a positive even integer.

We know that any positive even integer can be expressed as n = 2k

Therefore, n(n + 1)(n + 2) = 2k(2k + 1)(2k + 2)……eq (1)

Now, considering the following cases,

Case 1: k = 1

n(n + 1)(n + 2) = 2k(2k + 1)(2k + 2)

= 2 × 1(2 × 1 + 1)(2 × 1 + 2)

= 2 × (3) × (4)

= 24

Hence, n(n + 1)(n + 2) is divisible by 24. (From (1))

Case 2: k = 2

n(n + 1)(n + 2) = 2k(2k + 1)(2k + 2)

= 2 × 2(2 × 2 + 1)(2 × 2 + 2)

= 4 × (5) × (6)

= 24 × 5

Hence, n(n + 1)(n + 2) is divisible by 24. (from (1))

Case 3: k3

Here, k and k + 1 being even integers, one of them will always be odd and the other will be even.

So, 2k(2k + 1)(2k + 2) will always be divisible by 2.

Also as k≥ 3 is a positive integer, so for some l∈N,

k = 3a or k = 3a + 1 or k = 3a + 2

For k = 3a

k(k + 1)(2k + 1) = 3a(3a + 1)(2(3a) + 1)

= 3(a(3a + 1)(6a + 1)

Therefore, it is divisible by 3.

For k = 3a + 1

k(k + 1)(2k + 1) = (3a + 1)(3a + 1 + 1)(2(3a + 1) + 1)

= (3a + 1)(3a + 2)(6a + 3)

= 3(3a + 1)(3a + 2)(2a + 1)

Therefore, it is divisible by 3.

For k = 3a + 2

k(k + 1)(2k + 1) = (3a + 2)(3a + 2 + 1)(2(3a + 2) + 1)

= (3a + 2)(3a + 3)(6a + 5)

= 3(3a + 2)(a + 1)(6a + 5)

Therefore, it is divisible by 3.

We can see that in any case k(k + 1)(2k + 1) is divisible by 2 and 3.

Also, we know that 2 and 3 are mutually prime numbers.

Therefore the expression must be divisible by 2 × 3 = 6

So, k(k + 1)(2k + 1) is divisible by 24. (From (1))

Let 2a + 3b be divisible by 17.

Therefore, for some integer k, 2a + 3b = 17k….. eq (1)

Now,

9a + 5b = 17a + 17b – 4(2a + 3b)

= 17(a + b) – 4(17k) (from eq (1))

= 17(a + b – 4k)

Therefore, we can say that 9a + 5b is divisible by 17.

We know that from Euclid’s division lemma, for every natural number a if we take b = 5 then unique non-negative integers k and m can be obtained so that a = 5k + m

Where 0 ≤ m ≤ 5

Therefore, m = 0, 1, 2, 3, 4 and 5

So, we can say that a = 5k or a = 5k + 1 or a = 5k + 2 or a = 5k + 3 or a = 5k + 4 where k E N V {0}

Now, a = 5k + 3 = 5k + 5 – 2 = 5(k + 1) – 2

= 5k₁ – 2 where (k₁ = k + 1)

Therefore, a = 5k + 3 can be written as a = 5k – 2

Now, a = 5k + 4 = 5k + 5 – 1 = 5(k + 1) – 1

= 5k₁ – 1 where (k₁ = k + 1)

Therefore, a = 5k + 4 can be written as a = 5k – 1

So, every natural number can be written in the form 5k or 5k ± 1 or 5k ± 2, k E N V {0}.

Given here, there is no common factor between 6 and n.

Therefore, 6 and n are two distinct natural numbers.

We know that 6 has 2 and 3 as prime factors.

n can be written as n = 2k + 1 for all

Therefore, we can see that is divisible by 2.

Similarly, n can be written as n = 3k + 1 for all

Therefore, we can see that is divisible by 3.

Similarly, n can also be written as n = 3k–1 for all

Therefore, we can see that is divisible by 3.

So, n² — 1 is divisible by 2 and 3 both.

Since, 2 and 3 are prime numbers.

Therefore, n² — 1 is divisible by 2 × 3 = 6.

Let the four consecutive integers be n, (n + 1), (n + 2) and (n + 3).

Product = n(n + 1)(n + 2)(n + 3)

We know that every integer can be written in the form of 3k, 3k + 1 and 3k + 2.

For, n = 3k

Product = n(n + 1)(n + 2)(n + 3)

= 3k(3k + 1)(3k + 2)(3k + 3)

= 33k(3k + 1)(3k + 2)(k + 1)

Therefore it is divisible by 3.

For, n = 3k + 1

Product = n(n + 1)(n + 2)(n + 3)

= 3k + 1(3k + 1 + 1)(3k + 1 + 2)(3k + 1 + 3)

= 3(3k + 1)(3k + 2)(k + 1)(3k + 4)

Therefore it is divisible by 3.

For, n = 3k + 2

Product = n(n + 1)(n + 2)(n + 3)

= 3k + 2(3k + 2 + 1)(3k + 2 + 2)(3k + 2 + 3)

= 3(3k + 2)(k + 1)(3k + 4)(3k + 5)

Therefore it is divisible by 3.

n can also be expressed as 4p, 4p + 1, 4p + 2 and 4p + 2

For n = 4p

Product = n(n + 1)(n + 2)(n + 3)

= 4p(4p + 1)(4p + 2)(4p + 3)

= 24p(4p + 1)(2p + 1)(4p + 3)

= 8p(4p + 1)(2p + 1)(4p + 3)

Therefore it is divisible by 8.

