Belongs to the unit 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:**

(i) 135 and 225

Since 225 > 135, we apply the Euclid’s Division Lemma to 225 and 135 to obtain 225 = 135 × 1 + 90

Since remainder 90 ≠ 0, we apply the Euclid’s Division Lemma to 135 and 90 to obtain 135 = 90 × 1 + 45

We consider the new divisor 90 and new remainder 45, and apply the Euclid’s Division Lemma to obtain 90 = 2 × 45 + 0

**Since the remainder is zero, the process stops**. Since the divisor at this stage is 45, Therefore, the HCF of 135 and 225 is 45.

(ii) 196 and 38220

Since 38220 > 196, we apply the Euclid’s Division Lemma to 38220 and 196 to obtain 38220 = 196 × 195 + 0

**Since the remainder is zero, the process stops**. Since the divisor at this stage is 196, Therefore, HCF of 196 and 38220 is 196.

(iii) 867 and 255

Since 867 > 255, we apply the Euclid’s Division Lemma to 867 and 255 to obtain 867 = 255 × 3 + 102

Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102 to obtain 255 = 102 × 2 + 51.

We consider the new divisor 102 and new remainder 51, and apply the division lemma to obtain 102 = 51 × 2 + 0

**Since the remainder is zero, the process stops.** Since the divisor at this stage is 51, Therefore, 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 any positive integer and *b = 6*.

Then, by Euclid’s algorithm, a = 6q + r for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 6.

Therefore, *a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5 Also, 6q + 1 = 2 × 3q + 1 = 2k _{1} + 1,* where

*k*

_{1}is a positive integer.

*6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k _{2} + 1*, where

*k*is an integer.

_{2}*6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k _{3} + 1, where k_{3} *is an integer

Clearly, 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.

Therefore, any odd integer can be expressed in the form *6q + 1, or 6q + 3, or 6q + 5*

**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? **

**Solution:**

HCF (616, 32) will give the maximum number of columns in which they can march.

We can use Euclid’s algorithm to find the HCF.

616 = 32 × 19 + 8 32 = 8 × 4 + 0

The HCF (616, 32) is 8. Therefore, they can march in 8 columns each.

**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 the form 3m or 3m + 1.]

**Solution:**

Let a be any positive integer and b = 3.

Then a = 3q + r for some integer q ≥ 0 And r = 0, 1, 2 because 0 ≤ r < 3

Therefore, a = 3q or 3q + 1 or 3q + 2 Or,

𝑎^{2} = (3𝑞)^{2} 𝑜𝑟 (3𝑞 + 1)^{2} 𝑜𝑟 (3𝑞 + 2)^{2}

= (3𝑞)^{2} 𝑜𝑟 (9𝑞^{2} + 6𝑞 + 1) 𝑜𝑟 (9𝑞^{2} + 12𝑞 + 4)

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

= 3𝑘_{1} 𝑜𝑟 3𝑘_{2} + 1 𝑜𝑟 3𝑘_{3} + 1 Where k_{1}, k_{2}, and k_{3} are some positive integers

Hence, it can be said that the square of any positive integer is either of the form 3m or 3m + 1.

**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**

**Solution:**

Let *a* be any positive integer and *b = 3, 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. There are three cases.

** Case 1: **When *a = 3q,*

* a ^{3} = (3q)^{3} = 27q^{3} = 9(3q^{3} )= 9m, *where

*m*is an integer such that

*m = 3q*

^{3}**Case 2:** When *a = 3q + 1, a ^{3} = (3q +1)^{3} *

* a ^{3} = 27q^{3} + 27q^{2} + 9q + 1 *

*a ^{3} = 9(3q^{3} + 3q^{2} + q) + 1 *

*a ^{3} = 9m + 1*, where

*m*is an integer such that

*m = (3q*

^{3}+ 3q^{2}+ q)**Case 3:** When *a = 3q + 2, *

*a ^{3} = (3q +2)^{3} *

*a ^{3} = 27q^{3} + 54q^{2} + 36q + 8 *

*a ^{3} = 9(3q^{3} + 6q^{2} + 4q) + 8 *

*a ^{3} = 9m + 8,* where

*m*is an integer such that

*m = (3q 3 + 6q 2 + 4q)*

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

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