143-NCERT-10-1-Real Numbers - Ex-1.4

NCERT

10th Mathematics

Exercise 1.4

Topic: 1 Real Numbers

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EXERCISE 1.4

Q1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:

(i) 13/3125        (ii) 17/8        (iii) 64/455        (iv) 15/1600        (v) 29/343

(vi) 23/(2³5²)  (vii) 129/(2²5⁷7⁵)     (viii) 6/15    (ix) 35/50       (x) 77/210

Explanation:

1) Let x = p/q be a rational number, such that the prime factorisation of q is of form 2ⁿ 5^m, where n and m are non-negative integers. Then x has a decimal expansion which terminates.

Solution:

(i) 13/3125

1) Here the denominator is 3125.

2) Find the factors of 3125, we get,

3125 = 5 x 625

3125 = 5 x 5 x 125

3125 = 5 x 5 x 5 x 25

3125 = 5 x 5 x 5 x 5 x 5

3125 = 5⁵

3) Here our expression is 13/5⁵, and the denominator is 2ⁿ x 5^m where n=0 and m = 5, so the decimal expansion of 13/3125 is terminating.

(ii) 17/8

1) Here the denominator is 2³, which is of the form 2ⁿ x 5^m where n=3 and m = 0. 

2) So the decimal expansion of 17/8 is terminating.

(iii) 64/455

1) Here the denominator is 455.

2) Find the factors of 455, we get,

455 = 5 x 91

455 = 5 x 7 x 13

3) Here our expression is 13/(5 x 7 x 13), and the denominator is (5 x 7 x 13), which is not of the form 2ⁿ 5^m, so the decimal expansion of 64/455 is a non-terminating repeating decimal expansion.

(iv) 15/1600

1) Here the denominator is 1600.

2) Find the factors of 1600, we get,

1600 = 16 x 100

1600 = 16 x 4 x 25

1600 = 2⁶ x 5²

3) Here our expression is 15/(2⁶ x 5²) and the denominator is 2ⁿ x 5^m where n=6 and m = 2,  so the decimal expansion of 15/1600 is terminating.

(v) 29/343

1) Here the denominator can be written as 7³.

2) The decimal expansion of 29/7³, the denominator is (7³), which is not of form 2ⁿ 5^m, so the decimal expansion of 29/343 is a non-terminating repeating decimal expansion.

(vi) 23/(2³5²)

1) Here the denominator is of the form 2ⁿ x 5^m where n=3 and m = 2. 

2) The decimal expansion of 23/(2³5²) is terminating.

(vii) 129/(2²5⁷7⁵)

1) Here the denominator is not of the form 2ⁿ x 5^m and has 7⁵ as a factor. 

2) So the decimal expansion of 129/(2²5⁷7⁵) is a non-terminating repeating decimal expansion.

(viii) 6/15

1) Here the denominator is 15.

15 = 3 x 5

2) Here the denominator is (3 x 5), which is not of form 2ⁿ 5^m, so the decimal expansion of 6/15 is a non-terminating repeating decimal expansion.

(ix) 35/50

1) Here the denominator is 50.

2) Find the factors of 1600, we get,

50 = 2 x 25

50 = 2¹ x 5²

3) Here our expression is 35/(2¹ x 5²) and the denominator is of the form 2ⁿ x 5^m where n=1 and m = 2,  so the decimal expansion of 35/50 is terminating.

(x) 77/210

1) Here the denominator is 210.

210 = 21 x 10

210 = 3 x 7 x 2 x 5

2) Here the denominator is (3 x 7 x 2 x 5), which is not of the form 2ⁿ 5^m, so the decimal expansion of 77/210 is a non-terminating repeating decimal expansion.

Q2. Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.

Solution:

(i) 13/3125

13/3125 = (13 x 2)/(3125 x 2)

= (26)/(6250)

= (2.6)/(625)

= (2.6 x 2)/(625 x 2)

= (5.2)/(1250)

= (0.52)/(125)

= (0.52 x 2)/(125 x 2)

= (1.04)/(250)

= (0.104)/(25)

= (0.104 x 4)/(25 x 4)

= (0.416)/(100)

= (0.00416)

So 13/3125 = 0.00416

(ii) 17/8

So 17/8 = 2.125

(iii) 64/455.  It is non-terminating.