For n = 4p + 1

Product = n(n + 1)(n + 2)(n + 3)

= (4p + 1)(4p + 1 + 1)(4p + 1 + 2)(4p + 1 + 3)

= 24(4p + 1)(2p + 1)(4p + 3)(p + 1)

= 8p(4p + 1)(2p + 1)(4p + 3)(p + 1)

Therefore it is divisible by 8.

For n = 4p + 2

Product = n(n + 1)(n + 2)(n + 3)

= (4p + 2)(4p + 2 + 1)(4p + 2 + 2)(4p + 2 + 3)

= 24(2p + 1)(4p + 3)(p + 1)(4p + 5)

= 8p(2p + 1)(4p + 3)(p + 1)(4p + 5)

Therefore it is divisible by 8.

For n = 4p + 2

Product = n(n + 1)(n + 2)(n + 3)

= (4p + 2)(4p + 2 + 1)(4p + 2 + 2)(4p + 2 + 3)

= 24(2p + 1)(4p + 3)(p + 1)(4p + 5)

= 8p(2p + 1)(4p + 3)(p + 1)(4p + 5)

Therefore it is divisible by 8.

For n = 4p + 2

Product = n(n + 1)(n + 2)(n + 3)

= (4p + 2)(4p + 2 + 1)(4p + 2 + 2)(4p + 2 + 3)

= 24(2p + 1)(4p + 3)(p + 1)(4p + 5)

= 8p(2p + 1)(4p + 3)(p + 1)(4p + 5)

Therefore it is divisible by 8.

For n = 4p + 3

Product = n(n + 1)(n + 2)(n + 3)

= (4p + 3)(4p + 3 + 1)(4p + 3 + 2)(4p + 3 + 3)

= 24(4p + 3)(p + 1)(4p + 5)(2p + 3)

= 8p(4p + 3)(p + 1)(4p + 5)(2p + 3)

Therefore, it is divisible by 8.

Since it is divisible by both 3 and 8 and both are mutually prime numbers .

Therefore, it will be divisible by 38 = 24

So, product of four consecutive positive integers is divisible by 24.

144, 233

Here, 233 > 144

233 = 144 × 1 + 89

144 = 89 × 1 + 55

89 = 55 × 1 + 34

55 = 34 × 1 + 21

34 = 21 × 1 + 13

21 = 13 × 1 + 8

13 = 8 × 1 + 5

8 = 5 × 1 + 3

5 = 3 × 1 + 2

3 = 2 × 1 + 1

2 = 1 × 2 + 0

The last non- zero remainder is 1.

Therefore, g. c. d (144, 233) = 1

765, 65

Here 765 > 65

765 = 65 × 11 + 50

65 = 50 × 1 + 15

50 = 15 × 3 + 5

15 = 5 × 3 + 0

The last non- zero remainder is 5.

Therefore, g. c. d (765, 65) = 5

10211, 2517

Here, 10211 > 2517

10211 = 3517 × 4 + 143

2517 = 143 × 17 + 86

143 = 86 × 1 + 57

86 = 57 × 1 + 29

57 = 29 × 1 + 28

29 = 28 × 1 + 1

28 = 1 × 28 + 0

The last non- zero remainder is 1.

Therefore, g. c. d (10211, 2517) = 1

Here, 736 > 85

736 = 85 × 8 + 56

85 = 56 × 1 + 29

56 = 29 × 1 + 27

29 = 27 × 1 + 2

27 = 2 × 13 + 1

2 = 1 × 2 + 0

The last non- zero remainder is 1.

Therefore, g. c. d (736, 85) = 1

Given g.c.d. (a,b) = 1

From Euclid’s algorithm we can say that if b|a, then g.c.d(a,b) = b

Let g. c. d (a - b, a + b) = k

Therefore we can say that k is a factor of both (a–b) and (a + b).

We can write a–b = rk for some r N

And also a + b = sk for some sN

Now, (a + b) + (a–b) = rk + sk

a + a + b–b = k(r + s)

2a = k(r + s)………..eq(1)

(a + b) – (a – b) = rk – sk

a + b–a + b = k(r–s)

2b = k(r–s)………..eq(2)

Also, g. c. d (a, b) = 1

Therefore, 2 × g. c. d (a, b) = 2 × 1

g. c. d (2a, 2b) = 2

g. c. d [(r + s)k, (r – s)k] = 2 (from eq (1) and eq (2))

k × g. c. d(r + s, r – s) = 2

= 2 × 1

So, k × g. c. d(r + s, r – s) = 2

k × g. c. d(r + s, r – s) = 2 × 1 = 2 × g.c.d(a,b)

By comparing we get k = 2

We know that 1 and 2 are co–prime numbers.

Similarly, we get k = 1

So, g. c. d (a—b, a + b) = k = 2

or g. c. d (a—b, a + b) = k = 1

g. c. d (a—b, a + b) = 1 or 2

Here, 115 > 25

115 = 25 × 4 + 15

25 = 15 × 1 + 10

15 = 10 × 1 + 5

10 = 5 × 2 + 0

The last non- zero remainder is 5.

