Exercise 1.1

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

2.Use division algorithm to show that any positive odd integer is of the form 6q+1, or 6q+3 or 6q+5, where q is some integer.

Answer Let n be any positive integer.

By the division algorithm, for any integer n and divisor 6,

n=6q+r, where 0≤r<6 and q is an integer.

So, the possible remainders r are: 0, 1, 2, , ,

Thus, any positive integer n can be written as:
n=6q, 6q+1, 6q+2, 6q+, 6q+, or 6q+

Now,

• If r=0, n=6q even
• If r=1, n=6q+1 odd
• If r=2, n=6q+2
• If r=3, n=6q+3
• If r=4, n=6q+4
• If r=5, n=6q+5

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

3. Use division algorithm to show that the square of any positive integer is of the form 3p or 3p+1.

Answer
Let n be any positive integer.

By the division algorithm, for any integer n and divisor 3,
n=3q+r, where 0≤r<3 and q is an integer.

So, the possible remainders r are: , ,

Thus, any positive integer n can be written as:
n=3q, 3q+1, or 3q+2

Now,

• If r=0, n=3q, so n2=3q2=9q2=33q2 (of the form 3p)
• If r=1, n=3q+1, so n2=3q+12=9q2+6q+1=33q2+2q+1 (of the form 3p+1)
• If r=2, n=3q+2, so n2=3q+22=9q2+12q+4=33q2+4q+1+1 (of the form 3p+1)

Therefore, the square of any positive integer is of the form:
3p or 3p+1, where p is some integer.

4. Use division algorithm to show that the cube of any positive integer is of the form 9m, 9m+1 or 9m+8.

Answer
Let n be any positive integer.

By the division algorithm, for any integer n and divisor 3,
n=3q+r, where 0≤r<3 and q is an integer.

So, the possible remainders r are: , ,

Thus, any positive integer n can be written as:
n=3q, 3q+1, or 3q+2

Now,

• If r=0, n=3q, so n3=3q3=q3=3q3 (of the form 9m)
• If r=1, n=3q+1, so n3=3q+13=q3+q2+q+1=93q3+3q2+q+1 (of the form 9m+1)
• If r=2, n=3q+2, so n3=3q+23=q3+q2+q+8=9q3+q2+4q+8 (of the form 9m+8)

Therefore, the cube of any positive integer is of the form:
9m, 9m+, or m+8, where m is some integer.

5. Show that one and only one out of n, n+2 or n+4 is divisible by 3, where n is any positive integer.

Answer
Let n be any positive integer.

By the division algorithm, for any integer n and divisor 3,
n=3q+r, where 0≤r<3 and q is an integer.

So, the possible remainders r are: , ,

Thus, any positive integer n can be written as:
n=3q, 3q+1, or 3q+2

Now,

• If n=3q, then n is divisible by 3,
  n+2=3q+2,
  n+4=3q+4=3q+1 (since 4 mod 3 is 1)
  So, only n is divisible by 3.

• If n=3q+1, then
  n+2=3q+3=3q+1 (divisible by 3),
  n+4=3q+5=3q+2 (since 5 mod 3 is 2)
  So, only n+2 is divisible by 3.

• If n=3q+2, then
  n+2=3q+4=3q+1,
  n+4=3q+6=3q+2 (divisible by 3)
  So, only n+4 is divisible by 3.

Therefore, one and only one out of n, n+2 or n+4 is divisible by 3, for any positive integer n.