(iv) 15/1600

15/1600 = (15)/(16 x 100)

= (0.15)/(16)

= (0.15 x 5)/(16 x 5)

= (0.75)/(80)

= (0.075)/(8)

= (0.0375)/(4)

= (0.01875)/(2)

= (0.009375)

So 15/1600 = 0.009375

(v) 29/343.  It is non-terminating.

(vi) 23/(2³5²)

23/(2³5²) = (23)/(2¹2²5²)

= (23)/(2 x 10²)

= (23)/(2 x 100) 

= (11.5)/(100)

= (0.115)

So 23/(2³5²) = 0.115

(vii) 129/(2²5⁷7⁵).  It is non-terminating.

(viii) 6/15

6/15 = (6)/(15)

= (2)/(5)

= (2 x 2)/(5 x 2)

= (4)/(10)

= (0.4)

So 6/15 = 0.4

(ix) 35/50

       35/50 = (35 x 2)/(50 x 2)

                 = (70)/(100)

                 = (0.7)

So 35/50 = 0.7

(x) 77/210.  It is non-terminating.

Q3. The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form, p/q what can you say about the prime factors of q?

Explanation:

1) Let x = p/q be a rational number, such that the prime factorisation of q is of form 2ⁿ 5^m, where n and m are non-negative integers. Then x has a decimal expansion which terminates.

Solution:

(i) 43.123456789

1) 43.123456789 = (43123456789)/(10⁹)

= (43123456789)/(2⁹ x 5⁹)

2) Here the denominator is of the form 2ⁿ 5^m, so the number is rational.

(ii) 0.120120012000120000

1) Here the decimal expansion is neither terminating nor recurring. Therefore, the given number is an irrational number.

      ---------------

(iii) 43.123456789

1) Here the decimal expansion is non-terminating and recurring, so the given number is a rational number of the form p/q where q is not of form 2ⁿ 5^m. The prime factors of q will also have a factor other than 2 or 5.

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142-NCERT-10-1-Real Numbers - Ex-1.3

NCERT

10th Mathematics

Exercise 1.3

Topic: 1 Real Numbers

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EXERCISE 1.3

1. Prove that √5 is irrational.  

Explanation:

1) We can prove this using indirect proof. It is also known as proof by contradiction.

Solution:

1) Let if possible, consider that √5 be a rational number.

2) √5 = p/q where p and q are coprime and q ≠ 0. That is, there is no common factor other than 1.

(√5)² = (p/q)²

5 = (p)²/(q)²

q² = (p)²/5  ---------- equation 1

3) This shows that 5 divides (p)²  i.e. 5 also divides p.

4) So we have, p/5 = r (say).

(p)²/25 = (r)²

(p)² = 25 (r)²  ---------- equation 2

5) From Equation 1 and Equation 2, we have

(q)² = 25 (r)²/5

(q)² = 5 (r)²

(q)²/5 = (r)²

6) This shows that 5 divides (q)²  i.e. 5 also divides q.

7) This shows that 5 is a common factor of p and q, which contradicts our assumption that √5 is a rational number.

8) So, √5 is an irrational number.

Q2. Prove that 3 + 2√5 is irrational.

Explanation:

1) We can prove this using indirect proof. It is also known as proof by contradiction.

Solution:

1) Let 3 + 2√5 be a rational number.

2) So, 3 + 2√5 = p/q where p and q are co-primes and q ≠ 0. That is, there is no common factor other than 1.

3 + 2√5 = p/q 

2√5 = (p/q) - 3

2√5 = [(p-3q)/q]

√5 = (p-3q)/2q

3) As p and q are co-primes, (p-3q)/2q is rational, so √5 is also rational which contradicts the fact that √5 is irrational.

4) So, 3 + 2√5 is an irrational number.

Q3. Prove that the following are irrationals :

(i) 1/√2    (ii) 7√5    (iii) 6 + √2

Explanation:

1) We can prove this using indirect proof. It is also known as proof by contradiction.