Therefore, g. c. d(115, 25) = 5

Now, by Euclid’s Algorithm

g. c. d(a, b) × l.c.m.(a, b) = ab

Here, a = 115, b = 25

g. c. d(115, 25) × l.c.m.(115, 25) = 115 × 25

⇒ 5 × l.c.m.(115, 25) = 115 × 25

⇒ l.c.m.(115, 25) =

⇒ l.c.m.(115, 25) = 115 × 5

Therefore, l.c.m.(115, 25) = 575

7007 = 7 × 1001

= 7 × 7 × 143

= 7 × 7 × 11 × 13

= 7² × 11 × 13

Therefore, 7007 can be expressed as a product of prime numbers 7, 11 and 13.

7500 = 2 × 3750

= 2 × 2 × 1875

= 2 × 2 × 3 × 625

= 2 × 2 × 3 × 5 × 125

= 2 × 2 × 3 × 5 × 5 × 25

= 2 × 2 × 3 × 5 × 5 × 5 × 5

= 2² × 3 × 5⁴

Therefore, 7500 can be expressed as a product of prime numbers 2, 3 and 5.

10101 = 3 × 3367

= 3 × 7 × 481

= 3 × 7 × 13 × 37

Therefore, 10101 can be expressed as a product of prime numbers 3, 7, 13 and 37

15422 = 2 × 7711

= 2 × 11 × 701

Therefore, 15422 can be expressed as a product of prime numbers 2, 11 and 701

250 = 2 × 125

= 2 × 5 × 25

= 2 × 5 × 5 × 5 = 2 × 5⁵

336 = 2 × 168

= 2 × 2 × 84

= 2 × 2 × 2 × 42

= 2 × 2 × 2 × 2 × 21

= 2 × 2 × 2 × 2 × 3 × 7

= 2⁴ × 3 × 7

Therefore, g. c. d(250, 336) = 2

l.c.m.(250, 336) = 2⁴ × 3 × 7 × 5³

= 16 × 21 × 125

= 336 × 125 = 42000

4000 = 2 × 2000

= 2 × 2 × 1000

= 2 × 2 × 2 × 500

= 2 × 2 × 2 × 2 × 250

= 2 × 2 × 2 × 2 × 2 × 125

= 2 × 2 × 2 × 2 × 2 × 5 × 25

= 2 × 2 × 2 × 2 × 2 × 5 × 5 × 5

= 2⁵ × 5³

25 = 5 × 5

g. c. d(4000, 25) = 5 × 5 = 25

l.c.m.(4000, 25) = 2⁵ × 5³ = 4000

225 = 5 × 45 = 5 × 5 × 9 = 5 × 5 × 3 × 3

145 = 5 × 29

g. c. d(225, 145) = 5

l.c.m.(225, 145) = 5² × 3² × 29

= 25 × 9 × 29= 6525

175 = 5 × 35 = 5 × 5 × 7

1001 = 7 × 143 = 7 × 11 × 13

g. c. d(175, 1001) = 7

l.c.m.(175, 1001) = 5² × 7 × 11 × 13

= 25025

15 = 1 × 3 × 5

21 = 1 × 3 × 7

35 = 1 × 5 × 7

Therefore, g. c. d(15, 21, 35) = 1

l.c.m.(15, 21, 35) = 3 × 5 × 7 = 15 × 7 = 105

40 = 2 × 20 = 2 × 2 × 10

= 2 × 2 × 2 × 5

60 = 2 × 30 = 2 × 2 × 15 = 2 × 2 × 5 × 3

80 = 2 × 40 = 2 × 2 × 20 = 2 × 2 × 2 × 10

= 2 × 2 × 2 × 2 × 5

Therefore, g. c. d(40, 60, 80) = 2 × 2 × 5 = 20

l.c.m.(40, 60, 80) = 2⁴ × 3 × 5 = 16 × 15 = 240

49 = 7 × 7

42 = 2 × 21 = 2 × 3 × 7

91 = 7 × 13

Therefore, g. c. d(49, 42, 91) = 7

l.c.m.(49, 42, 91) = 7² × 2 × 3 × 13

= 49 × 78 = 3822

Let be a rational number.

Therefore,, where g. c. d(a, b) = 1 and a, b N.

Now, squaring both sides

……. eq (1)

5 | a²

5 | a

Now, let a = 5k, k N

So, b = 5k

5 | b

It is a contradiction as g. c. d (a, b) = 1

So, our assumption is wrong.

Therefore, is irrational.

Let be a rational number.

Therefore , where g. c. d(a, b) = 1 and a, b N.

Now, squaring both sides

……. eq (1)

15 | a²

15 | a

Now, let a = 15k, k N

So, b = 15k

15 | b

It is a contradiction as g. c. d(a, b) = 1

So, our assumption is wrong.

Therefore, is irrational.

Let k = is a rational number

Therefore, = 1 which is also a rational number

But, is irrational and we know that subtraction of a rational and irrational is always irrational.

So, our assumption is wrong.

is an irrational number.

Let k = is a rational number

Therefore,

is rational

is also rational

But, we know that is irrational.

So, our assumption is wrong.

is an irrational number.

Let k = is a rational number

Therefore, = is also a rational number

But, is irrational.

So, our assumption is wrong.

is an irrational number

Here, 105 > 91

105 = 91 × 1 + 14

91 = 14 × 6 + 7

14 = 7 × 2 + 0

Last non-negative remainder is 7.

