Real Numbers

Given numbers are 156 and 504

Here, 504 > 156

So, we divide 504 by 156

By using Euclid’s division lemma, we get

504 = 156 × 3 + 36

Here, r = 36 ≠ 0.

On taking 156 as dividend and 36 as the divisor and we apply Euclid’s division lemma, we get

156 = 36 × 4 + 12

Here, r = 12 ≠ 0

So, on taking 36 as dividend and 12 as the divisor and again we apply Euclid’s division lemma, we get

36 = 12 × 3 + 0

The remainder has now become 0, so our procedure stops. Since the divisor at this last stage is 12, the HCF of 156 and 504 is 12.

Given numbers are 135 and 225

Here, 225 > 135

So, we divide 225 by 135

By using Euclid’s division lemma, we get

225 = 135 × 1 + 90

Here, r = 90 ≠ 0.

On taking 135 as dividend and 90 as the divisor and we apply Euclid’s division lemma, we get

135 = 90 × 1 + 45

Here, r = 45 ≠ 0

So, on taking 90 as dividend and 45 as the divisor and again we apply Euclid’s division lemma, we get

90 = 45 × 2 + 0

The remainder has now become 0, so our procedure stops. Since the divisor at this last stage is 45, the HCF of 135 and 225 is 45.

Given numbers are 455 and 42

Here, 455 > 42

So, we divide 455 by 42

By using Euclid’s division lemma, we get

455 = 42 × 10 + 35

Here, r = 35 ≠ 0.

On taking 42 as dividend and 35 as the divisor and we apply Euclid’s division lemma, we get

42 = 35 × 1 + 7

Here, r = 7 ≠ 0

So, on taking 35 as dividend and 7 as the divisor and again we apply Euclid’s division lemma, we get

35 = 7 × 5 + 0

The remainder has now become 0, so our procedure stops. Since the divisor at this last stage is 7, the HCF of 455 and 42 is 7.

Given numbers are 8840 and 23120

Here, 23120 > 8840

So, we divide 23120 by 8840

By using Euclid’s division lemma, we get

23120 = 8840 × 2 + 5440

Here, r = 5440 ≠ 0.

On taking 8840 as dividend and 5440 as the divisor and we apply Euclid’s division lemma, we get

8840 = 5440 × 1 + 3400

Here, r = 3400 ≠ 0

On taking 5440 as dividend and 3400 as the divisor and again we apply Euclid’s division lemma, we get

5440 = 3400 × 1 + 2040

Here, r = 2040 ≠ 0.

On taking 3400 as dividend and 2040 as the divisor and we apply Euclid’s division lemma, we get

3400 = 2040 × 1 + 1360

Here, r = 1360 ≠ 0

So, on taking 2040 as dividend and 1360 as the divisor and again we apply Euclid’s division lemma, we get

2040 = 1360 × 1 + 680

Here, r = 680 ≠ 0

So, on taking 1360 as dividend and 680 as the divisor and again we apply Euclid’s division lemma, we get

1360 = 680 × 2 + 0

The remainder has now become 0, so our procedure stops. Since the divisor at this last stage is 680, the HCF of 8840 and 23120 is 680.

Given numbers are 4052 and 12576

Here, 12576 > 4052

So, we divide 12576 by 4052

By using Euclid’s division lemma, we get

12576 = 4052 × 3 + 420

Here, r = 420 ≠ 0.

On taking 4052 as dividend and 420 as the divisor and we apply Euclid’s division lemma, we get

4052 = 420 × 9 + 272

Here, r = 272 ≠ 0

On taking 420 as dividend and 272 as the divisor and again we apply Euclid’s division lemma, we get

420 = 272 × 1 + 148

Here, r = 148 ≠ 0

On taking 272 as dividend and 148 as the divisor and again we apply Euclid’s division lemma, we get

272 = 148 × 1 + 124

Here, r = 124 ≠ 0.

On taking 148 as dividend and 124 as the divisor and we apply Euclid’s division lemma, we get

148 = 124 × 1 + 24

Here, r = 24 ≠ 0

So, on taking 124 as dividend and 24 as the divisor and again we apply Euclid’s division lemma, we get

124 = 24 × 5 + 4

Here, r = 4 ≠ 0

So, on taking 24 as dividend and 4 as the divisor and again we apply

Euclid’s division lemma, we get

24 = 4 × 6 + 0

The remainder has now become 0, so our procedure stops. Since the divisor at this last stage is 4, the HCF of 4052 and 12576 is 4.

Given numbers are 3318 and 4661

Here, 4661 > 3318

So, we divide 4661 by 3318

By using Euclid’s division lemma, we get

4661 = 3318 × 1 + 1343

Here, r = 1343 ≠ 0.

On taking 3318 as dividend and 1343 as the divisor and we apply Euclid’s division lemma, we get

3318 = 1343 × 2 + 632

Here, r = 632 ≠ 0

So, on taking 1343 as dividend and 632 as the divisor and again we apply Euclid’s division lemma, we get

1343 = 632 × 2 + 79

Here, r = 79 ≠ 0

So, on taking 632 as dividend and 79 as the divisor and again we apply Euclid’s division lemma, we get

632 = 79 × 8 + 0

The remainder has now become 0, so our procedure stops. Since the divisor at this last stage is 79, the HCF of 3318 and 4661 is 79.

Given numbers are 250, 175 and 425

∴ 425 > 250 > 175

On applying Euclid’s division lemma for 425 and 250, we get

425 = 250 × 1 + 175

Here, r = 175 ≠ 0.