Solution:

(i) 1/√2

1) Let if possible, consider that 1/√2 be a rational number.

2) So, 1/√2 = p/q where p and q are co-primes and q ≠ 0. That is, there is no common factor other than 1.

1/√2 = p/q

√2 = q/p

3) As p and q are co-primes, q/p is rational, so √2 is also rational which contradicts the fact that √2 is irrational.

4) So, 1/√2 is an irrational number.

(ii) 7√5

1) Let 7√5 be a rational number.

2) So, 7√5 = p/q where p and q are co-primes and q ≠ 0. That is, there is no common factor other than 1.

7√5 = p/q

√5 = p/7q

3) As p and q are co-primes, p/q is rational, so √5 is also rational which contradicts the fact that √5 is irrational.

4) So, 7√5 is an irrational number.

(iii) 6 + √2

1) Let 6 + √2 be a rational number.

2) So, 6 + √2 = p/q where p and q are co-primes and q ≠ 0. That is, there is no common factor other than 1.

6 + √2 = p/q 

√2 = (p/q) - 6

√2 = (p-6q)/q

3) As p and q are co-primes, (p-6q)/q is rational, so √2 is also rational which contradicts the fact that √2 is irrational.

4) So, 6 + √2 is irrational number.

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141-NCERT-10-1-Real Numbers - Ex-1.2

NCERT

10th Mathematics

Exercise 1.2

Topic: 1 Real Numbers

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EXERCISE 1.2

Q1. Express each number as a product of its prime factors:

(i) 140     (ii) 156     (iii) 3825     (iv) 5005     (v) 7429

Explanation:

We know that every composite number can be expressed as a product of prime numbers. 

Solution:

(i) 140

1) Prime factors of 140:    Note: Apply divisibility test of 2.

 140 = 2 x 70

= 2 x 2 x 35        Note: Apply divisibility test of 5.

= 2 x 2 x 5 x 7

= 2² x 5 x 7

2) So the prime factors of 140 are 2² x 5 x 7

(ii) 156

1) Prime factors of 156:    Note: Apply divisibility test of 2.

 156 = 2 x 78

= 2 x 2 x 39       Note: Apply divisibility test of 3.

= 2 x 2 x 3 x 13

= 2² x 3 x 13

2) The prime factors of 156 are 2² x 3 x 13.

(iii) 3825

1) Prime factors of 3825:    Note: Apply divisibility test of 5.

3825 = 5 x 765

= 5 x 5 x 153        Note: Apply divisibility test of 3.

= 5 x 5 x 3 x 51

= 5 x 5 x 3 x 3 x 17

= 5² x 3² x 17

2) The prime factors of 3825 are 5² x 3² x 17.

(iv) 5005

1) Prime factors of 5005:    Note: Apply divisibility test of 5.

5005 = 5 x 1001            Note: Apply divisibility test of 7.

= 5 x 7 x 143         Note: Apply divisibility test of 11.

= 5 x 7 x 11 x 13

2) So the prime factors of 5005 are 5 x 7 x 11 x 13.

Note: Click here to know the details of the divisibility test of 7.

(v) 7429

Note:

a) 2 is not a factor of 7429 as it is not an even number. So 4, 8, 16, and so on

are also not the factors of 7429.

b) 3 is not a factor of 7429 as the sum of the digits is 22, which is not divisible

by 3. So 6, 9, 12, 15, and so on are also not factors 7429.

c) 5 is not a factor of 7429 as the unit placed digit is not 0 or 5. So 5, 10, 15,

and so on are also not factors 7429.

d) 7 is not a factor of 7429. So 14, 21, and so on are also not factors 7429.

Note: Click here to know the details of the divisibility test of 7.

1) Prime factors of 7429:        Note: Try directly dividing by 17.

7429 = 17 x 437               Note: Try directly dividing by 19.

= 17 x 19 x 23

2) So the prime factors of 7429 are 17 x 19 x 23.

Q2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.