Therefore, g. c. d(105, 91) = 7

Now, we know the relation

g. c. d(a, b) × l.c.m.(a, b) = ab

putting a = 105 and b = 91

g. c. d(105, 91) × l. c. d(105, 91) = 105 × 91

7 × l. c. d(105, 91) = 105 × 91

l. c. d(105, 91) =

l. c. d(105, 91) = 105 × 13 = 1365

Given ( + )( — ) = 4 and ( + ) is irrational.

(multiplying numerator and denominator by )

=

(we know that (a – b) × (a + b) = (a² – b²))

= =

Here, it is given that is irrational and we know that 4 is rational.

Therefore,

We know that rational divided by irrational is always irrational.

So, is irrational.

Given that the BRTS bus returns in 35 minutes and the express bus returns in 42 minutes.

So, we can say that the BRTS returns in the time intervals which are multiples of 35 and the express bus returns in the time intervals which are multiples of 42.

Therefore, both the buses meet at the time interval of l.c.m. of 35 and 42.

35 = 5 × 7

42 = 2 × 3 × 7

l.c.m.(35, 42) = 5 × 2 × 3 × 7 = 210 minutes

So, both the buses will meet after 210 minutes at the initial spot.

Convert the given rational number into lowest divisible terms.

If the denominator is in the power of 2 OR/AND 5, then the given number is terminating else non- terminating.

Prime factorization of 625 = 5 × 5 × 5 × 5 = 5⁴ × 1 = 5⁴ × 2⁰

i.e. it is in the power of 5

So, the given rational number is terminating in nature.

Convert the given rational number into lowest divisible terms.

If the denominator is in the power of 2 OR/AND 5, then the given number is terminating else non- terminating.

Prime factorization of 3125 = 5 × 5 × 5 × 5 × 5 = 5⁵

i.e. it is in the power of 5

So, the given rational number is terminating in nature.

Convert the given rational number into lowest divisible terms.

If the denominator is in the power of 2 OR/AND 5, then the given number is terminating else non- terminating.

So, the given rational number is terminating in nature.

Prime factorization of 6250 = 5 × 5 × 5 × 5 × 5 × 2 = 5⁵ × 2¹

i.e. it is in the power of 5 & 2

Convert the given rational number into lowest divisible terms.

If the denominator is in the power of 2 OR/AND 5, then the given number is terminating else non- terminating.

Prime factorization of 15625 = 5 × 5 × 5 × 5 × 5 × 5 × 5 = 5⁷

i.e. it is in the power of 5

So, the given rational number is terminating in nature

Convert the given rational number into lowest divisible terms.

If the denominator is in the power of 2 OR/AND 5, then the given number is terminating else non- terminating.

Prime factorization of 500 = 5 × 5 × 5 × 2 × 2 =

i.e. it is in the power of 5 & 2

So, the given rational number is terminating in nature

Convert the given rational number into lowest divisible terms.

If the denominator is in the power of 2 OR/AND 5, then the given number is terminating else non- terminating.

Prime factorization of 1600 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 =

i.e. it is in the power of 5 & 2

So, the given rational number is terminating in nature

Convert the given rational number into lowest divisible terms.

If the denominator is in the power of 2 OR/AND 5, then the given number is terminating else non- terminating.

can be written as

Prime factorization of 26 = 2 × 13

Here, denominator has got a factor other than 2 &5

So, it is non-terminating in nature.

Convert the given rational number into lowest divisible terms.

If the denominator is in the power of 2 OR/AND 5, then the given number is terminating else non- terminating.

can be written as

As, its denominator is itself 5. So, it is terminating in nature .

Convert the given rational number into lowest divisible terms.

If the denominator is in the power of 2 OR/AND 5, then the given number is terminating else non- terminating.

Prime factorization of 343 = 7 × 7 × 7

i.e. it is not in the power of 5 & 2

So, the given rational number is non- terminating in nature.

Convert the given rational number into lowest divisible terms.

If the denominator is in the power of 2 OR/AND 5, then the given number is terminating else non- terminating.

Prime factorization of 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2⁷

i.e. it is in the power of 2.

So, the given rational number is terminating in nature .

The given decimal number is non-recurring and non-terminating Therefore, it’s an irrational number.

Let × = 3. 456789123456…………eq(1)

As there are 9 repeating digits. So we multiply by 10⁹

10⁹ × = 3456789123. 45678912345678…. …..eq(2)

Subtracting (1)from(2):

99999999 × = 3456789120

This is a terminating number so its rational for

Let 10 × = 23. 12121212…. ….eq(1)

As there are 2 repeating digits. So we multiply by 10²

1000 × = 2312. 121212……….eq(2)

Subtracting (1) from (2):

990 × = 2289

Let × = 0. 142857142857142…. …eq(1)

As there are 6 repeating digits. So we multiply by 10⁶

10⁶ × = 142857. 14285714…. …..eq(2)

Subtracting (1) from (2):

99999 × = 142857

Let × = 0. 9999…. ….eq(1)

As there are repeating digits.