So, again applying Euclid’s division lemma with new dividend 250 and new divisor 175, we get

250 = 175 × 1 + 75

Here, r = 75 ≠ 0

So, on taking 175 as dividend and 75 as the divisor and again we apply Euclid’s division lemma, we get

175 = 75 × 2 + 25

Here, r = 25 ≠ 0.

So, again applying Euclid’s division lemma with new dividend 75 and new divisor 25, we get

75 = 25 × 3 + 0

Here, r = 0 and divisor is 25.

So, HCF of 425 and 225 is 25.

Now, applying Euclid’s division lemma for 175 and 25, we get

175 = 25 × 7 + 0

Here, remainder = 0

So, HCF of 250, 175 and 425 is 25.

Given numbers are 4407, 2938 and 1469

∴ 4407 > 2938 > 1469

On applying Euclid’s division lemma for 4407 and 2938, we get

4407 = 2938 × 1 + 1469

Here, r = 1469 ≠ 0.

So, again applying Euclid’s division lemma with new dividend 2938 and new divisor 1469, we get

2938 = 1469 × 2 + 0

The remainder has now become 0, so our procedure stops. Since the divisor at this stage is 1469, the HCF of 4407 and 2938 is 1469.

Now, applying Euclid’s division lemma for 1469 and 1469, we get

1469 = 1469 × 1 + 0

Here, remainder = 0

So, HCF of 4407, 2938 and 1469 is 1469.

Let a and b be any two positive integers, such that a > b.

Then, a = bq + r, 0 ≤ r < b …(i) [by Euclid’s division lemma]

On putting b = 2 in Eq. (i), we get

a = 2q + r, 0 ≤ r < 2 …(ii)

r = 0 or 1

When r = 0, then from Eq. (ii), a = 2q, which is divisible by 2

When r = 1, then from Eq. (ii), a = 2q + 1, which is not divisible by 2.

Thus, every positive integer is either of the form 2q or 2q + 1.

That means every positive integer is either even or odd. So, if a is a positive even integer, then a is of the form 2q and if a, is a positive odd integer, then a is of the form 2q + 1.

Let a be any positive odd integer. We apply the division algorithm with a and b = 4.

Since 0 ≤ r < 4, the possible remainders are 0,1,2 and 3.

i.e. a can be 4q, or 4q + 1, or 4q + 2, or 4q + 3, where q is the quotient.

As we know a is odd, a can’t be 4q or 4q + 2 because they both are divisible by 2.

Therefore, any positive odd integer is of the form 4q + 1 or 4q + 3.

Given the capacities of the two containers are 250 L and 425 L.

Here, 425 > 250

Now, we divide 425 by 250.

We used Euclid’s division lemma.

425 = 250 × 1 + 175

Here, remainder r = 175 ≠ 0

So, the new dividend is 250 and the new divisor is 175, again we apply Euclid division algorithm.

250 = 175 × 1 + 75

Here, remainder r = 75 ≠ 0

On taking the new dividend is 175 and the new divisor is 75, we apply Euclid division algorithm.

175 = 75 × 2 + 25

Here, remainder r = 25 ≠ 0

On taking new dividend is 75 and the new divisor is 25, again we apply Euclid division algorithm.

75 = 25 × 3 + 0

Here, remainder is zero and divisor is 25.

So, the HCF of 425 and 250 is 25.

Hence, the maximum capacity of the required container is 25 L.

Given length and breadth are 4661 m and 3318 m respectively.

Here, 4661 > 3318

So, we divide 4661 by 3318

By using Euclid’s division lemma, we get

4661 = 3318 × 1 + 1343

Here, r = 1343 ≠ 0.

On taking 3318 as dividend and 1343 as the divisor and we apply Euclid’s division lemma, we get

3318 = 1343 × 2 + 632

Here, r = 632 ≠ 0

So, on taking 1343 as dividend and 632 as the divisor and again we apply Euclid’s division lemma, we get

1343 = 632 × 2 + 79

Here, r = 79 ≠ 0

So, on taking 632 as dividend and 79 as the divisor and again we apply Euclid’s division lemma, we get

632 = 79 × 8 + 0

The remainder has now become 0, so our procedure stops. Since the divisor at this last stage is 79, the HCF of 3318 and 4661 is 79.

Hence, the maximum length of such tiles is 79 meters.

Firstly, we find the length of the largest tile so for that we have to find the HCF of 1658 and 832.

Here, 1658 > 832

So, we divide 1658 by 832

By using Euclid’s division lemma, we get

1658 = 832 × 1 + 826

Here, r = 826 ≠ 0.

On taking 832 as dividend and 826 as the divisor and we apply Euclid’s division lemma, we get

832 = 826 × 1 + 6

Here, r = 6 ≠ 0

So, on taking 826 as dividend and 6 as the divisor and again we apply Euclid’s division lemma, we get

826 = 6 × 137 + 4

Here, r = 4 ≠ 0

So, on taking 6 as dividend and 4 as the divisor and again we apply Euclid’s division lemma, we get

6 = 4 × 1 + 2

Here, r = 2 ≠ 0

So, on taking 4 as dividend and 2 as the divisor and again we apply Euclid’s division lemma, we get

4 = 2 × 2 + 0

The remainder has now become 0, so our procedure stops. Since the divisor at this last stage is 79, the HCF of 1658 and 832 is 2.

