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

(i) 135 and 225

(ii) 196 and 38220

(iii) 867 and 255

(i) 225 > 135 we always divide greater number with smaller one.

Divide 225 by 135 we get 1 quotient and 90 as remainder so that

225= 135 × 1 + 90

Divide 135 by 90 we get 1 quotient and 45 as remainder so that

135= 90 × 1 + 45

Divide 90 by 45 we get 2 quotient and no remainder so we can write it as

90 = 2 × 45+ 0

As there are no remainder so divisor 45 is our HCF.

(ii) 38220 > 196 we always divide greater number with smaller one.

Divide 38220 by 196 then we get quotient 195 and no remainder so we can write it as

38220 = 196 × 195 + 0

As there is no remainder so divisor 196 is our HCF.

(iii) 867 > 255 we always divide greater number with smaller one.

Divide 867 by 255 then we get quotient 3 and remainder is 102 so we can write it as

867 = 255 × 3 + 102

Divide 255 by 102 then we get quotient 2 and remainder is 51 so we can write it as

255 = 102 × 2 + 51

Divide 102 by 51 we get quotient 2 and no remainder so we can write it as

102 = 51 × 2 + 0

As there is no remainder so divisor 51 is our HCF.

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

Let take a as any positive integer and b = 6.

Then using Euclid’s algorithm we get a = 6q + r here r is remainder and value of q is more than or equal to 0 and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < b and the value of b is 6

So total possible forms will 6q + 0 , 6q + 1 , 6q + 2,6q + 3, 6q + 4, 6q + 5

6q + 0

6 is divisible by 2 so it is a even number

6q + 1

6 is divisible by 2 but 1 is not divisible by 2 so it is a odd number

6q + 2

6 is divisible by 2 and 2 is also divisible by 2 so it is a even number

6q +3

6 is divisible by 2 but 3 is not divisible by 2 so it is a odd number

6q + 4

6 is divisible by 2 and 4 is also divisible by 2 it is a even number

6q + 5

6 is divisible by 2 but 5 is not divisible by 2 so it is a odd number

So odd numbers will in form of 6q + 1, or 6q + 3, or 6q + 5

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?

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.

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?

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.