So we multiply by 10

10 × = 9. 9999…. ……..eq(2)

Subtracting (1) from (2):

9 × = 9

× = 1

As the given number is non-terminating and non-recurring so its rational number can be shown as,

As the given number is non-terminating and non-recurring so its rational number can be shown as,

As the given number is non-terminating and non-recurring so its rational number can be shown as,

Let =

Now, analysing × and y such that × + y = 5 and × y = 6

Therefore, it can be found that × = 3 and y = 2

So, =

Let =

Now, analysing × and y such that × + y = 9 and × y = 14

Therefore, it can be found that × = 7 and y = 2

So, =

Let =

Now, analysing × and y such that × + y = 2 and × y =

Therefore, it can be found that × = 3 and y = 2

So, =

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

= ………….. eq (1)

=

=

=

= ………….. eq (2)

=

=

=

= ………………eq (3)

Now, subtracting eq (3) from the addition of eq (1) and eq (2).

+ =

+

Now, rationalising the terms

+

= +

= +

= +

= +

= +

=

= =

25 = 5 × 5

35 = 5 7

Therefore, g. c. d(25, 35) = 5

And l.c.m. (25, 35) = 5 × 7 × 5 = 175

105 = 3 × 35 = 3 × 5 × 7

125 = 5 × 25 = 5 × 5 × 5

Therefore, g. c. d(105, 125) = 5

And l.c.m.(105, 125) = 3 × 5 × 5 × 5 × 7 = 2675

220 = 2 × 110 = 2 × 2 × 55 = 2 × 2 × 5 × 11

132 = 2 × 66 = 2 × 2 × 33 = 2 × 2 × 3 × 11

Therefore, g. c. d(220, 132) = 2 × 2 × 11 = 44

And l.c.m.(220, 132) = 2 × 2 × 3 × 11 × 5 = 660

3125 = 5 × 625 = 5 × 5 × 125 = 5 × 5 × 5 × 5 × 5

625 = 5 × 125 = 5 × 5 × 5 × 5

Therefore, g. c. d(3125, 625) = 5 × 5 × 5 × 5 = 625

And l.c.m.(3125, 625) = 5 × 5 × 5 × 5 × 5 = 3125

15625 = 5 × 5 × 5 × 5 × 5 × 5

35 = 5 × 7

Therefore, g. c. d(15625, 625) = 5

And l.c.m.(15625, 625) = 5 × 5 × 5 × 5 × 5 × 5 × 7 = 109375

15 = 5 × 3

25 = 5 × 5

35 = 5 × 7

Therefore, g. c. d(15, 25, 35) = 5

And l.c.m.(15, 25, 35) = 5 × 5 × 3 × 7 = 525

18 = 3 × 3 × 2

12 = 3 × 2 × 2

16 = 2 × 2 × 2 × 2

Therefore, g. c. d(18, 12, 16) = 2

And l.c.m.(18, 12, 16) = 2 × 2 × 2 × 2 × 3 × 3 = 144

16 = 2 × 2 × 2 × 2

24 = 2 × 2 × 2 × 3

36 = 2 × 3 × 2 × 3

Therefore, g. c. d(16, 24, 36) = 2 × 2 = 4

And l.c.m.(16, 24, 36) = 2 × 2 × 2 × 2 × 3 × 3 = 144

35 = 5 × 7

28 = 2 × 2 × 7

63 = 3 × 3 × 7

Therefore, g. c. d(35, 28, 63) = 7

And l.c.m.(35, 28, 63) = 3 × 3 × 2 × 2 × 7 × 5 = 1260

For L.C.M or G.C.D we first need to find the prime factors of the numbers

112 = 2 × 2 × 2 × 2 × 7

128 = 2 × 2 × 2 × 2 × 2 × 2 × 2

144 = 2 × 2 × 2 × 2 × 3 × 3

Therefore, g. c. d(112, 128, 144) = 2 × 2 × 2 × 2 = 16

Let k = is a rational number

Therefore,

Squaring both sides we have,

is rational

is also rational

But, we know that is irrational.

So, our assumption is wrong.

is an irrational number.

Let k = is a rational number

Therefore, = is also a rational number

But, is irrational.

So, our assumption is wrong.

is an irrational number

Let is a rational number

Therefore, k =

Squaring both sides we have,

is rational

is also rational

But, we know that is irrational.

So, our assumption is wrong.

is an irrational number.

So, is also an irrational number.

Let k = is a rational number

Therefore,

Squaring both sides we have,

is rational

is also rational

But, we know that is irrational.

So, our assumption is wrong.

is an irrational number.

Let k = is a rational number

Therefore, is a rational number

But, we know that is irrational.

So, our assumption is wrong.

is an irrational number.

Let k = is a rational number

Therefore,

Squaring both sides we have,

is rational

is also rational

But, we know that is irrational.

So, our assumption is wrong.

is an irrational number.

Let k =

Squaring both sides we have,

is rational

is also rational

But, we know that is irrational.

So, our assumption is wrong.

Therefore, is an irrational number.

Let k = is a rational number

Therefore, = is also a rational number

But, is irrational.

So, our assumption is wrong.

is an irrational number.

So, is an irrational number.

Let k = is a rational number

Therefore, = is also a rational number

But, is irrational.

So, our assumption is wrong.

is an irrational number.

So, is also an irrational number.

We know that 37 is a prime number.

Therefore is an irrational number as it has only two factors 37 and 1.

Here, the denominator is 125.

125 = 5 × 5 × 5

We can see that that 125 contains 5 as its prime factors.

Also, 211 is a prime number.

So, g. c. d(211, 125) = 1

Therefore,
has a terminating decimal expansion.

Here, the denominator is 625.