So, the length of the largest tile is 2 cm

Area of each tile = 2 × 2 = 4cm²

The required number of tiles =

=

= 344864

Least number of square tiles are required are 344864

Given number is 4320

Factorization of 4320 is

Hence, 4320 = 2⁵ × 3³ × 5 (Product of its prime factors)

Given number is 7560

Factorization of 7560 is

Hence, 7560 = 2³ × 3³ × 5 × 7 (Product of its prime factors)

Given number is 140

Factorization of 140 is

Hence, 140 = 2² × 5 × 7 (Product of its prime factors)

Given number is 5005

Factorization of 5005 is

Hence, 5005 = 5 × 7 × 11 × 13 (Product of its prime factors)

Given number is 32760

Factorization of 32760 is

Hence, 32760 = 2³ × 3² × 5 × 7 × 13 (Product of its prime factors)

Given number is 156

Factorization of 156 is

Hence, 156 = 2² × 3 × 13 (Product of its prime factors)

Given number is 729

Factorization of 729 is

Hence, 729 = 3 × 3 × 3 × 3 × 3 × 3 = 3⁶ (Product of its prime factors)

To find the highest power of 5 in 23750, we have to factorize 23570

Hence, 23750 = 2 × 5⁴ × 19

So, the highest power of 5 in 23750 is 4.

Given number is 1440

Factorization of 1440 is

Factors of 1440 = 2⁵ × 3² × 5

So, the highest power of 2 is 5.

We have to factorize the 6370 to find the value of m, n, k and p

Factorization of 6370 is

6370 = 2 × 5 × 7² × 13

On Comparing, we get

6370 = 2^m .5ⁿ.7^k.13^p = 2¹ × 5¹ × 7² × 13¹

m = 1

n = 1

k = 2

p = 1

So, m + n + k + p = 5

Given numbers are 32 and 62

For pairs to be co-primes there should be no common factor except 1

Factorization of 32 and 62 are

Factors of

Factors of

Here, we can see that 2 is the common factor. So, (32,62) is not co-prime.

Given numbers are 18 and 25

For pairs to be co-primes there should be no common factor except 1

Factorization of 18 and 25 are

Factors of 32 = 2 × 3 × 3

Factors of 62 = 5 × 5

Therefore, there is no common factor. So, (18,25) is co-prime.

Given numbers are 31 and 93

For pairs to be co-primes there should be no common factor except 1

Factorization of 31 and 93 are

Factors of

Factors of

Here, we can see that 31 is the common factor. So, (31, 93) is not co-prime.

Here, a = 2520; b = 2; c = 315; d = 3; x = 3; y = 5

Here, a = 15015; b = 5005; c = 5; d = 143; x = 13

Here, a = 18380; b = 2; c = 1365; d = 3; x = 5; y = 13

Here, a = 3; b = 147407; c = 11339; d = 667; x = 29

Given numbers are 96 and 404.

The prime factorization of 96 and 404 gives:

Here, 2² is the smallest power of the common factor 2.

Therefore, the H.C.F of these two integers is 2 × 2 = 4

2⁵ × 3¹ × 101¹ are the greatest powers of the prime factors 2, 3 and 101 respectively involved in the given numbers.

Now, L.C.M of 96 and 404 is 2 × 2 × 2 × 2 × 2 × 3 × 101 = 9696

Given numbers are 6 and 20

The prime factorization of 6 and 20 gives:

Here, 2¹ is the smallest power of the common factor 2.

Therefore, the H.C.F of these two integers = 2

2² × 3¹ × 5¹ are the greatest powers of the prime factors 2, 3 and 5 respectively involved in the given numbers.

Now, L.C.M of 6 and 20 = 2 × 2 × 3 × 5 = 60

Given numbers are 26 and 91.

The prime factorization of 26 and 91 gives:

Here, 13¹ is the smallest power of the common factor 13.

Therefore, the H.C.F of these two integers = 13

2¹ × 7¹ × 13¹ are the greatest powers of the prime factors 2, 7 and 13 respectively involved in the given numbers.

Now, L.C.M of 6 and 21 is 2 × 7 × 13 = 182

Given numbers are 87 and 145.

The prime factorization of 87 and 145 gives:

Here, 29¹ is the smallest power of the common factor 29.

Therefore, the H.C.F of these two integers = 29

3¹ × 5¹ × 29¹ are the greatest powers of the prime factors 3, 5 and 29 respectively involved in the given numbers.

Now, L.C.M of 87 and 145 = 3 × 5 × 29 = 435

Given numbers are 1485 and 4356.

The prime factorization of 1485 and 4356 gives:

Here, 3²× 11 is the smallest power of the common factors 3 and 11.

Therefore, the H.C.F of these two integers = 3 × 3 × 11 = 99

2² × 3³× 5¹ × 11² are the greatest powers of the prime factors 2, 3 and 7 respectively involved in the given numbers.

Now, L.C.M of 1485 and 4356 = 2 × 2 × 3 × 3 × 3 × 5 × 11 × 11 = 65430

Given numbers are 1095 and 1168.

The prime factorization of 1095 and 1168 gives:

Here, 73¹ is the smallest power of the common factor 73.

Therefore, the H.C.F of these two integers = 73

2⁴ × 3¹× 5¹ × 73¹ are the greatest powers of the prime factors 2, 3, 5 and 73 respectively involved in the given numbers.

Now, L.C.M of 1485 and 4356 = 2 × 2 × 2 × 2 × 3 × 5 × 73= 17520

Given numbers are 6 and 21.

The prime factorization of 6 and 21 gives:

Here, 3¹ is the smallest power of the common factor 3.

Therefore, the H.C.F of these two integers = 3

2¹ × 3¹ × 7¹ are the greatest powers of the prime factors 2, 3 and 7 respectively involved in the given numbers.

