Question

An odd positive integer is of the form 4q+1 (or) 4q+3 where q i some integerthen find the possible remainders

Answer 1

Hey,

Let a be any positive integer 
by EDL  
a = bq +r 
0 ≤ r < b
possible remainders are 0, 1, 2 , 3 

 this shows that a can be in the form of 4q, 4q+1, 4q+2, 4q+3 q is quotient 
as a is odd a  can't be the form of 4q or 4q+2 as they are even
so a will be in the form of 4q + 1 or 4q+3...
Hence proved.

HOPE IT HELPS:-))

Answer 2

Step-by-step explanation:

Let a be the positive integer.

And, b = 4 .

Then by Euclid's division lemma,

We can write a = 4q + r ,for some integer q and 0 ≤ r < 4 .

°•° Then, possible values of r is 0, 1, 2, 3 .

Taking r = 0 .

a = 4q .

Taking r = 1 .

a = 4q + 1 .

Taking r = 2

a = 4q + 2 .

Taking r = 3 .

a = 4q + 3 .

But a is an odd positive integer, so a can't be 4q , or 4q + 2 [ As these are even ] .

•°• a can be of the form 4q + 1 or 4q + 3 for some integer q .

Hence , it is solved .