625 = 5 × 5 × 5 × 5 × 2⁰

We can see that that 625 contains 5 as its prime factors.

Also, 156 = 2 × 2 × 3 × 13

So, g. c. d(156, 625) = 1

Therefore,
has a terminating decimal expansion.

Here, the denominator is 35.

35 = 5 × 7 × 2⁰

We can see that that 35 contains 7 as its prime factors other than 5 and 2.

Also, 337 is a prime number.

So, g. c. d(337, 35) = 1

Therefore,
does not have a terminating decimal expansion.

Here, the denominator is 49.

49 = 7 × 7 × 2⁰ × 5⁰

We can see that that 49 contains 7 as its prime factors other than 2 and 5.

Also, 132 = 2 × 2 × 3 × 11

So, g. c. d(132, 49) = 1

Therefore,
does not have a terminating decimal expansion

Here, the denominator is 16.

16 = 2 × 2 × 2 × 2 × 5⁰

We can see that that 16 contains 2 as its prime factors.

Also, 235 = 5 × 47

So, g. c. d(235, 16) = 1

Therefore,
has a terminating decimal expansion.

Let =

Now, analysing x and y such that x + y = 12 and xy = 35

Therefore, it can be found that x = 7 and y = 5

So, =

Let =

Now, analysing x and y such that x + y = 8 and xy = 7

Therefore, it can be found that x = 7 and y = 1

So, =

Let =

Now, analysing x and y such that x + y = 6 and xy = 5

Therefore, it can be found that x = 5 and y = 1

So, =

Let =

= =

Now, analysing x and y such that x + y = 14 and xy = 45

Therefore, it can be found that x = 9 and y = 5

So, =

=

=

=

+

Now, rationalising the terms

= +

= +

= +

=

=

=

=

=

=

=

=

=

=

=

Let k be the largest number dividing 230 and 142 leaving remainders 5 and 7 respectively.

Therefore, we can write 230 = ka + 5 and 142 = kb + 7

Where, a, b

230 = ka + 5 142 = kb + 7

⇒ ka = 230 – 5 ⇒ kb = 142 – 7

ka = 225 kb = 135

But, k is the largest number.

So, k becomes the largest divisor, g. c. d of 225 and 135

225 = 5 × 5 × 3 × 3

135 = 5 × 3 × 3 × 3

Therefore, g. c. d(225, 135) = k

g. c. d(225, 135) = 5 × 3 × 3 = 45

k = 45

Therefore, the largest number dividing 230 and 142 leaving remainders 5 and 7 respectively is 45.

Let k be the largest number dividing 110, 62 and 92 leaving remainders 5, 6 and 1 respectively.

Therefore, we can write 110 = ka + 5, 62 = kb + 6 and 92 = kc + 1

Where, a, b and c

110 = ka + 5 62 = kb + 6 92 = kc + 1

⇒ ka = 110 – 5 ⇒ kb = 62 – 6 ⇒ kc = 92 – 1

ka = 105 kb = 56 kc = 91

But, k is the largest number.

So, k becomes the largest divisor, g. c. d of 105, 56 and 91

105 = 5 × 3 × 7

56 = 7 × 2 × 2 × 2

91 = 7 × 13

Therefore, g. c. d(105, 56, 91) = k

g. c. d(105, 56, 91) = 7

k = 7

Therefore, the largest number dividing 110, 62 and 92 leaving remainders 5, 6 and 1 respectively is 7.

The length of the largest scale measuring instrument which can measure all the three dimensions is the g. c. d of their measures.

735 = 5 × 3 × 7 × 7

635 = 5 × 5 × 5 × 5

415 = 5 × 83

Therefore, g. c. d(735, 625, 415) = 5

So, the length of the largest scale is 5 cm.

For packing 150 litres of milk of higher fat and 240 litres of milk of lower fat in tins of equal capacity, we need to find the g. c. d of 150 and 240.

150 = 5 × 3 × 5 × 2

240 = 3 × 2 × 2 × 2 × 2 × 5

Therefore, g. c. d(150, 240) = 5 × 3 × 2 = 30

So, the capacity of the tin should be 30 litres.

The smallest number which is a multiple of 125 and 225 is the l.c.m. of them.

125 = 5 × 5 × 5

225 = 5 × 5 × 3 × 3

Therefore, l.c.m.(125, 225) = 5 × 5 × 5 × 3 × 3 = 1125

Now, we have to find the smallest number which is decreased by 15.

Therefore, the required number = 1125 + 15 = 1140

To find the smallest number of six digits divisible by 18, 24 and 30 we find the l.c.m. of them.

18 = 3 × 3 × 2

24 = 2 × 2 × 2 × 3

30 = 3 × 2 × 5

Therefore, l.c.m.(18, 24, 30) = 2 × 2 × 2 × 3 × 3 × 5 = 360

We know that the smallest six digit number is 100000.

Dividing 100000 by 18 = = 277. 77

So, the integer bigger and nearest to 277. 77 is 278.

Therefore, required integer = 278 × 360 = 100080

So, the smallest number of six digits divisible by 18, 24 and 30

is 100080.

Let us suppose 3 is a factor of a and b.

Therefore, we can write a = 3k and b = 3m

Now, a² + b² = (3k)² + (3m)²

= 9k² + 9m²

= 3(3k² + 3m²)

Therefore it is divisible by 3.