(i) 26 and 91     (ii) 510 and 92     (iii) 336 and 54

Explanation:

1) To get the LCM and HCF of the given numbers, we need to find all prime factors. 

2) HCF means Highest Common Factor, so find the product of all smallest power of common factor in the numbers.

3) LCM means Least Common Multiple, so find the product of the greatest power of each prime factor in the numbers.

4) HCF (p, q) × LCM (p, q) = p × q, where p and q are any two numbers that are positive integers.

Solution:

(i) 26 and 91

1) Now we will find the prime factors of 26.

26 = 2 x 13

2) Now we will find the prime factors of 91.

91 = 7 x 13

3) Here common factor is 13.

4) So HCF of 26 and 91 is 13.

5) Simmilarly, LCM = 2 x 7 x 13 = 14 x 13 = 182.

6) LCM = 182

7) Here, 26 x 91 = 2366 -------- equation 1

8) Here, HCF x LCM = 13 x 182 = 2366 -------- equation 2.

9) From equation 1 & 2, we can say that HCF(26,91) x LCM(26,91) = 26 x 91. 

(ii) 510 and 92

1) Now we will find the prime factors of 510.

510 = 2 x 255

510 = 2 x 3 x 85

510 = 2 x 3 x 5 x 17

2) Now we will find the prime factors of 92.

92 = 2 x 46

92 = 2 x 2 x 23

3) Here common factor is 2.

4) So HCF of 510 and 92 is 2.

5) Simmilarly, LCM = 2 x 2 x 3 x 5 x 17 x 23 = 23460.

6) LCM = 23460

7) Here, 510 x 92 = 46920 -------- equation 1

8) Here, HCF x LCM = 2 x 23460 = 46920 -------- equation 2.

9) From equation 1 & 2, we can say that HCF(510,92) x LCM(510,92) = 510 x 92.

(iii) 336 and 54

1) Now we will find the prime factors of 336.

336 = 2 x 168

336 = 2 x 2 x 84

336 = 2 x 2 x 2 x 42

336 = 2 x 2 x 2 x 2 x 21

336 = 2 x 2 x 2 x 2 x 3 x 7

2) Now we will find the prime factors of 54.

54 = 2 x 27

54 = 2 x 3 x 9

54 = 2 x 3 x 3 x 3

3) Here common factor is 2 x 3 = 6.

4) So HCF of 336 and 54 is 6.

5) Simmilarly, LCM = 2 x 2 x 2 x 2 x 3 x 3 x 3 x 7 = 3024.

6) LCM = 3024

7) Here, 336 x 54 = 18144 -------- equation 1

8) Here, HCF x LCM = 6 x 3024 = 18144 -------- equation 2.

9) From equation 1 & 2, we can say that HCF(336, 54) x LCM(336, 54) = 336 x 54.

Q3. Find the LCM and HCF of the following integers by applying the prime factorisation method.

(i) 12, 15, and 21     (ii) 17, 23, and 29     (iii) 8, 9 and 25

Explanation:

1) Find all prime factors of the given numbers. 

2) HCF means Highest Common Factor, so find the product of all smallest power of common factor in the numbers.

3) LCM means Least Common Multiple, so find the product of the greatest power of each prime factor in the numbers.

Solution:

(i) 12, 15 and 21

1) Now we will find the prime factors of 12.

12 = 2 x 6

12 = 2 x 2 x 3

2) Now we will find the prime factors of 15.

15 = 3 x 5

3) Now we will find the prime factors of 21.

21 = 3 x 7

4) Here common factor is 3.

5) So HCF of 12, 15, and 21 is 3.

6) Simmilarly, LCM = 2 x 2 x 3 x 5 x 7 = 420.

7) LCM = 420.

(ii) 17, 23 and 29

1) Now we will find the prime factors of 17.

17 = 1 x 17

2) Now we will find the prime factors of 23.

23 = 1 x 23

3) Now we will find the prime factors of 29.

29 = 1 x 29

4) Here common factor is 1.

5) So HCF of 17, 23, and 29 is 1.

6) Simmilarly, LCM = 1 x 17 x 23 x 29 = 11339.

