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

i. 135 and 225

ii. 196 and 38220

iii. 867 and 255

i. 135 and 225

In 135 and 225, 225 is greater than 135. Therefore, by Euclid’s division algorithm, we have,

• 225 = 135 × 1 + 90, where 135 is the divisor, 1 is the quotient and 90 is the remainder. Since remainder 90 \not =0, apply division lemma for 135 and 90.
• 135 = 90 × 1 + 45. Since remainder 45 \not =0, apply division lemma for 90 and 45.
• 90 = 45 × 2 + 0. Since remainder is zero here, we stop our procedure.

Hence, the HCF of 225 and 135 is 45.

ii. 196 and 38220

Here 38220 is greater than 196, so by Euclid’s division algorithm, we have

• 38220 = 196 × 195 + 0. Here the remainder is zero. So we stop the procedure here.

Hence, the HCF of 196 and 38220 is 196.

iii. 867 and 255

Here 867 is greater than 255, so by Euclid’s division algorithm, we have

• 867 = 255 × 3 + 102. Since remainder 102 \not =0, apply division lemma for 255 and 102.
• 255 = 102 × 2 + 51. Since remainder 51 \not =0, apply division lemma for 102 and 51.
• 102 = 51 × 2 + 0. Since remainder is equal to zero, we stop the process 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.

Let’s first assume that any positive integer a of the form 6q + r, where q is some integer.

i.e. a = bq + r;  0 \leb

Here a = 6q +r ; 0 \le r < 6.

Here  0 ≤ r < 6 which means the possible remainders are 0, 1, 2, 3, 4, 5.

Therefore, possible values of a are 6q + 0, 6q + 1, 6q + 2, 6q + 3, 6q + 4, 6q + 5.

If a = 6q, 6q+2, 6q+4, then a is an even number, which is divisible by 2.

Thus we proved that any positive odd integer is of the form of 6q+1, 6q+3 and 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 number of columns. What is the maximum number of columns in which they can march?

To find the maximum number of columns in which army contingent can march can be found using HCF of 616 and 32. 

Since, 616>32, therefore,

• 616 = 32 × 19 + 8. Here 8 \not =0. We continue the process with 32 and 8.
• 32 = 8 × 4 + 0. Here remainder = 0. 

Hence HCF of 616 and 32 is 8. Hence, the maximum number of columns in which they can march is 8.

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.

Let a be any positive integer and it is in the form 3q, 3q+1, 3q+2.

If a =3q

By squaring both side we get (a)^2=(3q)^2.

a^2=9q^2=3(3q^2)=3m

where m =(3q^2) \\ a^2 = 3m……..(i)

If a = 3q + 1

By squaring both sides we get 

(a)^2=(3q+1)^2 \\ a^2 = 9q^2 + 1 + 2\times3q\times1 \\ a^2=3(3q^2+2q)+1=3m+1 \\where \space m=3q^2+2q\\i.e. \space a^2=3m+1……..(ii)

If a = 3q + 2

By squaring both sides we get

a^2=(3q+2)^2=9q^2+12q+4\\3(3q^2+4q+1)+1=3m+1\\where \space m=3q^2+4q+1\\ \therefore a^2=3m+1….(iii)

Hence, from equation (i), (ii) and (iii), we proved that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

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.

Let x be any positive integer and it is in the form 9m, 9m+1, 9m+8

If a = 3q

a^3=27q^3 \\ a^3=9\times3q^3=9m……(i) \\where \space m=3q^3

If a = (3q+1)

Cubing both sides we get

a^3=(3q+1)^3 \\ We \space know \space that \space (a+b)^3=a^3+b^3+3a^2b+3ab^2 \\ a^3=(3q)^3+1^3+3 \times(3q)^2\times1+3\times1\times3q \\a^3=27q^3+27q^2+9q+1 \\a^3=9(3q^3+3q^2+q)+1 \\a^3=9m+1……(ii) \\ where m=3q^3+3q^2+q

If a = 3q+2

$$$$a^3=(3q+2)^3 \\a^3=(3q)^3+2^3+3\times(3q)^2\times2+3\times(2)^2\times3q \\a^3=27q^3+54q^2+36q+8 \\a^3=9(3q^3+6q^2+4q)+8 \\a^3=93+8……(iii)

Hence, from equation (i), (ii) and (iii), we proved that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.