Question

show that any positive integer is of the form 3q, 3q + 1 or 3 q + 2 where q is some integer

Answer 1

let x be the integer

x=3q

x²=9q²

x²=3(3q²)

x²=3q [let 3q² be q]

============================================

x=3q+1

x²=(3q+1)²

x²=9q²+6q+1

x²=3(3q²+2q)+1

x²=(3q+1) [let 3q²+2q be q]

============================================

x=3q+2

x²=(3q+2)²

x²=9q²+12q+4

x²=3(3q²+4q+1)+1

x²=(3q+1) [3q²+4q+1 be q]

=============================================

it proves q² is in the form of 3q and (3q+1) but not in the form of (3q+2)

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Answer 2

$\huge \bf{ \red{hey}}$

$\huge{ \mathfrak{ \blue{here \: is \: answer}}}$

let a be any positive integer

then

b= 3

a= bq+r

0≤r<b

0≤r<3

r= 0,1,2

case 1.

r=0

a= bq+r

3q+0

3q

case 2.

r=1

a= 3q+1

3q+1



case3.

r=2

a=3q+2






hence from above it is proved that any positive integer is of the form 3q,3q+1 and 3q+2

$\huge \boxed{ \boxed{ \green{HOPE\: IT \: HELPS}}}$

$\huge{ \pink{thanks}}$