7) LCM = 11339.

(iii) 8, 9 and 25

1) Now we will find the prime factors of 8.

8 = 1 x 2 x 4

8 = 1 x 2 x 2 x 2

2) Now we will find the prime factors of 9.

9 = 1 x 3 x 3

3) Now we will find the prime factors of 25.

25 = 1 x 5 x 5

4) Here common factor is 1.

5) So HCF of 8, 9, and 25 is 1.

6) Simmilarly, LCM = 1 x 2 x 2 x 2 x 3 x 3 x 5 x 5 = 1800.

7) LCM = 1800.

Q4. Given that HCF (306, 657) = 9, find LCM (306, 657).

Explanation:

1) We know that HCF (p, q) × LCM (p, q) = p × q, where p and q are any two numbers that are positive integers.

2) Out of the above 3 values, if 2 values are given, then we can find the 3rd value.

Solution:

1) Here HCF (306, 657) = 9.

2) So, HCF of 306 and 657 = 9.

3) Products of 306 and 657 are 306 x 657

4) We know that HCF (p, q) × LCM (p, q) = p × q

So, 

[LCM (306, 657)] = [306 x 657] / HCF (306, 657)

[LCM (306, 657)] = [306 x 657] / 9

[LCM (306, 657)] = [34 x 657]

[LCM (306, 657)] = [22338]

5) So, LCM (306, 657) = 22338.

Q5. Check whether 6ⁿ can end with the digit 0 for any natural number n.

Explanation:

1) If any number ends with the digit 0, then that number must be divisible by 2 and 5 simultaneously.

2) To check whether 6ⁿ ends with 0, we must find whether 6ⁿ has 2 and 5 as prime factors or not.

Solution:

1) Factors of 6ⁿ = (2 x 3)ⁿ  

2) Here 2 is the factor of 6ⁿ, but 5 is not a factor of 6ⁿ.

3) So 6ⁿ can't end with 0 for any natural number n.

Q6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Explanation:

1) Prime numbers have only 2 factors. 1 and the number itself. e.g. the prime number 5 has two factors. 1 and 5 itself.

2) The positive Number, which has factors other than 1, and itself is known as a composite number.

Solution:

1) The first expression is (7 × 11 × 13) + 13. We will simplify it.

(7 × 11 × 13 + 13) = 13 ((7 x 11 x 1) + 1)

= 13 (77 + 1) 

= 13 (78)

= 13 (2 x 39)

= 13 (2 x 3 x 13)

2) The first expression has factors as (2 x 3 x 13 x 13)

3) The number obtained from (7 × 6 × 5 × 4 × 3 × 2 × 1) + 5 is a composite number.

4) The second expression is given as (7 × 11 × 13) + 13. We will simplify it.

(7 × 6 × 5 × 4 × 3 × 2 × 1) + 5 = 5 ((7 x 6 x 4 x 3 x 2 x 1) + 1)

= 5 (1008 + 1) 

= 5 (1009)

5) The second expression has factors as (5 x 1009)

6) The number obtained from (7 × 6 × 5 × 4 × 3 × 2 × 1) + 5 is a composite number.

Q7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time and go in the same direction. After how many minutes will they meet again at the starting point?

Explanation:

1) Sonia takes 18 minutes and Ravi takes 12 minutes to complete one round. 

2) Now, we need to find out how many minutes will they meet again at the same point. 

3) Here we will find LCM of 18 and 12 to get the time when both meet again at the starting point.

Solution:

1) We will find LCM of 18 and 12

2) Now we will find the prime factors of 18.

18 = 2 x 9

18 = 2 x 3 x 3

3) Now we will find the prime factors of 12.

12 = 2 x 6

12 = 2 x 2 x 3

4) Here, the common factor is 2 x 3 = 6.

5) Simmilarly, LCM = 2 x 2 x 3 x 3 = 36.

6) LCM = 36

7) After 36 minutes Sonia and Ravi will meet again at the starting point.

