Real Numbers

Let a be any positive integer.

Then a = 8m + 1 where m is some integer

a = 8m + 1

⟹ a² = (8m + 1)²

= 64m² + 16m + 1

= 8q + 1

Where q = 8m² + 2m

Hence square of an odd integer is of the form 8q + 1 where q is a positive integer.

According to the Euclid’s Division Lemma,

Let a be any positive integer and b = 9

Then by Euclid’s Lemma we have,

a = 9q + r

Where r = 0, 1, 2, 3, 4, 5, 6, 7, 8 [since 0 ≤ r ≤ b] and here value of b is 9]

So all the possible forms of a are as follows:

9q

9q + 1

9q + 2

9q + 3

9q + 4

9q + 5

9q + 6

9q + 7

9q + 8

Now to find the cubes of these values we have the following expansion formula:

(a + b)³ = a³ + 3a²b + 3ab² + b³

(9q + b)³ = 729q³ + 243q²b + 27qb² + b³

Now when we divide the above equation by 9 we get the quotient as

81q³ + 27q²b + 3qb² and the remainder is b³.

So we have to consider the value of b³

So to put b = 0

We get 9m + 0 = 9m

So to put b = 1

We get 1³ = 1 so we get = 9m + 1

So to put b = 2

We get 2³ = 8 so we get = 9m + 8

So to put b = 3

We get 3³ = 27 and it is divisible by 9, so we get = 9m

So to put b = 4

We get 4³ = 64 and when divided by 9 we get 1 as remainder, so we get = 9m + 8

So to put b = 5

We get 5³ = 125 and when divided by 9 we get 8 as remainder, so we get = 9m + 8

So to put b = 6

We get 6³ = 216 and when divided by 9 we get 0 as remainder, so we get = 9m

So to put b = 7

We get 7³ = 343 and when divided by 9 we get 1 as remainder, so we get = 9m + 1

So to put b = 8

We get 8³ = 512 and when divided by 9 we get 8 as remainder, so we get = 9m + 8

So it is proved that all the values are in the form of 9m, 9m + 1 and 9m + 8.

Let take a as any positive integer and b = 6.
Then using Euclid’s algorithm we get a = 6q + r here r is remainder and value of q is more
than or equal to 0 and r = 0, 1, 2, 3, 4, 5 because 0 ≤r < b and the value of b is 6

So total possible forms will 6q + 0 , 6q + 1 , 6q + 2,6q + 3,6q + 4,6q + 5
6q + 0, 6 is divisible by 2 so it is an even number

6q + 1, 6 is divisible by 2 but 1 is not divisible by 2 so it is an odd number

6q + 2, 6 is divisible by 2 and 2 is also divisible by 2 so it is an even number

6q + 3, 6 is divisible by 2 but 3 is not divisible by 2 so it is an odd number

6q + 4, 6 is divisible by 2 and 4 is also divisible by 2 it is an even number

6q + 5, 6 is divisible by 2 but 5 is not divisible by 2 so it is an odd number

So odd numbers will in form of 6q + 1, or 6q + 3, or 6q + 5

From the integers given in the question 210 and 55, it is observed that 210 > 55. So by Euclid’s Division Lemma we get the following:

210 = 55 × 3 + 45

Her the remainder is 45 which is not equal to zero. So applying Euclid’s Division Lemma on divisor 55 and remainder 45.

55 = 45 × 1 + 10

Here the remainder is 10 which is not equal to zero. So applying Euclid’s Division Lemma on divisor 45 and remainder 10.

45 = 10 × 4 + 5

Here the remainder is 5 which is not equal to zero. So applying Euclid’s Division Lemma on divisor 10 and remainder 5.

10 = 5 × 2 + 0

So from the above relation is seen that remainder zero is obtained.

So the HCF of 210 and 55 is 5.

The entire process can be expressed in the following way:

From the integers given in the question 420 and 130, it is observed that 420 > 130. So by Euclid’s Division Lemma we get the following:

420 = 130 × 3 + 30

Here the remainder is 30 which is not equal to zero. So applying Euclid’s Division Lemma on divisor 130 and remainder 30.

