Question

show that any positive integer is in form 3q, 3q+1,3q+2 where q is an integer

Answer 1

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Sol.

Let a be any positive integer and b = 3
Than 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

a² = (3q)² or (3q+1)² or (3q+2)²
a² = (9q)² or 9q²+6q+1 or 9q²+12q+4

=> 3 ×(3q)² or 3(3q²+2q)+1 or 3(3q²+4q+1) +1
=>3k1 or 3k2+1 or 3k3+1

where , k1 ,k2 , k3 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

Hope it's helps you

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Answer 2

Let a be any positive integer where due to Euclid's Lemma,

a = 3q + r where 0 ≤ r < 3

Thus, r is equal or greater than 0 but less than 3.

So exepected values of r can be

• r =  0
• r = 1
• r = 2

So, a = 3q + 0 or a = 3q

a = 3q + 1 and a = 3q + 2

So for any positive integer, 3q or 3q + 1 or 3q + 2 are possible values.