**Question** 1:

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

**Answer**:

(i) 135 and 225

Since 225 > 135, we apply the division lemma to 225 and 135 to obtain

225 = 135 × 1 + 90

Since remainder 90 ≠ 0, we apply the 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 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 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 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.

**Question** 2:

Show that any positive odd integer is of the form , or , or , where *q* is some integer.

**Answer**:

Let *a* be any positive integer and *b* = 6. Then, by Euclid’s algorithm,

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

Therefore, *a* = 6*q* or 6*q* + 1 or 6*q* + 2 or 6*q + *3 or 6*q* + 4 or 6*q* + 5

Also, 6*q* + 1 = 2 × 3*q* + 1 = 2*k*_{1} + 1, where* k*_{1} is a positive integer

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

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

Clearly, 6*q* + 1, 6*q* + 3, 6*q* + 5 are of the form 2*k* + 1, where *k* is an integer.

Therefore, 6*q* + 1, 6*q* + 3, 6*q* + 5 are not exactly divisible by 2. Hence, these expressions of numbers are odd numbers.

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

or 6*q* + 5

**Question** 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?

**Answer**:

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.

**Question** 4:

Use Euclid’s division lemma to show that the square of any positive integer is either of form 3*m* or 3*m* + 1 for some integer *m*.

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

**Answer**:

Let *a* be any positive integer and *b* = 3.

Then *a* = 3*q* + *r* for some integer *q* ≥ 0

And *r* = 0, 1, 2 because 0 ≤ *r* < 3

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

Or,

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 3*m* or 3*m* + 1.

**Question** 5:

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

**Answer**:

Let *a* be any positive integer and *b* = 3

*a =* 3*q* + *r*, where *q *≥ 0 and 0 ≤ *r* < 3

Therefore, every number can be represented as these three forms. There are three cases.

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

Where *m* is an integer such that *m* =

**Case 2**: When *a* = 3*q* + 1,

*a**3 **= (*3*q +*1*)**3*

*a**3** = *27*q**3 **+ *27*q**2 **+ *9*q + *1

*a**3* = 9(3*q**3 **+ *3*q**2 **+ q*) + 1

*a**3 **= *9*m* + 1

Where* m* is an integer such that *m* = (3*q**3 **+ *3*q**2 **+ q)*

**Case 3**: When *a* = 3*q* + 2,

*a**3 **= (*3*q +*2*)**3*

*a**3** = *27*q**3 **+ *54*q**2 **+ *36*q + *8

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

*a**3 **= *9*m* + 8

Where *m* is an integer such that *m* = (3*q**3 **+ *6*q**2 **+ *4*q)*

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

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