So, 3 is a factor of a² + b²

Similarly, the reverse condition is also true.

(as a²–b² = (a + b)(a–b))

….. eq (1)

Now, n > 1

n – 1 > 0

Also,

Therefore, and are distinct positive integers.

So, we can say that

n–1 > 0

Similarly, n + 1 > 0

Therefore, are also distinct.

Thus, from eq(1) n⁴ + 4 has two distinct factors and 1 as a factor.

So, n⁴ + 4 is a composite number for n > 1.

Steps of the man measure = 90 cm

Steps of the woman measures = 80 cm

Steps of the child measures = 60 cm

To find the minimum distance each man, woman and child should walk we have to find the l.c.m. of their steps.

90 = 2 × 5 × 3 × 3

80 = 2 × 2 × 2 × 2 × 5

60 = 2 × 5 × 2 × 3

Therefore, l.c.m.(90, 80, 60) = 2 × 2 × 2 × 2 × 3 × 3 × 5 = 720

So, the minimum distance each should walk to cover the distance in complete steps is 720 cm.

To find the required number we must find the l.c.m. of 16, 24 and 40.

16 = 2 × 2 × 2 × 2

24 = 2 × 2 × 2 × 3

40 = 2 × 2 × 2 × 5

l.c.m.(16, 24, 40) = 2 × 2 × 2 × 2 × 3 × 5 = 240

To find the number between 24001 and 25000,

Divide 24001 by 240

= 100. 004

So, the integer nearer to and greater than it is 101.

Hence, a number divisible by 240 and between 24001 and 25000 = 101 × 240 = 24240

Let the four consecutive integers be n, (n + 1), (n + 2) and (n + 3).

Product = n(n + 1)(n + 2)(n + 3)

We know that every integer can be written in the form of 3k, 3k + 1 and 3k + 2.

For, n = 3k

Product = n(n + 1)(n + 2)(n + 3)

= 3k(3k + 1)(3k + 2)(3k + 3)

= 33k(3k + 1)(3k + 2)(k + 1)

Therefore it is divisible by 3.

For, n = 3k + 1

Product = n(n + 1)(n + 2)(n + 3)

= 3k + 1(3k + 1 + 1)(3k + 1 + 2)(3k + 1 + 3)

= 3(3k + 1)(3k + 2)(k + 1)(3k + 4)

Therefore it is divisible by 3.

For, n = 3k + 2

Product = n(n + 1)(n + 2)(n + 3)

= 3k + 2(3k + 2 + 1)(3k + 2 + 2)(3k + 2 + 3)

= 3(3k + 2)(k + 1)(3k + 4)(3k + 5)

Therefore it is divisible by 3.

n can also be expressed as 4p, 4p + 1, 4p + 2 and 4p + 2

For n = 4p

Product = n(n + 1)(n + 2)(n + 3)

= 4p(4p + 1)(4p + 2)(4p + 3)

= 24p(4p + 1)(2p + 1)(4p + 3)

= 8p(4p + 1)(2p + 1)(4p + 3)

Therefore it is divisible by 8.

For n = 4p + 1

Product = n(n + 1)(n + 2)(n + 3)

= (4p + 1)(4p + 1 + 1)(4p + 1 + 2)(4p + 1 + 3)

= 24(4p + 1)(2p + 1)(4p + 3)(p + 1)

= 8p(4p + 1)(2p + 1)(4p + 3)(p + 1)

Therefore it is divisible by 8.

For n = 4p + 2

Product = n(n + 1)(n + 2)(n + 3)

= (4p + 2)(4p + 2 + 1)(4p + 2 + 2)(4p + 2 + 3)

= 24(2p + 1)(4p + 3)(p + 1)(4p + 5)

= 8p(2p + 1)(4p + 3)(p + 1)(4p + 5)

Therefore it is divisible by 8.

For n = 4p + 2

Product = n(n + 1)(n + 2)(n + 3)

= (4p + 2)(4p + 2 + 1)(4p + 2 + 2)(4p + 2 + 3)

= 24(2p + 1)(4p + 3)(p + 1)(4p + 5)

= 8p(2p + 1)(4p + 3)(p + 1)(4p + 5)

Therefore it is divisible by 8.

For n = 4p + 2

Product = n(n + 1)(n + 2)(n + 3)

= (4p + 2)(4p + 2 + 1)(4p + 2 + 2)(4p + 2 + 3)

= 24(2p + 1)(4p + 3)(p + 1)(4p + 5)

= 8p(2p + 1)(4p + 3)(p + 1)(4p + 5)

Therefore it is divisible by 8.

For n = 4p + 3

Product = n(n + 1)(n + 2)(n + 3)

= (4p + 3)(4p + 3 + 1)(4p + 3 + 2)(4p + 3 + 3)

= 24(4p + 3)(p + 1)(4p + 5)(2p + 3)

= 8p(4p + 3)(p + 1)(4p + 5)(2p + 3)

Therefore it is divisible by 8.

Since it is divisible by both 3 and 8 and both are mutually prime numbers .

Therefore, it will be divisible by 38 = 24

So, product of four consecutive positive integers is divisible by 24.

which is an integer.

l.c.m. of two numbers =

g. c. d of two numbers =

=

p₁ and p₂ are distinct prime numbers.

So, 1 is the only common factor among them.

Therefore, g. c. d of them is 1.

p,q and r are distinct primes.