Now, L.C.M of 6 and 21 is 2 ×3 × 7 = 42

Given numbers are 96 and 404

The prime factorization of 96 and 404 gives:

96 = 2⁵ × 3 and 404 = 2² × 101

Therefore, the H.C.F of these two integers = 2² = 4

Now, the L.C.M of 96 and 404 = 2 × 2 × 2 × 2 × 2 × 3 × 101 = 9696

Now, we have to verify

L.H.S = L.C.M × H.C.F = 9696 × 4 = 38784

R.H.S = Product of two numbers = 96 × 404 = 38784

Hence, L.H.S = R.H.S

So, the product of two numbers is equal to the product of their HCF and LCM.

Given numbers are 852 and 1491

The prime factorization of 852 and 1491 gives:

852 = 2 × 2 × 3 × 71 and 1491 = 3 × 7 × 71

Therefore, the H.C.F of these two integers = 3 × 71 = 213

Now, the L.C.M of 96 and 404 = 2 × 2 × 3 × 7 × 71 = 5964

Now, we have to verify

L.H.S = L.C.M × H.C.F = 5964 × 213 = 1270332

R.H.S = Product of two numbers = 852 × 1491 = 1270332

Hence, L.H.S = R.H.S

So, the product of two numbers is equal to the product of their HCF and LCM.

Given numbers are 777 and 1147

The prime factorization of 777 and 1147 gives:

777 = 3 × 7 × 37 and 1147 = 31 × 37

Therefore, the H.C.F of these two integers = 37

Now, the L.C.M of 96 and 404 = 3 × 7 × 31 × 37 = 24087

Now, we have to verify

L.H.S = L.C.M × H.C.F = 24087 × 37 = 891219

R.H.S = Product of two numbers = 777 × 1147 = 891219

Hence, L.H.S = R.H.S

So, the product of two numbers is equal to the product of their HCF and LCM.

Given numbers are 36 and 64

The prime factorization of 36 and 64 gives:

36 = 2 × 2 × 3 × 3 and 64 = 2⁶

Therefore, the H.C.F of these two integers = 2 × 2 = 4

Now, the L.C.M of 36 and 64 = 3 × 3 × 2⁶ = 576

Now, we have to verify

L.H.S = L.C.M × H.C.F = 576 × 4 = 2304

R.H.S = Product of two numbers = 36 × 64 = 2304

Hence, L.H.S = R.H.S

So, the product of two numbers is equal to the product of their HCF and LCM.

Given numbers are 32 and 80

The prime factorization of 32 and 80 gives:

32 = 2⁵ and 80 = 2⁴ × 5

Therefore, the H.C.F of these two integers = 2⁴ = 16

Now, the L.C.M of 32 and 80 = 5 × 2⁵ = 160

Now, we have to verify

L.H.S = L.C.M × H.C.F = 160 × 16 = 2560

R.H.S = Product of two numbers = 32 × 80 = 2560

Hence, L.H.S = R.H.S

So, the product of two numbers is equal to the product of their HCF and LCM.

Given numbers are 902 and 1517

The prime factorization of 902 and 1517 gives:

902 = 2 × 11 × 41 and 1517 = 37 × 41

Therefore, the H.C.F of these two integers = 41

Now, the L.C.M of 902 and 1517 = 2 ×

11 × 37 × 41 = 33374

Now, we have to verify

L.H.S = L.C.M × H.C.F = 33374 × 41 = 1368334

R.H.S = Product of two numbers = 902 × 1517 = 1368334

Hence, L.H.S = R.H.S

So, the product of two numbers is equal to the product of their HCF and LCM.

Given numbers are 6, 72 and 120

Factorization of 6, 72 and 120

Here, 2¹ × 3¹ are the smallest powers of the common factors 2 and 3, respectively.

So, HCF (6, 72, 120) = 2 × 3 = 6

2³ × 3² × 5¹ are the greatest powers of the prime factors 2, 3 and 5 respectively involved in the given three numbers .

LCM of these three integers = 2 × 2 × 2 × 3 × 3 × 5 = 360

Given numbers are 8, 9 and 25

Factorization of 8, 9 and 25

8 = 2 × 2 × 2 × 1 = 2³ × 1

9 = 3 × 3 × 1 = 3² × 1

25 = 5 × 5 × 1 = 5² × 1

Here, 1¹ is the smallest power of the common factor 1.

So, HCF (8, 9, 25) = 1

2³ × 3² × 5² are the greatest powers of the prime factors 2, 3 and 5 respectively involved in the given three numbers .

LCM of these three integers = 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800

Given numbers are 12, 15 and 21

Factorization of 12, 15 and 21

Here, 3¹ is the smallest power of the common factor 3.

So, HCF (12, 15, 21) = 3

2² × 3¹ × 5¹ × 7¹ are the greatest powers of the prime factors 2, 3, 5 and 7 respectively involved in the given three numbers .

LCM of these three integers = 2 × 2 × 3 × 5 × 7 = 420

Given numbers are 36, 45 and 72

Factorization of 36, 45 and 72

Here, 3² is the smallest power of the common factor 3.

So, HCF (36, 45, 72) = 3 × 3 = 9

2³ × 3² × 5¹ are the greatest powers of the prime factors 2, 3 and 5 respectively involved in the given three numbers .

LCM of these three integers = 2 × 2 × 2 × 3 × 3 × 5 = 360

Given numbers are 42, 63 and 140

Factorization of 42, 63 and 140

Here, 7¹ is the smallest power of the common factor 7.

So, HCF (42, 63, 140) = 7

2² × 3² × 5¹ × 7¹ are the greatest powers of the prime factors 2, 3, 5 and 7 respectively involved in the given three numbers .

LCM of these three integers = 2 × 2 × 3 × 3 × 5 × 7 = 1260

Given numbers are 48, 72 and 108

Factorization of 48, 72 and 108

Here, 2² × 3¹ are the smallest powers of the common factors 2 and 3 respectively.

