Question

show that any positive odd integer is in the form 6q+1 or 6q+3 or 6q+5 where q is some integer
​please answer fast

Answer 1

Solution :-

Let 'a'be any positiveinteger and b = 6

Then by euclid's division lemma there exist integers 'q' and 'r' such that a = 6q + r where 0 ≤ r < 6 i.e r = 0, 1, 2,3,4,5

The positive integers will be of the form

• a = 6q + 0
• a = 6q + 1
• a = 6q + 2
• a = 6q + 3
• a = 6q + 4
• a = 6q + 5

6q, 6q + 2, 6q + 4 can be written as 2(3q), 2(3q + 1), 2(3q + 2) respectively which are of the general form of even numbers

Whereas, 6q + 1 , 6q + 3, 6q + 5 can be written as

• 6q + 1 = 2(3q) + 1
• 6q + 3 = 6q + 2 + 1 = 2(3q + 1) + 1
• 6q + 5 = 6q + 4 + 1 = 2(3q + 2) + 1

These are according to the general form of an odd number 2n + 1

Therefore 'a' is an odd positive integer of theform 6q + 1, 6q + 3, 6q + 5

Answer 2

$\Huge{\underline{\underline{\mathfrak{ Solution \colon }}}}$

Let a be a positive integer & b =6

By euclid's algorithm

a = 6q +r and r =0,1,2,3,4,5

Therefore, a= 6q or 6q+1 or 6q+2 or 6q+3 or 6q+4 or 6q+5

Also

6q + 1= 2 × 3q + 1

=2k1 + 1,where k1 is a positive integer

6q + 3= ( 6q + 2 ) + 1 =2 ( 3q +1 ) +1

=2k2 + 1, where k2 is an integer

6q + 5 = ( 6q + 4 ) + 1 = 2 ( 3q + 2 ) + 1

=2k3 + 1 ,where k3 is an integer

Now

6q + 1 ,6q + 3 ,6q + 5 are of the form 2k+1 ,where k is an integer

therefore,

6q +1 ,6q +3 ,6q +5 are not exactly divisible by 2.

Add therefore,any odd integer can be expressed in the form of 6q + 1 ,6q + 3 or 6q+ 5