**Download NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.1 – Real Numbers**

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

**(i) 135 and 225 **

**(ii) 196 and 38220 **

**(iii) 867 and 255**

Solution: Euclid’s division lemma state that “Given positive integers and , there exist unique integers and satisfying a = bq + r, 0≤r<b”

If , r = 0 then b, is called HCF of a and b.

**(i)** Here 225 > 135. Applying division lemma to 225 and 135, we get

225 = 135 x 1 + 90

Since the remainder 90 ≠ 0, again apply the division lemma to 90 and 135

135 = 90 x 1 + 45

Since now remainder is 45 ≠ 0, apply division algorithm to 45 and 90

90 = 45 x 2 + 0

The remainder is now become zero and the divisor is 45 here.

Hence the HCF of 135 and 225 is 45

**(ii)** Since 38220 > 196, Applying division lemma to 38220 and 196, we get

38220 = 196 x 195 + 0

The remainder is become zero and the divisor is 196.

Hence the HCF of 38220 and 196 is 196.

**(iii) **Here 867 > 255. Applying division lemma to 867 and 255, we get

867 = 255 x 3 + 102

Since the remainder 102 ≠ 0, again apply the division lemma to 102 and 255

255 = 102 x 2 + 51

Since now remainder is 51 ≠ 0, apply division algorithm to 102 and 51

102 = 51 x 2 + 0

The remainder is now become zero and the divisor is 51 here.

Hence the HCF of 867 and 255 is 51

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

**Solution:** Let a be positive odd integer which gives q as quotient and r as remainder when it is divide by 6.

Then using Euclid’s division lemma a = bq + r, 0 ≤ r < b we get,

a = 6q + r, 0 ≤ r < 6,

So possible values of r are 0, 1, 2, 3, 4 and 5.

Now, when r = 0, a = 6q = 2 x 3q, which is divisible by 2 so 6q is even positive number.

when, r = 1, a = 6q + 1, not divisible by 2, hence 6q + 1 is odd positive number.

when, r = 2, a = 6q + 2 = 2(3q + 1 ) i.e., divisible by 2, hence 6q + 2 is even positive number.

when, r = 3, a = 6q + 3 = 3(2q + 1), not divisible by 2, so 6q + 3 is odd positive number.

when, r = 4, a = 6q + 4 = 2(3q + 2), divisible by 2, hence 6q + 4 is even positive number.

when, r = 5, a = 6q + 5, not divisible by 2, so it is an odd positive integer.

Hence only odd positive integers are 6q + 1, 6q + 3, 6q + 5.

Therefore, any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5, where q is some integer.

**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 numbers of columns. What is the maximum number of column in which they can march?**

**Solution:** To find the maximum number of columns we have to find the HCF of 616 and 32.

Use the Euclid’s division algorithm to 616, 32 we have

616 = 32 x 19 + 8

32 = 8 x 4 + 0

Since remainder is 0 and divisor is 8 hence HCF of is 616, 32 is 8

Therefore, the maximum number of columns in which the army can march is 8.

**4. Use Euclid’s division lemma two show that the square of any positive integer is either of the form or for some integer .**

**Solution:** Let x be any positive integer, which gives q as quotient and r as remainder when it is divided by 3.

Then using Euclid’s division lemma a = bq + r, 0 ≤ r < b we get,

a = 3q + r, 0 ≤ r < 3,

So possible values of r are 0, 1, 2.

Now, when r = 0, x = 3q

Therefore, x^{2 } = (3q)^{2} = 9q^{2} = 3 x 3q^{2 }= 3m, where m= 3q^{2
}When, r = 1, x = 3q + 1

x^{2} = (3q + 1)^{2} = 9q^{2 }+ 6q + 1 = 3(3q^{2} + 2q) + 1 = 3m + 1 where m = 3q^{2} + 2q

And when r = 2, x = 3q + 2

Hence x^{2} = (3q + 2)^{2
}= 9q^{2 }+ 12q + 4

= 9q^{2} + 12q + 3 + 1

= 3(3q^{2 }+ 4q + 1) + 1

= 3m + 1, where m= 3q^{2 }+ 4q + 1) + 1

Therefore, the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

**5. Use Euclid division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8**

**Solution:** Let x be any positive integer, which gives q as quotient and r as remainder when it is divided by 9.

Then using Euclid’s division lemma a = bq + r, 0 ≤ r < b we get,

a = 9q + r, 0 ≤ r < 9 we get,

So possible values of r are 0, 1, 2, 3, 4, 5, 6, 7, 8.

Now, when r = 0, x = 9q

Therefore, x^{3 }= (9q)^{3} = 9 x 81q^{3} = 9m, where m = 81q^{3
}When r = 1, x = 9q + 1

x^{3} = (9q + 1)^{3} = (9q)^{3} + 3.(9q)^{2}.1 + 3.9q.1^{2} + 1^{3
}9(81q^{3} + 3 x 9q^{2 }+ 3q) + 1

= 9m + 1, where m = 81q^{3} + 3 x 9q^{2 }+ 3q

When r = 2, x = 9q + 2

Hence x^{3} = (9q + 2)^{3
}= (9q)^{3} + 3.(9q)^{2}.2 + 3 x 9q.2^{2} + 2^{3
}= 9(81q^{3} + 3.9.2q^{2} + 3q.2^{2}) + 8

= 9m + 8, where m = 81q^{3} + 3.9.2q^{2} + 3q.2^{2}

When r = 3, x = 9q + 3

Hence x^{3} = (9q + 3)^{3
}= (9q)^{3} + 3.(9q)^{2}.3 + 3.9q.3^{2} + 3^{3
}= 9(81q^{3} + 3.9.q^{2}.3 + 3q.3^{2}) + 27

= 9(81q^{3} + 3.9.q^{2}.3 + 3q.3^{2} + 3)

= 9m, where m = 81q^{3} + 3.9.q^{2}.3 + 3q.3^{2} + 3

When r = 4, x = 9q + 4

Hence x^{3} = (9q + 4)^{3
}= (9q)^{3} + 3.(9q)^{2}.4 + 3.9q.4^{2} + 4^{3
}= 9(81q^{3} + 3.9.q^{2}.4 + 3q.4^{2}) + 64

= 9(81q^{3} + 3.9.q^{2}.4 + 3q.4^{2} + 7) + 1

= 9m + 1, where m = 81q^{3} + 3.9.q^{2}.4 + 3q.4^{2} + 7

Continuing this process for r = 5, 6, 7 and 8, we get that x^{3} is of the form 9m or 9m + 1 or 9m + 8.

Therefore, the cube of any positive integer is either of the form 9m or 9m + 1 or 9m + 8, for some integer m.

**Download NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.1 – Real Numbers**