So, HCF (48, 72, 108) = 2 × 2 × 3 = 12

2⁴ × 3³ are the greatest powers of the prime factors 2 and 3 respectively involved in the given three numbers .

LCM of these three integers = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432

Given: HCF (96 , 404) = 4

To Find: LCM (96, 404)

We use the formula

L.C.M (a,b) × H.C.F (a,b) = Product of two numbers (a×b)

LCM (96, 404) × HCF (96, 404) = 96 × 404

⇒ LCM (96, 404) × 4 = 96 × 404 [HCF (96, 404) = 4]

⇒ LCM (96, 404) =

⇒ LCM (96, 404) = 9696

Given: LCM (72, 126) = 504

To Find: HCF (72, 126)

We use the formula

L.C.M (a,b) × H.C.F (a,b) = Product of two numbers (a×b)

LCM (72, 126) × HCF (72, 126) = 72 × 126

⇒ 504 × HCF (72, 126) = 72 × 126 [LCM(72,126)=504]

⇒ HCF (72, 126) =

⇒ HCF (72 , 126) = 18

Given: HCF (18, 504) = 18

To Find: LCM (18, 504)

We use the formula

L.C.M (a,b) × H.C.F (a,b) = Product of two numbers (a×b)

LCM (18, 504) × HCF (18, 504) = 18 × 504

⇒ LCM (18, 504) × 18 = 18 × 504 [HCF(18, 504) = 18]

⇒ LCM (18, 504) =

⇒ LCM (18, 504) = 504

Given: LCM (96, 168) = 672

To Find: HCF (96, 168)

We use the formula

L.C.M (a,b) × H.C.F (a,b) = Product of two numbers (a×b)

LCM (96, 168) × HCF (96, 168) = 96 × 168

⇒ 672 × HCF (96, 168) = 96 × 168 [LCM(96, 168)=672]

⇒ HCF (96, 168) =

⇒ HCF (96, 168) = 24

Given: HCF (306, 657) = 9

To Find: LCM (306, 657)

We use the formula

L.C.M (a,b) × H.C.F (a,b) = Product of two numbers (a×b)

LCM (306, 657) × HCF (306, 657) = 306 × 657

LCM (306, 657) × 9 = 306 × 657 [HCF(306,657)= 9]

LCM (306, 657) =

⇒ LCM (306, 657) = 22338

Given: HCF (36, 64) = 4

To Find: LCM (36, 64)

We use the formula

L.C.M (a,b) × H.C.F (a,b) = Product of two numbers (a×b)

LCM (36, 64) × HCF (36, 64) = 36 × 64

LCM (36, 64) × 4 = 36 × 64 [HCF (36, 64)= 4]

LCM (36, 64) =

LCM (36, 64) = 576

If (15)ⁿ end with the digit 0, then the number should be divisible by 2 and 5.

As 2 × 5 = 10

This means the prime factorization of 15ⁿ should contain prime factors 2 and 5.

But (15)ⁿ = (3 × 5)ⁿ and it does not have the prime factor 2 but have 3 and 5.

, 2 is not present in the prime factorization, there is no natural number nor which 15ⁿ ends with digit zero.

So, 15ⁿ cannot end with digit zero.

If (24)ⁿ end with the digit 5, then the number should be divisible by 5.

This means the prime factorization of 24ⁿ should contain prime factors 5.

But (24)ⁿ = (2³ × 3)ⁿ and it does not have the prime factor 5 but have 3 and 2.

, 5 is not present in the prime factorization, there is no natural number nor which 24ⁿ ends with digit 5.

So, 24ⁿ cannot end with digit 5.

If (21)ⁿ end with the digit 0, then the number should be divisible by 2 and 5.

As 2 × 5 = 10

This means the prime factorization of 21ⁿ should contain prime factors 2 and 5.

But (21)ⁿ = (3 × 7)ⁿ and it does not have the prime factor 2 and 5 but have 3 and 7.

, 2 and 5 is not present in the prime factorization, there is no natural number nor which 21ⁿ ends with digit zero.

So, 21ⁿ cannot end with digit zero.

If (8)ⁿ end with the digit 5, then the number should be divisible by 5.

This means the prime factorization of 8ⁿ should contain prime factor 5.

But (8)ⁿ = (2³)ⁿ and it does not have the prime factor 5 but have 2. So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorization of 8ⁿ.

, 5 is not present in the prime factorization, there is no natural number nor which 8ⁿ ends with digit 5.

So, 8ⁿ cannot end with digit 5.

If (4)ⁿ end with the digit 0, then the number should be divisible by 5.

As 2 × 5 = 10

This means the prime factorization of 4ⁿ should contain prime factor 5.

This is not possible because (4)ⁿ = (2²ⁿ), so the only prime in the factorization of 4ⁿ is 2. So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorization of 4ⁿ.

, 5 is not present in the prime factorization, there is no natural number nor which 4ⁿ ends with digit zero.

So, 4ⁿ cannot end with digit zero.

If (7)ⁿ end with the digit 5, then the number should be divisible by 5.

This means the prime factorization of 7ⁿ should contain prime factor 5.

But (7)ⁿ does not have the prime factor 5. So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorization of 7ⁿ.

, 5 is not present in the prime factorization, there is no natural number nor which 7ⁿ ends with digit 5.

So, 7ⁿ cannot end with digit 5.

Composite number: A whole number that can be divided exactly by numbers other than 1 or itself.

