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
•
Total views: 306.9k
•
Views today: 4.84k
Answer
306.9k+ 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 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.
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 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.
Recently Updated Pages
Paulings electronegativity values for elements are class 11 chemistry CBSE

For a particle executing simple harmonic motion the class 11 physics CBSE

Does Nichrome have high resistance class 12 physics CBSE

The function f satisfies the functional equation 3fleft class 12 maths JEE_Main

Write a letter to the Principal of your school to plead class 10 english CBSE

Look at the handout below Write a letter to the organizers class 11 english CBSE

Trending doubts
What was the capital of Kanishka A Mathura B Purushapura class 7 social studies CBSE

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Tropic of Cancer passes through how many states? Name them.

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

Name the Largest and the Smallest Cell in the Human Body ?