130 = 30 × 4 + 10

Her the remainder is 10 which is not equal to zero. So applying Euclid’s Division Lemma on divisor 30 and remainder 10.

30 = 10 × 3 + 0

So from the above relation is seen that remainder zero is obtained.

So the HCF of 420 and 130 is 10.

The entire process can be expressed in the following way:

From the integers given in the question 243 and 75, it is observed that 243 > 75. So by Euclid’s Division Lemma we get the following:

243 = 75 × 3 + 18

Here the remainder is 18 which is not equal to zero. So applying Euclid’s Division Lemma on divisor 75 and remainder 18.

75 = 18 × 4 + 3

Here the remainder is 3 which is not equal to zero. So applying Euclid’s Division Lemma on divisor 18 and remainder 3.

18 = 3 × 6 + 0

So from the above relation is seen that remainder zero is obtained.

So the HCF of 243 and 75 is 3.

The entire process can be expressed in the following way:

From the integers given in the question 225 and 135, it is observed that 225 > 135. So by Euclid’s Division Lemma we get the following:

225 = 135 × 1 + 90

Here the remainder is 90 which is not equal to zero. So applying Euclid’s Division Lemma on divisor 135 and remainder 90.

135 = 90 × 1 + 45

Here the remainder is 45 which is not equal to zero. So applying Euclid’s Division Lemma on divisor 90 and remainder 45.

90 = 45 × 2 + 0

So from the above relation is seen that remainder zero is obtained.

So the HCF of 135 and 225 is 45.

The entire process can be expressed in the following way:

From the integers given in the question 38220 and 196, it is observed that 38220 > 196. So by Euclid’s Division Lemma we get the following:

38220 = 196 × 195 + 0

So from the above relation is seen that remainder zero is obtained.

So the HCF of 38220 and 196 is 195.

The entire process can be expressed in the following way:

From the integers given in the question 867 and 255, it is observed that 867 > 255. So by Euclid’s Division Lemma we get the following:

867 = 255 × 3 + 102

Here the remainder is 102 which is not equal to zero. So applying Euclid’s Division Lemma on divisor 255 and remainder 102.

255 = 102 × 2 + 51

Here the remainder is 51 which is not equal to zero. So applying Euclid’s Division Lemma on divisor 102 and remainder 51.

102 = 51 × 2 + 0

So from the above relation is seen that remainder zero is obtained.

So the HCF of 867 and 255 is 51.

The entire process can be expressed in the following way:

If the HCF of 408 and 1032 can be written as

By Euclid’s division algorithm,

1032 = 408×2 + 216

408 = 216×1 + 192

216 = 192×1 + 24

192 = 24×8 + 0

Since the remainder becomes 0 here, so HCF of 408 and 1032 is 24

Now,

1032x - 408*5 = HCF of these numbers

⟹ 1032x - 2040 = 24

⟹ 1032x = 24 + 2040

⟹ x = 2064/1032

⟹ x = 2

So value of x is 2.

The number 468 can be expressed as follows:

468 = 2 × 2 × 3 × 3 × 13

The number 945 can be expressed as follows:

945 = 5 × 7 × 3 × 3 × 3

The number 140 can be expressed as follows:

140 = 2 × 2 × 5 × 7

The number 3825 can be expressed as follows:

3825 = 17 × 5 × 5 × 3 × 3

The number 20570 can be expressed as follows:

20570 = 17 × 5 × 2 × 11 × 11

The numbers are expressed as follows:

96 = 2 × 2 × 2 × 2 × 2 × 3

404 = 2 × 2 × 101

So for finding the HCF we may write as follows:

So HCF = 4

For finding the LCM we may write as follows:

So LCM = 2⁵ × 3 × 101

= 9696

So Product of HCM × LCM = 9696 × 4

= 38784

So product of numbers = 404 × 96

= 38784

The numbers are expressed as follows:

336 = 2 × 2 × 2 × 2 × 3 × 7

54 = 2 × 3 × 3 × 3

So for finding the HCF we may write as follows:

So HCF = 2 × 3

= 6

For finding the LCM we may write as follows:

So LCM = 2⁴ × 3³ × 7

= 3024

So Product of HCM × LCM = 3024 × 6

= 18144

So product of numbers = 54 × 336

= 18144

The numbers are expressed as follows:

336 = 2 × 5 × 3 × 3

144 = 2 × 2 × 2 × 2 × 3 × 3

So for finding the HCF we may write as follows:

So HCF = 2 × 3²

= 18

For finding the LCM we may write as follows:

So LCM = 2⁴ × 3² × 5

= 720

So Product of HCM × LCM = 720 × 18

= 12960

So product of numbers = 90 × 144

= 12960

The numbers are expressed as follows:

12 = 2 × 2 × 3

15 = 3 × 5

21 = 7 × 3

So for finding the HCF we may write as follows:

So HCF = 3

= 3

For finding the LCM we may write as follows:

So LCM = 2² × 3 × 5 × 7

= 420

The numbers are expressed as follows:

24 = 2 × 2 × 2 × 3

15 = 3 × 5

36 = 2 × 2 × 3 × 3

So for finding the HCF we may write as follows:

So HCF = 3

For finding the LCM we may write as follows:

So LCM = 2³ × 3² × 5

= 360

The numbers can be expressed as follows:

17 = 17 × 1

23 = 23 × 1

29 = 29 × 1

So HCF = 1

For finding the LCM we may write as follows:

So LCM = 17 × 23 × 29 × 1

= 11339

The numbers can be expressed as follows:

6 = 2 × 3

72 = 2 × 2 × 2 × 3 × 3

120 = 2 × 2 × 2 × 3 × 5

So for finding the HCF we may write as follows:

So HCF = 2 × 3

= 6

For finding the LCM we may write as follows:

So LCM = 2³ × 3² × 5

= 360

The numbers can be expressed as follows:

40 = 2 × 2 × 2 × 5

36 = 2 × 2 × 3 × 3

126 = 2 × 3 × 3 × 7

So for finding the HCF we may write as follows:

So HCF = 2

For finding the LCM we may write as follows:

So LCM = 2³ × 3² × 5 × 7

= 2520

The numbers can be expressed as follows:

8 = 2 × 2 × 2 × 1

9 = 3 × 3 × 1

25 = 5 × 5 × 1

So for finding the HCF we may write as follows:

So HCF = 1

For finding the LCM we may write as follows:

So LCM = 2³ × 3² × 5² × 1

= 1800

In the question it is mentioned that Anupriya takes 12 minutes and Raman takes 18 minutes. Since Anupriya takes lesser time as compared to Raman, so they both will meet again at the same time when Anupriya completes one round with respect to Raman. So the total time taken will be the LCM of 12 and 18 minutes.

12 = 2 × 2 × 3

18 = 2 × 3 × 3S

So LCM of 12 and 18 can be calculated as follows:

So LCM = 2² × 3²

= 36

So they both will meet after 36 minutes.

Given:

Number of participants for Hindi = 60

Number of participants for English = 84

Number of participants for Mathematics = 108

Number of participants for each room will be the HCF of 60, 84 and 108.

60 = 2 × 2 × 3 × 5

84 = 2 × 2 × 3 × 7

108 = 2 × 2 × 3 × 3 × 3

So for finding the HCF we may write as follows:

So HCF = 2² × 3

= 12

So total number of participants = 60 + 54 + 108

= 252

Number of rooms required = 252 / 12

= 21

So total 21 rooms are required.

Let us consider that 5 - √3 is a rational number.

Let 5 - √3 = a/b for b ≠ 0 ……………… (i)

Where a and b are co-prime integer numbers.

From (i) we can write as follows:

5 – a/b = √3

Now from the LHS we know that a and b are integral numbers and therefore the fraction will also be rational number which implies that √3 has to be a rational number. But we know that √3 is an irrational number.

So our assumption at the beginning of the problem is proved false.

So 5 - √3 is an irrational number.