Given 7 x 11 x 13 x 17 +17
17 (7 x 11 x 13 x 17 +1)

17 (7 x 11 x 13 x 17 +1)

17 (17017 + 1)

17 (17018)

17 (2 × 8509)

17 × 2 × 8509

So, given number is the composite number because it is the product of more than one prime numbers.

Composite number: A whole number that can be divided exactly by numbers other than 1 or itself.

5 x 7 x 13 + 5

5 (1 x 7 x 13 +1)

5 (91 +1)

5 (92)

5 (2² × 23)

5 × 2 × 2 × 23

So, given number is the composite number because it is the product of more than one prime numbers.

Composite number: A whole number that can be divided exactly by numbers other than 1 or itself.

5 x 7 x 11 x 13 + 55

5 (1 x 7 × 11 x 13 +11)

5 × 11 (91 +1)

5 × 11 (92)

5 × 11 (2² × 23)

5 × 11 × 2 × 2 × 23

So, given number is the composite number because it is the product of more than one prime numbers.

Lengths of three measuring rods = 64cm, 80cm and 96cm

Least Length of cloth that can be measured = LCM (64, 80, 96)

64 = 2⁶

80 = 2³ × 3 × 5

96 = 2⁵ × 3

So, 2⁶ × 3 × 5 are the greatest powers of the prime factors 2, 3 and 5

LCM (64, 80, 96) = 2⁶ × 3 × 5 = 960

Least Length of cloth that can be measured is 960 cm

Milk in three containers = 27L, 36L, 72L

Biggest measure which can exactly measure the milk = HCF (27, 36, 72)

27 = 3³

36 = 2² × 3²

72 = 2³ × 3²

Here, 3² is the smallest power of the common factor of the prime 3

HCF (27, 36, 72) = 9

So, biggest measure which can exactly measure the milk = 9L

Mixtures of milk and water in three containers = 403kg, 434kg,

465kg

Biggest measure which can exactly measure different quantities = HCF (403, 434, 465)

Here, 31¹ is the smallest power of the common factor.

HCF (403, 434, 465) = 31

So, biggest measure which can exactly measure the milk = 31L

Let us assume that √2 is rational. So, we can find integers p and q (≠ 0) such that √2 = .

Suppose p and q have a common factor other than 1.

Then, we divide by the common factor to get √2 = , where a and b are coprime.

So, b√2 = a.

Squaring on both sides, we get

2b² = a²

Therefore, 2 divides a².

Now, by Theorem which states that Let p be a prime number. If p divides a² , then p divides a, where a is a positive integer,

2 divides a².

So, we can write a = 2c for some integer c

Substituting for a, we get 2b² = 4c² ,i.e. b² = 2c² .

This means that 2 divides b², and so 2 divides b (again using the above Theorem with p = 2). Therefore, a and b have at least 2 as a common factor.

But this contradicts the fact that a and b have no common factors other than 1.

This contradiction has arisen because of our incorrect assumption that 2 is rational.

So, we conclude that √2 is irrational.

Let us assume that √3 is rational.

Hence, √3 can be written in the form

where a and b (≠ 0) are co-prime (no common factor other than 1).

Hence, √3 =

So, b√3 = a.

Squaring on both sides, we get

3b² = a²

= b²

Hence, 3 divides a².

By theorem: Let p is a prime number and p divides a² , then p divides a, where a is a positive integer,

3 divides a also …(1)

Hence, we can say a = 3c for some integer c

Now, we know that 3b² = a²

Putting a = 3c

3b² = (3c)²

3b² = 9c²

b² = 3c²

Hence, 3 divides b²

By theorem: Let p is a prime number and p divides a², then p divides a, where a is a positive integer,

So, 3 divides b also …(2)

By (1) and (2)

3 divides both a and b

Hence, 3 is a factor of a and b

So, a and b have a factor 3

Therefore, a and b are not co-prime.

Hence, our assumption is wrong

Therefore, by contradiction √3 is irrational.

Let us assume that be a rational number.

Then, it will be of the form where a and b are co-prime and b≠0.

Now, =

=

=

Since, 5a is an integer and b is also an integer

So, is a rational number

is a rational number

But this contradicts to the fact that is an irrational number.

Therefore, our assumption is wrong.

Hence, is an irrational number.

Suppose 6^1/3 is rational.

Then, 6^1/3 = for some integers n and m which are co-prime.

So, 6 =

6m³ = n³

So, n³ must be divisible by 6

n must be divisible by 6.

Let n = 6p for some integer p

This gives

6 =

1 =

m³ is divisible by 6

Hence, m must be divisible by 6.

But n and m where co-prime.

So, we have a contradiction.

Hence, (6)^1/3 is irrational

Let us assume that 3 be a rational number.

Then, it will be of the form where a and b are co-prime and b≠0.

Now, =3

=

Since, a is an integer and 3b is also an integer (3b ≠ 0)

So, is a rational number

is a rational number

But this contradicts to the fact that is an irrational number.

Therefore, our assumption is wrong.

Hence, 3 is an irrational number.

Let us assume that 5 be a rational number.

Then, it will be of the form where a and b are co-prime and b≠0.

Now, =5

=

Since, a is an integer and 3b is also an integer (5b ≠ 0)

So, is a rational number

is a rational number

But this contradicts to the fact that is an irrational number.

Therefore, our assumption is wrong.

Hence, 5 is an irrational number.

Let us assume 6 + is rational

6 + can be written in the form where a and b are co-prime.

Hence, 6 + =

= – 6

=

=

Since, rational ≠ irrational

This is a contradiction.

, Our assumption is incorrect.

Hence, 6 + is irrational.

Let us assume 5 - is rational

5 - can be written in the form where a and b are co-prime.

Hence, 5 - =

- = – 5

- =

=-

Since, rational ≠ irrational

This is a contradiction.

