CBSE Class 10 - Maths - Real Numbers (NCERT Ex 1.1)

Real Numbers

(NCERT Ex 1.1)

1. Use Euclid’s division algorithm to find the HCF of :
 
(i) 135 and 225 
(ii) 196 and 38220 
(iii) 867 and 255

Answer:
(i) 135 and 225
Step 1 : Since 225 > 135, apply Euclid’s division lemma to 225 and 135 to get
                                225 = 135 × 1 + 90

Step 2: Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to get
                                135 = 90 × 1 + 45

Step 3: Now the remainder r = 45 ≠ 0 so we apply Euclid’s division lemma to 90 and 45 to have
                                 90 = 2 × 45 + 0

Step 4: The remainder has now become zero. Since the divisor at this stage is 45,
            ∴ HCF(135,225) =45

(ii) 196 and 38220
Step 1: Since 38220 > 196, apply the division lemma to 38220 and 196 to get
                            38220 = 196 × 195 + 0

Step 2: Since remainder is zero, the process stops. Since the divisor at this stage is 196,
             ∴ HCF(38220, 196) = 196

(iii) 867 and 255

Step 1: Since 867 > 255, we apply the division lemma to 867 and 255 to get
                   867 = 255 × 3 + 102
Step 2: Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102 to have
                   255 = 102 × 2 + 51
Step 3: Now the new divisor  is 102 and new remainder is 51. Apply the division lemma to get
                   102 = 51 × 2 + 0
Step 4: Since the remainder is zero, the process stops. Since the divisor at this stage is 51,
              ∴ HCF(867, 255) = 51



Q2: Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

Answer: Let a be any odd positive integer and b = 6.
By applying Euclid’s algorithm, we have
      a = 6q + r          for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 6.

∴ a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5
Since a is an odd integer, it cannot take 6q, 6q + 2 and 6q + 4 as values.

Also, 6q + 1 = 2 × 3q + 1 = 2k1 + 1, where k₁ is a positive integer
6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k₂ + 1, where k₂ is an integer
6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k₃ + 1, where k₃ is an integer

It is clear from above three statements that, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.

Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2.
Hence, these expressions of numbers are odd numbers.
∴ any odd integer can be expressed in the form 6q + 1, or 6q + 3, or 6q + 5

Q3: 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?

Answer: HCF (616, 32) will give the maximum number of columns in which the Army contingent and the band can march.

Applying Euclid’s algorithm to find  HCF (616, 32),

Step 1: 616 = 32 × 19 + 8
Step2: 32 = 8 × 4 + 0
Step3: Since the remainder is zero, the divisor at this stage is 8. ∴ HCF (616, 32) = 8

Thus, they can march in 8 columns each.

Q4: 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 the form 3m or 3m + 1.]

Answer: Let a be any positive integer and b = 3.
Applying Euclid's division lemma to a and b=3, we get

a = 3q + r for some integer q ≥ 0  and r = 0, 1, 2
⇒  a = 3q or 3q + 1 or 3q + 2
Squaring 'a' i.e.

      a² = (3q)²  or  (3q +1 )²  or (3q + 2)²
⇒  a² = 9q²     or  (9q² + 6q + 1 )  or (9q² + 12q + 4)
⇒  a² = 3(3q²)    or  3(3q² + 2q ) + 1   or  3(3q² + 4q + 1) + 1
⇒  a² = 3 k₁     or  3k₂  + 1   or  3k₃  + 1
where  k₁= 3q² ,  k₂= 3q² + 2q  and  k₃=  3q² + 4q + 1
⇒ Since q = 1,2,3... are integers. Their products and sums will also be integers. ∴ k₁, k₂, and k₂ are also integers.

Thus we can say that the square of any positive integer is either of the form 3m or 3m + 1 for any integer m.

Q5: Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Answer: Let a be any positive integer and b = 3. Using Euclid’s division lemma we have
 a = 3q + r, where q ≥ 0 and 0 ≤ r < 3
 ∴ a = 3q or 3q +1 or 3q + 2
Therefore, every number can be represented as these three forms. Therefore three cases arise:

Case 1: When a = 3q,
      a³ = (3q)³ = 27q³ = 9(3q³) = 9m
      Where m is an integer such that m =  3q³

Case 2: When a = 3q + 1,
     a³ = (3q + 1)³
     a³ = (27q³ + 27q² + 9q + 1)  = 9(3q³ + 3q² + q) + 1
     a³ = 9m + 1
     where m is an integer such that m = 3q³ + 3q² + q

Case 3: When a = 3q + 2,
     a³ = (3q + 2)³
     a³ = (27q³ + 53q² + 36q + 8)  = 9(3q³ + 6q² + 4q) + 8
     a³ = 9m +8
     Where m is an integer such that m =  3q³ + 6q² + 4q

Thus the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.

Similar Questions asked in Examinations

Q6: Prove that 6 divides n(n + 1)(2n + 1) for any n ∈ N

Answer: If n = 1 or 2, ⇒ values of expression n(n + 1)(2n + 1) are 6 or 30 which are divisible by 6.
 ∴ n ≥ 3.
Also n and n+1 are consecutive numbers and either of them is an even number.
∴ n or n+1 is divisible by 2
or 2 divides  n(n + 1)(2n + 1)

Also n = 3k or 3k +1 or 3k + 2 for some k ∈ N

Case 1: when n = 3k,
       n(n + 1)(2n + 1) = 3k(3k+1)(6k+1) which is divisible by 3

Case 2: when n = 3k + 1,
       n(n + 1)(2n + 1)= (3k +1)(3k + 2)(6k + 3) = 3(3k +1)(3k + 2)(2k + 1) is divisible by 3.

Case 3: when n = 3k + 2,
       n(n + 1)(2n + 1) = (3k + 2)(3k + 3)(6k +5) = 3(3k + 2)(k + 1)(6k +5) is divisible by 3.

∴ In all cases, 2 and  3 divides n(n +1)(2n + 1)
Since 2 and 3 have no common factor, thus we an say n(n + 1)(2n + 1)is divisible by 6.