Let us consider that 1/√2 is a rational number.

Let 1/√2 = a/b for b ≠ 0 ……………… (i)

Where a and b are co-prime integer numbers.

From (i) we can write as follows:

b = a√2

b² = 2a² …………………………….. (i)

Now since 2a² is divisible by 2, so b² has to be divisible by 2. From theorem 2.3 we can clearly states that if b² is divided by 2 so b is also divisible by 2. So we conclude that 2 divides b.

Now we can write the integer b in following format,

b = 2c

b² = 4c² …………………………… (ii)

By comparing (i) and (ii) we can state as follows:

4c² = 2a²

a² = 2c²

From the above equation we can conclude that a² is divisible by 2 and also by a.

From the values of a and b it is seen that a and b has a common factor and it is clearly indicates that 2 is a common factor. But it is assumed in the beginning that a and b has no common factors.

So our assumption made at the beginning of the problem is wrong.

Hence it is proved that 1/√2 is an irrational number.

Let us consider that 6 + √2 is a rational number.

Let 6 + √2 = a/b for b ≠ 0 ……………… (i)

Where a and b are co-prime integer numbers.

From (i) we can write as follows:

a/b - 6 = √2

Now from the LHS we know that a and b are integral numbers and therefore the fraction will also be rational number which implies that √2 has to be a rational number. But we know that √2 is an irrational number.

So our assumption at the beginning of the problem is proved false.

So 6 + √2 is an irrational number.

Let us consider that 1/√2 is a rational number.

Let 3√2 = a/b for b ≠ 0 ……………… (i)

Where a and b are co-prime integer numbers.

From (i) we can write as follows:

a = 3b√2

a² = 18b² …………………………….. (i)

Now since 18b² is divisible by 18, so a² has to be divisible by 18. From theorem 2.3 we can clearly states that if a² is divided by 18 so a is also divisible by 18. So we conclude that a divides 18.

Now we can write the integer b in following format,

a = 18c

a² = 324c² …………………………… (ii)

By comparing (i) and (ii) we can state as follows:

324c² = 18b²

b² = 18c²

From the above equation we can conclude that b² is divisible by 18 and also by 18.

From the values of a and b it is seen that a and b has a common factor and it is clearly indicates that 18 is a common factor. But it is assumed in the beginning that a and b has no common factors.

So our assumption made at the beginning of the problem is wrong.

Hence it is proved that 3√2 is an irrational number.

Let √p + √q be a rational number.

Let √p + √q = x where x is integral number.

Squaring both sides.

(√p + √q)² = x²

p + q + 2√pq = x2

Since p and q are co-prime positive integers. So root of p and root of q will definitely be an irrational numbers as they are not perfect squares. So √pq has to be an irrational number.

So the assumption made at the beginning of the problem is false.

So it is proved that √p + √q is an irrational number.

(i)

Hence we can see that the denominator is of the form 2^m × 5ⁿ and hence decimal expansion of 15/1600 is of terminating type.

(ii)

Hence we can see that the denominator is of the form 2^m × 5ⁿ and hence decimal expansion of 13/3125 is of terminating type.

(iii)

Hence we can see that the denominator is of the form 2^m × 5ⁿ and hence decimal expansion of 23/200 is of terminating type.

(iv)

Hence we can see that the denominator is not of the form 2^m × 5ⁿ and hence decimal expansion of 17/6 is of non-terminating type.

(v)

Hence we can see that the denominator is not of the form 2^m × 5ⁿ and hence decimal expansion of 129 / (2² × 5⁷ × 7⁵) is of non-terminating type.

(vi)

Hence we can see that the denominator is of the form 2^m × 5ⁿ and hence decimal expansion of 35/50 is of terminating type.

(vii)

Hence we can see that the denominator is of the form 2^m × 5ⁿ and hence decimal expansion of 7/80 is of terminating type.

(i)

= 0.104

Hence we can see that the denominator is of the form 2^m × 5ⁿ and hence decimal expansion of 13/125 is of terminating type.