, Our assumption is incorrect.

Hence, 5 - is irrational.

Let us assume 2 + is rational

2 + can be written in the form where a and b are co-prime.

Hence, 2 + =

= – 2

=

=

Since, rational ≠ irrational

This is a contradiction.

, Our assumption is incorrect.

Hence, 2 + is irrational.

Let us assume 3 + is rational

3 + can be written in the form where a and b are co-prime.

Hence, 3 + =

= – 3

=

=

Since, rational ≠ irrational

This is a contradiction.

, Our assumption is incorrect.

Hence, 3 + is irrational.

Let us assume - is rational

Let, - =

Squaring both sides, we get

² =

5 - 2 =

2 = – 5

2 =

=

Since, rational ≠ irrational

This is a contradiction.

, Our assumption is incorrect.

Hence, is irrational.

Let us assume - is rational

Let, - =

Squaring both sides, we get

² =

12 - 2 =

2 = – 12

2 =

=

Since, rational ≠ irrational

This is a contradiction.

, Our assumption is incorrect.

Hence, is irrational.

Given rational number is

is terminating if

a) p and q are co-prime &

b) q is of the form of 2ⁿ 5^m where n and m are non-negative integers.

Firstly, we check co-prime

17 = 17 × 1

8 = 2 × 2 × 2

17 and 8 have no common factors

Therefore, 17 and 8 are co-prime.

Now, we have to check that q is in the form of 2ⁿ5^m

8 = 2³

= 1 × 2³

= 5⁰ × 2³

So, denominator is of the form 2ⁿ5^m where n = 3 and m = 0

Thus, is a terminating decimal.

Given rational number is

is terminating if

a) p and q are co-prime &

b) q is of the form of 2ⁿ 5^m where n and m are non-negative integers.

Firstly, we check co-prime

3 = 3 × 1

8 = 2 × 2 × 2

3 and 8 have no common factors

Therefore, 3 and 8 are co-prime.

Now, we have to check that q is in the form of 2ⁿ5^m

8 = 2³

= 1 × 2³

= 5⁰ × 2³

So, denominator is of the form 2ⁿ5^m where n = 3 and m = 0

Thus, is a terminating decimal.

Given rational number is

is terminating if

a) p and q are co-prime &

b) q is of the form of 2ⁿ 5^m where n and m are non-negative integers.

Firstly, we check co-prime

29 = 29 × 1

343 = 7 × 7 × 7

29 and 343 have no common factors

Therefore, 29 and 343 are co-prime.

Now, we have to check that q is in the form of 2ⁿ5^m

343 = 7³

So, denominator is not of the form 2ⁿ5^m

Thus, is a non-terminatingrepeating decimal.

Given rational number is

is terminating if

a) p and q are co-prime &

b) q is of the form of 2ⁿ 5^m where n and m are non-negative integers.

Firstly, we check co-prime

13 = 13 × 1

125 = 5 × 5 × 5

13 and 125 have no common factors

Therefore, 13 and 125 are co-prime.

Now, we have to check that q is in the form of 2ⁿ5^m

125 = 5³

= 1 × 2³

= 2⁰ × 5³

So, denominator is of the form 2ⁿ5^m where n = 0 and m = 3

Thus, is a terminating decimal.

Given rational number is

is terminating if

a) p and q are co-prime &

b) q is of the form of 2ⁿ 5^m where n and m are non-negative integers.

Firstly, we check co-prime

27 = 3 × 3 × 3

8 = 2 × 2 × 2

27 and 8 have no common factors

Therefore, 27 and 8 are co-prime.

Now, we have to check that q is in the form of 2ⁿ5^m

8 = 2³

= 1 × 2³

= 5⁰ × 2³

So, denominator is of the form 2ⁿ5^m where n = 3 and m = 0

Thus, is a terminating decimal.

Given rational number is

is terminating if

a) p and q are co-prime &

b) q is of the form of 2ⁿ 5^m where n and m are non-negative integers.

Firstly, we check co-prime

7 = 7 × 1

80 = 2 × 2 × 2 × 2 × 5

7 and 80 have no common factors

Therefore, 7 and 80 are co-prime.

Now, we have to check that q is in the form of 2ⁿ5^m

80 = 2⁴ × 5

So, denominator is of the form 2ⁿ5^m where n = 4 and m = 1

Thus, is a terminating decimal.

Given rational number is

is terminating if

a) p and q are co-prime &

b) q is of the form of 2ⁿ 5^m where n and m are non-negative integers.

Firstly, we check co-prime

64 = 2⁶

455 = 5 × 7 × 13

64 and 455 have no common factors

Therefore, 64 and 455 are co-prime.

Now, we have to check that q is in the form of 2ⁿ5^m

455 = 5 × 7 × 13

So, denominator is not of the form 2ⁿ5^m

Thus, is a non-terminating repeating decimal.

Given rational number is

=

is terminating if

a) p and q are co-prime &

b) q is of the form of 2ⁿ 5^m where n and m are non-negative integers.

Firstly, we check co-prime

2 and 5 have no common factor

Therefore, 2 and 5 are co-prime.

Now, we have to check that q is in the form of 2ⁿ5^m

5 = 5¹ × 1

= 5¹ × 2⁰

So, denominator is of the form 2ⁿ5^m where n = 0 and m = 1

Thus, is a terminating decimal.

Given rational number is

=

is terminating if

a) p and q are co-prime &

b) q is of the form of 2ⁿ 5^m where n and m are non-negative integers.

Firstly, we check co-prime

7 = 1 × 7

10 = 2 × 5

7 and 10 have no common factor

Therefore, 7 and 10 are co-prime.