#simple method

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140-NCERT-10-1-Real Numbers - Ex-1.1

NCERT

10th Mathematics

Exercise 1.1

Topic: 1 Real Numbers

EXERCISE 1.1

 Q-1. Use Euclid’s division algorithm to find the HCF of :

        (i) 135 and 225          (ii) 196 and 38220           (iii) 867 and 255

The steps to use Euclid’s division algorithm are as follows.

To obtain the HCF of two positive integers, say p and q, with p > q, follow the steps below:  

Step 1: Apply Euclid’s division lemma, to p and q. So, we find whole numbers,

r and s such that p = qr + s, 0 ≤ s < q.      

Step 2: If s = 0, q is the HCF of p and q. If s ≠ 0, apply the division lemma to q and s

Step 3: Continue the process till the remainder is zero. The divisor at this stage

will be the required HCF.

Solution:

(i) 135 and 225

1) Step 1: Since 225 > 135, apply the division lemma to 135 and 225,

225 = (135 × 1) + 90

2) Step 2: Since the remainder is 90 ≠ 0, apply the division lemma to 135 and 90,

135 = (90 × 1) + 45

3) Step 3: Here the remainder of 45 ≠ 0, apply the division lemma to 90 and 45, 

90 = (45 × 2) + 0.

4) Here, the reminder s = 0, and 45 is the divisor, so 45 is the HCF of

135 and 225.

(ii) 196 and 38220

1) Step 1: Since 38220 > 196, apply the division lemma to 196 and 38220,

38220 = (196 × 195) + 0

2) Here, the reminder s = 0, and 195 is the divisor, so 195 is the HCF of 

196 and 38220.

(iii) 867 and 255

1) Step 1: Since 867 > 255, apply the division lemma to 867 and 255, to get

867 = (255 × 3) + 102

2) Step 2: Since the remainder is 102 ≠ 0, apply the division lemma to 255 and

102, to get

255 = (102 × 2) + 51

3) Step 3: Here the remainder of 51 ≠ 0, apply the division lemma to 102 and 51,

102 = (51 × 2) + 0.

4) Here, the reminder s = 0, and 51 is the divisor, we can say that 51 is the HCF

of 867 and 255.

Q-2. Show that any positive odd integer is of the form 6q + 1, 6q + 3, or 6q + 5,

where q is some integer.

Explanation:

1) As per Euclid’s division algorithm. 

2) Let, for any positive integer ‘p’ in the form of 6q + s, where q is some integer 

(as given in the problem), and 's' be a reminder. This means, 0 ≤ s < 6 i.e. 's' may be any integer between 0 and 6, i.e. s = 0 or 1 or 2 or 3 or 4 or 5 but it can’t be 6 because s < 6. 

3) So, by Euclid’s division lemma, possible values for p will be 6q, 6q+1, 6q+2,

6q+3, 6q+4, or 6q+5.

Solution:

1) Let ‘p’ be any positive integer. So according to Euclid’s algorithm, 

for a = 6q + s some integer q ≥ 0 and s = 0, 1, 2, 3, 4, 5 as 0 ≤ s < 6.

2)  ∴ p can be (6q +0), (6q +1), (6q + 2), (6q +3), (6q + 4), or (6q +5).

3) Since p is odd, it can't be (6q +0), (6q + 2), or (6q + 4), as they all are divisible

by 2.

4) We know that odd numbers are of the form (2n+1), for any integer n.

5) Here we can write (6q +1), (6q +3), (6q +5) in the form of (2n+1) as follows:

(6q+1) = 2(3q) + 1, taking 3q = n_1, so we have

   = 2( n₁) + 1 which is an odd number. 

(6q+3) = 2(3q) + 3, 

   = 2(3q) + 2 + 1

   = 2(3q + 1) + 1 taking (3q + 1) = n_2, so we have 

   = 2( n₂) + 1 which is an odd number.    

(6q+5) = 2(3q) + 5, 

   = 2(3q) + 4 + 1

   = 2(3q + 2) + 1 taking (3q + 2) = n_3, so we have 

   = 2( n₃) + 1 which is an odd number.

6) Therefore, any odd integer is of the form (6q +1), (6q +3), (6q +5).