(ii)

= 23.3408

Hence we can see that the denominator is of the form 2^m × 5ⁿ and hence decimal expansion of 14588/625 is of terminating type.

(iii)

= 0.098

Hence we can see that the denominator is of the form 2^m × 5ⁿ and hence decimal expansion of 49/500 is of terminating type.

(i) Since the given decimal expansion has non-terminating expansion which implies that it is an irrational number.

(ii) We know the general form of a rational number is p/q

Now let us find out the prime factorization of the denominator.

Hence the denominator is in the form of 2^m × 5ⁿ.

(iii) Since the given decimal expansion has non-terminating expansion which implies that it is an irrational number.

The Square of 14 is 196.

196 = 14 × 14

= 2 × 2 × 7 × 2

= 2² × 7²

Power of 2 = 2

Power of 7 = 2

Sum of powers = 2 + 2

= 4

So correct answer is C.

m = pq³

n = p³q²

So for finding the HCF we may write as follows:

Common Factor Least Power

p 1

q 2

So HCF = q² × p

= pq²

So correct answer is B.

The numbers can be expressed as follows:

95 = 5 × 19

152 = 2 × 2 × 2 × 19

So for finding the HCF we may write as follows:

Common Factor Least Power

19 1

So HCF = 19

So correct answer is B.

We know the basic following relation:

HCF × LCM = 1^st Number × 2^nd Number

30 × LCM = 1080

LCM = 1080 / 30

LCM = 36

So correct answer is C.

From the question we can see that the denominator is not of the form 2^m × 5ⁿ and hence decimal expansion of 441 / (2² × 5⁷ × 7²) is of non-terminating repeating type.

So correct answer is B.

⟹ 0.086

So it is clearly seen that after three decimal places the rational number is terminating

So the correct answer is C.

When 3 will be multiplied with √27 it would be 3√27 which is equal to 9√3 which is not rational number.

When √3 will be multiplied with √27 then it would result in √81 which is 9 and a rational number.

When 9 will be multiplied with √27 it would be 9√27 which is equal to 27√3 which is not rational number.

When 3√3 will be multiplied with √27 then it would result in 3√81 which is 27 and a rational number.

Since the number to be multiplied has to be the smallest the correct answer is √3.

So correct answer is B.

If the LCM = HCF, then the two numbers has to be equal.

So the correct answer is B.

We know the basic following relation:

HCF × LCM = 1^st Number × 2^nd Number

1^st Number = a

2^nd Number = 18

36 × 2 = 18 × a

a = 72 / 18

a = 4

So the correct answer is D.

Any power of 6 will have 6 at its units place and any power of 5 will have 5 at its units place. So when we subtract the units digits that is 6 – 5 we get 1 as the answer.

So the correct answer is A.

For the decimal to be of terminating nature q has to be of the form 2^m × 5ⁿ.

⟹ 6

So the answer is a rational number.

Let a be any odd integer and b = 4. So by applying Euclid’s Division Lemma,

So we have a = 4q + r

Where r is in the range of 0 to 4.

Case I:

For r = 0

a = 4q

Case II:

For r = 1

a = 4q + 1

Case III:

For r = 2

a = 4q + 2

a = 2 (2q + 1)

Case IV:

For r = 3

a = 4q + 3

Since 4q and 4q + 2 is multiple of 2. Thus they are even numbers.

Hence any positive odd integers is in the form of 4q + 1 and 4q + 3.

Hence Proved.

Let n and n-1 be the 2 positive integers.

Product = n × (n-1)

= n² – n

Case I (when n is even):

Let n = 2q

n² – n = (2q) ² - 2q

= 4q² - 2q

= 2q × (2q-1)

Hence the product n² – n is divisible by 2

Case II (when n is odd):

Let n be 2q + 1

n² – n = (2q + 1)²- (2q + 1)

= 4q² + 4q + 1 - 2q – 1

= 4q² + 2q

= 2q × (2q + 1)

Hence the product n² – n is divisible by 2

Hence Proved

It is mentioned in the question that that on dividing 2053 by the required number there is a remainder of 5 which means that 2053 - 5 = 2048 is exactly divisible by required number . In the similar manner,

967 - 7 = 960 the required number is the largest number satisfying the above property.