Now, we have to check that q is in the form of 2ⁿ5^m

10 = 5¹ × 2¹

So, denominator is of the form 2ⁿ5^m where n = 1 and m = 1

Thus, is a terminating decimal.

Given rational number is

is terminating if

a) p and q are co-prime &

b) q is of the form of 2ⁿ 5^m where n and m are non-negative integers.

Firstly, we check co-prime

129 = 3 × 43

Denominator = 2² ×5⁷ ×7⁵

129 and 2² ×5⁷ ×7⁵ have no common factors

Therefore, 129 and 2² ×5⁷ ×7⁵ are co-prime.

Now, we have to check that q is in the form of 2ⁿ5^m

Denominator = 2² ×5⁷ ×7⁵

So, denominator is not of the form 2ⁿ5^m

Thus, is a non-terminating repeating decimal.

Given rational number is

is terminating if

a) p and q are co-prime &

b) q is of the form of 2ⁿ 5^m where n and m are non-negative integers.

Firstly, we check co-prime

28 = 7 × 2²

625 = 5⁴

28 and 625 have no common factors

Therefore, 28 and 625 are co-prime.

Now, we have to check that q is in the form of 2ⁿ5^m

625 = 5⁴ × 1

= 5⁴ × 2⁰

So, denominator is of the form 2ⁿ5^m where n = 0 and m = 4

Thus, is a terminating decimal.

Given rational number is

is terminating if

a) p and q are co-prime &

b) q is of the form of 2ⁿ 5^m where n and m are non-negative integers.

Firstly, we check co-prime

29 = 29 × 1

243 = 3⁵

29 and 243 have no common factors

Therefore, 29 and 243 are co-prime.

Now, we have to check that q is in the form of 2ⁿ5^m

243 = 3⁵

So, the denominator is not of the form 2ⁿ5^m

Thus, is a non- terminatingrepeating decimal.

We know, =

Multiplying and dividing by 5³

=

=

=

=

=

= 2.125

We know, =

Multiplying and dividing by 5³

=

=

=

=

=

= 0.375

We know, =

Given rational number is

is terminating if

a) p and q are co-prime &

b) q is of the form of 2ⁿ 5^m where n and m are non-negative integers.

Firstly we check co-prime

29 = 29 × 1

343 = 7 × 7 × 7

29 and 343 have no common factors

Therefore, 29 and 343 are co-prime.

Now, we have to check that q is in the form of 2ⁿ5^m

343 = 7³

So, the denominator is not of the form 2ⁿ5^m

Thus, is a non-terminatingrepeating decimal.

We know, =

Multiplying and dividing by 2³

=

=

=

=

=

= 0.104

We know, =

Multiplying and dividing by 5³

=

=

=

=

=

= 3.375

We know, =

Multiplying and dividing by 5³

=

=

=

=

=

= 0.0875

We know, =

Since the denominator is not of the form 2ⁿ5^m

has a non-terminating repeating decimal expansion.

We know, = =

Multiplying and dividing by 2¹

=

=

=

= 0.4

We know,

=

=

=

= 0.7

Given rational number is

Since the denominator is not of the form 2ⁿ5^m

has a non-terminating repeating decimal expansion.

We know, =

Multiplying and dividing by 2⁶

=

=

=

=

=

= 0.0448

Given rational number is

=

Since the denominator is not of the form 2ⁿ5^m.

has a non-terminating repeating decimal expansion.

(i) 7.2345

Here, 7.2345 has terminating decimal expansion.

So, it represents a rational number.

i.e. 7.2345 = =

Thus, q = 10⁴, those factors are 2³ × 5³

(ii)

is non-terminating but repeating.

So, it would be a rational number.

In a non-terminating repeating expansion of ,

q will have factors other than 2 or 5.

(iii) 23.245789

23.245789 is terminating decimal expansion

So, it would be a rational number.

i.e. 23.245789 = =

Thus, q = 10⁶, those factors are 2⁵ × 5⁵

In a terminating expansion of , q is of the form 2ⁿ5^m

So, prime factors of q will be either 2 or 5 or both.

(iv)

is non-terminating but repeating.

So, it would be a rational number.

In a non-terminating repeating expansion of ,

q will have factors other than 2 or 5.

(v) 0.120120012000120000…

0.120120012000120000… is non-terminating and non-repeating.

So, it is not a rational number as we see in the chart.

(vi) 23.142857

23.142857 is terminating expansion.

So, it would be a rational number.

i.e. 23.142857 = =

Thus, q = 10⁶, whose factors are 2⁵ × 5⁵

In a terminating expansion of , q is of the form 2ⁿ5^m

So, prime factors of q will be either 2 or 5 or both.

(vii) 2.313313313331…

2.313313313331… is non-terminating and non-repeating.

So, it is not a rational number as we see in the chart.

(viii) 0.02002000220002…

0.02002000220002… is non-terminating and non-repeating.

So, it is not a rational number as we see in the chart.

(ix) 3.300030000300003…

3.300030000300003… is non-terminating and non-repeating.

So, it is not a rational number as we see in the chart.

(x) 1.7320508…

1.7320508… is non-terminating and non-repeating.

So, it is not a rational number as we see in the chart.

(xi) 2.645713

2.645713 is terminating expansion

So, it would be a rational number.

i.e. 2.645713 = =

Thus, q = 10⁶, those factors are 2⁵ × 5⁵

In a terminating expansion of , q is of the form 2ⁿ5^m

So, prime factors of q will be either 2 or 5 or both.

(xii) 2.8284271…

2.8284271… is non-terminating and non-repeating.

So, it is not a rational number as we see in the chart.