Question

show that any positive odd integer is of the form of 6q+ 1 or 6q + 3 or 6q + 5 where Q is the some integer

Answer 1

HEY

HERE IS UR ANS

By euclid's division algorithm for two integers 'a' and 'b', we have

a = bq + r where

$0 \leqslant r < b$

Let, b = 6

so, r = 0, 1, 2, 3, 4, 5

Now, a = 6q , a = 6q + 1 , a = 6q + 2

a = 6q + 3 , a = 6q + 4 , a = 6q + 5

clearly, a = 6q , a = 6q + 2 , a = 6q + 4 are even as they are divisible by 2

But, 6q + 1 , 6q + 3 & 6q + 5 are odd as they are not divisible by 2

Therefore, any positive odd integer is of the form 6q + 1 , 6q + 3 or 6q + 5

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Answer 2

HEY FRIEND HERE IS UR ANSWER,

Let a be any positive integer and b = 6. Then, by Euclid’s algorithm, a = 6q + r, for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5, because 0≤r<6.

Now substituting the value of r, we get,

If r = 0, then a = 6q

Similarly, for r= 1, 2, 3, 4 and 5, the value of a is 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5, respectively.

If a = 6q, 6q+2, 6q+4, then a is an even number and divisible by 2. A positive integer can be either even or odd Therefore, any positive odd integer is of the form of 6q+1, 6q+3 and 6q+5, where q is some integer.

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