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

Answer

Verified

363.9k+ views

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.

Last updated date: 30th Sep 2023

•

Total views: 363.9k

•

Views today: 10.63k

Recently Updated Pages

What do you mean by public facilities

Paragraph on Friendship

Slogan on Noise Pollution

Disadvantages of Advertising

Prepare a Pocket Guide on First Aid for your School

10 Slogans on Save the Tiger

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

What is meant by shramdaan AVoluntary contribution class 11 social science CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

An alternating current can be produced by A a transformer class 12 physics CBSE

What is the value of 01+23+45+67++1617+1819+20 class 11 maths CBSE

Give 10 examples for herbs , shrubs , climbers , creepers