# Show that any positive odd integer is of the form $6q+1,6q+3$$,$ or $6q+5$ where q is some integer.

Last updated date: 26th Mar 2023

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Answer

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Hint: Use the Euclid’s Division Lemma here which states that if we have two positive integers a and b , then there exist unique integers q and r which satisfy the condition $a=bq+r$ where $0\le r

Complete step-by-step answer:

We will use Euclid’s Division Lemma here which states that if we have two positive integers a and b, then there exist unique integers q and r which satisfy the condition $a=bq+r$ where $0\le rComplete step-by-step answer:

We will use Euclid’s Division Lemma here which states that if we have two positive integers a and b, then there exist unique integers q and r which satisfy the condition $a=bq+r$ where $0\le r

Now for our question, let a be any odd integer and let $b=6$ . Then according to the Euclid’s Division Lemma, we have the following:

$a=6q+r$ , where since r is the remainder, so $0\le r<6$ .

r can either be 1, 2, 3, 4 or 5.

Now, in the equation, $a=6q+r$ we have \[6q=\] even integer as 6 is an even integer.

If \[r=1\] , our equation becomes $a=6q+1$ .

This is an odd integer as $6q$ is an even number and 1 is an odd number. The sum of an even number and an odd number is always an odd number.

Hence, $a=6q+1$ is an odd integer …(1)

If $r=2$ , our equation becomes $a=6q+2$ $$

This is an even integer as $6q$ is an even number and 2 is an even number. The sum of two even numbers is always an even number.

Hence, $a=6q+2$ is an even integer

If $r=3$ , our equation becomes $a=6q+3$

This is an odd integer as $6q$ is an even number and 3 is an odd number. The sum of an even number and an odd number is always an odd number.

Hence, $a=6q+3$ is an odd integer …(2)

If $r=4$ , our equation becomes $a=6q+4$ $$

This is an even integer as $6q$ is an even number and 4 is an even number. The sum of two even numbers is always an even number.

Hence, $a=6q+4$ is an even integer

If $r=5$, our equation becomes $a=6q+5$

This is an odd integer as $6q$ is an even number and 5 is an odd number. The sum of an even number and an odd number is always an odd number.

Hence, $a=6q+5$ is an odd integer …(3)

Therefore, from (1), (2), and (3) any positive odd integer is of the form $6q+1,6q+3$ or $6q+5$ ,where q is some integer.

Note: In this question the facts that the sum of an even number and an odd number is always an odd number and that the sum of two even numbers is always an even number is very important.

Also the value of r cannot exceed b as r is the remainder when a is divided by b and the remainder cannot exceed the divisor which is b in this case.

Now for our question, let a be any odd integer and let $b=6$ . Then according to the Euclid’s Division Lemma, we have the following:

$a=6q+r$ , where since r is the remainder, so $0\le r<6$ .

r can either be 1, 2, 3, 4 or 5.

Now, in the equation, $a=6q+r$ we have \[6q=\] even integer as 6 is an even integer.

If \[r=1\] , our equation becomes $a=6q+1$ .

This is an odd integer as $6q$ is an even number and 1 is an odd number. The sum of an even number and an odd number is always an odd number.

Hence, $a=6q+1$ is an odd integer …(1)

If $r=2$ , our equation becomes $a=6q+2$ $$

This is an even integer as $6q$ is an even number and 2 is an even number. The sum of two even numbers is always an even number.

Hence, $a=6q+2$ is an even integer

If $r=3$ , our equation becomes $a=6q+3$

This is an odd integer as $6q$ is an even number and 3 is an odd number. The sum of an even number and an odd number is always an odd number.

Hence, $a=6q+3$ is an odd integer …(2)

If $r=4$ , our equation becomes $a=6q+4$ $$

This is an even integer as $6q$ is an even number and 4 is an even number. The sum of two even numbers is always an even number.

Hence, $a=6q+4$ is an even integer

If $r=5$, our equation becomes $a=6q+5$

This is an odd integer as $6q$ is an even number and 5 is an odd number. The sum of an even number and an odd number is always an odd number.

Hence, $a=6q+5$ is an odd integer …(3)

Therefore, from (1), (2), and (3) any positive odd integer is of the form $6q+1,6q+3$ or $6q+5$ ,where q is some integer.

Note: In this question the facts that the sum of an even number and an odd number is always an odd number and that the sum of two even numbers is always an even number is very important.

Also the value of r cannot exceed b as r is the remainder when a is divided by b and the remainder cannot exceed the divisor which is b in this case.

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