Q-3. An army contingent of 616 members is to march behind an army band of

32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

Explanation:

1) We know that HCF means: HCF is the highest common factor, which can be

divided exactly into two or more numbers.

2) Here we have to find HCF of 616 and 32 to get the maximum number of columns

in which, they can march.

Solution:

1) Here we will find HCF of (616, 32)

2) Using Euclid’s algorithm,

Step 1: Since 616 > 32,

616 = (32 × 19) + 8

Step 2: Since the remainder, 8 ≠ 0,

32 = (8 × 4) + 0

Step 3: Here the remainder is 0, so HCF of 616 and 32 is 32.

3) The maximum number of columns they can march is 32.

Q-4. Use Euclid’s division lemma to show that the square of any positive

integer is either of the form 3m or 3m + 1 for some integer m.

[Hint: Let x be any positive integer then it is of the form 3q, 3q + 1, or 3q + 2. Now square each of these and show that they can be rewritten in form 3m or 3m + 1.]

Explanation:

1) Let 'c' be a positive integer. Using Euclid’s lemma, for positive integers c and d,

there exist unique integers q and s, such that c = dq + s, 0 ≤ s < d.

2) If we keep the value of d = 3, then 0 ≤ s < 3, i.e. s = 0, 1, or 2. It can’t be 3

because s<3. 

3) So, the possible values for c = 3q, (3q + 1), or (3q + 2). 

4) Now, find the square of all the possible values of c. 

5) If q is any positive integer, its square will also be a positive integer. 

Solution:

1) Let 'c' be any positive integer and d = 3.

2) So, c = 3q + s for some integer q ≥ 0 and s = 0,1,2 as 0 ≤ s < 3.

3) ∴ c = (3q + 0), (3q + 1), (3q + 2)

4) Now we will find c² , for c = (3q + 0), (3q + 1), (3q + 2)

c² = (3q + 0)²

= 9 q² 

= 3 (3 q²)  take (3 q²) = m

= (3 m)

c² = (3q + 1)²

= 9 q² + 6q + 1

= 3 (3 q² + 2q) + 1   take (3 q² + 2q) = m

= (3 m +1)

c² = (3q + 2)²

= 9 q² + 12q + 4

= 3 (3 q² + 4q +1) + 1   take (3 q² + 2q + 1) = m

= (3 m +1)

5) So, the square of any positive integer is of form 3m or (3m + 1) for some integer m.

Q-5. Use Euclid’s division lemma to show that the cube of any positive integer

is of the form 9m, 9m + 1, or 9m + 8.

Explanation:

1) Let 'c' be a positive integer. Using Euclid’s lemma, for positive integers c and d,

there exist unique integers q and s, such that c = dq + s, 0 ≤ s < d.

2) If we keep the value of d = 3, then 0 ≤ s < 3, i.e. s = 0, 1, or 2. It can’t be 3

because s<3. 

3) So, the possible values for c = 3q, (3q + 1), or (3q + 2). 

4) Now, find the cube of all the possible values of c. 

5) If q is any positive integer, its square will also be a positive integer. 

Solution:

1) Let 'c' be any positive integer and d = 3.

2) So, c = 3q + s for some integer q ≥ 0 and s = 0,1,2 as 0 ≤ s < 3.

3) ∴ c = (3q + 0), (3q + 1), (3q + 2)

4) Now we will find c³ , for c = (3q + 0), (3q + 1), (3q + 2)

c³ = (3q + 0)³

= 27 q³ 

= 9 (3 q³)  take (3 q³) = m

= (9 m)

c³ = (3q + 1)³

= 27 q³ +27 q² + 9q + 1

= 9 (3 q³ + 3 q² + q) + 1   take (3 q² + 3 q² + q) = m

= (9 m +1)

c³ = (3q + 2)³

= 27 q³ +54 q² + 36q + 8

= 9 (3 q³ + 6 q² + 4q) + 8   take (3 q³ + 6 q² + 4q) = m

= (9 m + 8)

5) So, the cube of any positive integer is of the form (9m), (9m + 1), or (9m + 8) for some integer m.

# simple methods

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