So, they have no common factor other than 1.

Then their g. c. d is 1.

Therefore, l.c.m. of two numbers =

= = pqr

15 = 5 × 3 × 1

24 = 2 × 2 × 2 × 3 × 1

40 = 2 × 2 × 2 × 5 × 1

Therefore, g. c. d(15, 24, 40) = 1

15 = 3 × 5

24 = 2 × 2 × 2 × 3

40 = 2 × 2 × 2 × 5

Therefore, l.c.m.(15, 24, 40) = 2 × 2 × 2 × 3 × 5 = 120

It is repeating and terminating.

=

=

=

=

Therefore, x = 9, y =

x² – 4y = = 81 – 141 = – 60

x² – 4y can not be the square of a rational number as we know that square of a rational number is always positive.

Therefore, the given expression does not exist as a binomial surd.

136 = 2 × 2 × 2 × 17

221 = 13 × 17

391 = 17 × 23

So, g. c. d. (136, 221, 391) = 17

136 = 2 × 2 × 2 × 17

221 = 13 × 17

391 = 17 × 23

So, l.c.m.. (136, 221, 391) = 2 × 2 × 2 × 23 × 13 × 17 = 40664

We know that g. c. d. (a, b) × l.c.m.. (a, b) = ab

8 × 64 = ab

ab = 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

Given, g. c. d (a, b) = 8 = 2 × 2 × 2

So, greatest divisor is 8.

Therefore, one of the number is 8.

Therefore the other number is

Given g.c.d. (a,b) = 1

From Euclid’s algorithm we can say that if b|a, then g.c.d(a,b) = b

Let g. c. d (a—b, a + b) = k

Therefore we can say that k is a factor of both (a–b) and (a + b).

We can write a–b = rk for some r N

And also a + b = sk for some sN

Now, (a + b) + (a–b) = rk + sk

a + a + b–b = k(r + s)

2a = k(r + s)………..eq(1)

(a + b) – (a – b) = rk – sk

a + b–a + b = k(r–s)

2b = k(r–s)………..eq(2)

Also, g. c. d (a, b) = 1

Therefore, 2 × g. c. d (a, b) = 2 × 1

g. c. d (2a, 2b) = 2

g. c. d [(r + s)k, (r – s)k] = 2 (from eq (1) and eq (2))

k × g. c. d(r + s, r – s) = 2

= 2 × 1

So, k × g. c. d(r + s, r – s) = 2

k × g. c. d(r + s, r – s) = 2 × 1 = 2 × g.c.d(a,b)

By comparing we get k = 2

We know that 1 and 2 are co–prime numbers.

Similarly, we get k = 1

So, g. c. d (a—b, a + b) = k = 2

or g. c. d (a—b, a + b) = k = 1

g. c. d (a—b, a + b) = 1 or 2

(as a²–b² = (a + b)(a–b))

….. eq (1)

Now, n > 1

n – 1 > 0

Also,

Therefore, and are distinct positive integers.

So, we can say that

n–1 > 0

Similarly, n + 1 > 0

Therefore, are also distinct.

Thus, from eq(1) n⁴ + 4 has two distinct factors and 1 as a factor.

So, n⁴ + 4 is a composite number for n > 1.

Except 48 all the numbers are divisible by 18.

5⁵ means the decimal expansion will terminate after 5 digits.

Any power of 5 ends with 5.

2^m5ⁿ = 2^{m – 1} × 5^{n – 1} × 2 × 5 = 2^{m – 1} × 5^{n – 1} × 10

(m, n N so (m – 1), (n – 1) N – {0} )

2^m × 5ⁿ ends with 0

(5k + 1)² = 25k² + 10k + 1 = 5(5k² + 2k) + 1

(5k + 1)² leaves remainder 1 when divided by 5

From Euclid’s Division lemma, for any positive integer aN.

a = 6m + r where, 0<r<6 and m N

a = 6m, a = 6m + 1, a = 6m + 2, a = 6m + 3, a = 6m + 4 or a = 6m + 5

Now, a = 6m

Then a² = 6²m²

= 36m² = 6 × (6m)² + 0

Here, remainder is 0.

If a = 6m + 1

Then a² = (6m + 1)²

= 36m² + 12m + 1

= 6(6m² + 2m) + 1

Here, remainder is 1.

If a = 6m + 2

Then a² = (6m + 2)²

= 36m² + 24m + 4

= 6(6m² + 4m) + 4

Here, remainder is 4.

If a = 6m + 3

Then a² = (6m + 3)²

= 36m² + 32m + 9

= 6(6m² + 6m + 1) + 3

Here, remainder is 3.

If a = 6m + 4

Then a² = (6m + 4)²

= 36m² + 48m + 16

= 6(6m² + 8m + 2) + 4

Here, remainder is 4.

If a = 6m + 5

Then a² = (6m + 5)²

= 36m² + 60m + 25

= 6(6m² + 10m + 4) + 1

Here, remainder is 1 = .

Thus in any case, if a² is divided by 6, then the remainder is 0, 1, 3 or 4 but not 5.

Product of three consecutive integers = odd even odd

Or Product of three consecutive integers = even odd even

In both the cases atleast one is even.

So, It is divisible by 2

Also, in three consecutive integers one of them is a multiple of 3.

Also, 2&3 are mutually prime numbers.

So, product of three consecutive integers is divisible by 6.