Therefore, it is the HCF of 2048 & 960

2048 = 64 × 32

= 2⁶ × 2⁵

= 2¹¹

960 = 5 × 2 × 2 × 2 × 2 × 2 × 2 × 3

So for finding the HCF we may write as follows:

Common Factor Least Power

2 6

So HCF = 2⁶

= 64

HCF of 2048 & 960 is 64

Hence required number is 64

By definition,

A composite number is a positive integer that has a factor other than 1 and itself. Now considering your numbers,

7×11×13 + 13 may be written as, i.e. 13 * (78). So other than 1 and the number itself, 13 and 78 are also the factors of the number. Further, 78 = 39 x 2. So, 39 and 2 are also its factors. So this number is definitely not prime. Hence its composite number.

Similarly, 7×6×5×4×3×2×1 + 5 can be written as , i.e. 5 * (1009). So, other than the number and 1, it have 5 and 1009 as its factors too. So it is also a composite number.

We know the basic relation:

HCF × LCM = 1^st Number × 2^nd Number

1^st Number = 306

2^nd Number = 657

HCF = 9

LCM = (306 × 657) / 9

= 34 × 657 [as 306 / 9 = 34]

= 22,338

So LCM = 22338

Length of rectangular verandah = 1872 cm.

Breadth of courtyard = 1320 cm

HCF of the above measurements:

1872 = 1320 × 1 + 552

1320 = 552 × 2 + 216

552 = 216 × 2 + 120

216 = 120 × 1 + 96

120 = 96 × 1 + 24

96 = 24 × 4 + 0

HCF of 1872 and 1320 = 24 cm

Number of tiles = Area of courtyard / areas of side of tiles

= (1872 × 1320) / (24 × 24)

= 4290

Hence total number of tiles required is 4290.

Let us consider that 5 - √3 is a rational number.

Let 5√2 = a/b for b ≠ 0 ……………… (i)

Where a and b are co-prime integer numbers.

From (i) we can write as follows:

a/5b = √2

Now from the LHS we know that a and b are integral numbers and therefore the fraction will also be rational number which implies that √2 has to be a rational number. But we know that √2 is an irrational number.

So our assumption at the beginning of the problem is proved false.

So 5√2 is an irrational number.

Let us consider that 2/√7 is a rational number.

Let 2/√7 = a/b for b ≠ 0 ……………… (i)

Where a and b are co-prime integer numbers.

From (i) we can write as follows:

2b/a = √7

Now from the LHS we know that a and b are integral numbers and therefore the fraction will also be rational number which implies that √7 has to be a rational number. But we know that √7 is an irrational number.

So our assumption at the beginning of the problem is proved false.

So 2/√7 is an irrational number.

Let us consider that 3/2√5 is a rational number.

Let 3/2√5 = a/b for b ≠ 0 ……………… (i)

Where a and b are co-prime integer numbers.

From (i) we can write as follows:

3b/2a = √5

Now from the LHS we know that a and b are integral numbers and therefore the fraction will also be rational number which implies that √5 has to be a rational number. But we know that √5 is an irrational number.

So our assumption at the beginning of the problem is proved false.

So 3/2√5 is an irrational number.

us consider that 2 + √2 is a rational number.

Let 2 + √2 = a/b for b ≠ 0 ……………… (i)

Where a and b are co-prime integer numbers.

From (i) we can write as follows:

a/b – 2 = √2

Now from the LHS we know that a and b are integral numbers and therefore the fraction will also be rational number which implies that √2 has to be a rational number. But we know that √2 is an irrational number.

So our assumption at the beginning of the problem is proved false.

So 2 + √2 is an irrational number.

(i) We know the general form of a rational number is p/q

Now let us find out the prime factorization of the denominator.

Hence the denominator is in the form of 2^m × 5ⁿ.

(ii) Since the above given number has non-terminating decimal expansion, the given number is an irrational number. So it is not possible to find the prime factorization of the given number.