✕

Find the best tutors and institutes for Class 10 Tuition

Find Best Class 10 Tuition

✕

Search for topics

Follow 5

3 Answers

Let a be any positive integer and b = 6. Then, by Euclid’s algorithm, a = 6q + rfor some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 6. Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5 Also, 6q + 1 = 2 × 3q + 1 = 2k1 + 1, where k1 is a positive integer 6q... read more

Let *a* be any positive integer and *b* = 6. Then, by Euclid’s algorithm,

*a* = 6*q* + *r*for some integer *q*** ≥ **0, and *r* = 0, 1, 2, 3, 4, 5 because 0 **≤ ***r* < 6.

Therefore, *a* = 6*q* or 6*q* + 1 or 6*q* + 2 or 6*q + *3 or 6*q* + 4 or 6*q* + 5

Also, 6*q* + 1 = 2 × 3*q* + 1 = 2*k*_{1} + 1, where* k*_{1} is a positive integer

6*q* + 3 = (6*q* + 2) + 1 = 2 (3*q* + 1) + 1 = 2*k*_{2} + 1, where* k*_{2} is an integer

6*q* + 5 = (6*q* + 4) + 1 = 2 (3*q* + 2) + 1 = 2*k*_{3} + 1, where* k*_{3} is an integer

Clearly, 6*q* + 1, 6*q* + 3, 6*q* + 5 are of the form 2*k* + 1, where *k* is an integer.

Therefore, 6*q* + 1, 6*q* + 3, 6*q* + 5 are not exactly divisible by 2. Hence, these expressions of numbers are odd numbers.

And therefore, any odd integer can be expressed in the form 6*q* + 1, or 6*q* + 3,

or 6*q* + 5

0

Comments

Let 'a' be any +ve integer. Also suppose there exists two +ve integers q and r as quotient and remainder respectively when 'a' is divided by 6. Then , By using Euclid's Division lemma, we get a = 6q + r .................( A ) , where 0 ≤ r < 6. Obviously, r = 0 or 1 or 2 or 3 or... read more

Let 'a' be any **+ve** integer.

Also suppose there exists two +ve integers **q and** r as quotient and remainder respectively when 'a' is divided by **6. **

**Then , **

**By using**** Euclid's Division lemma, we get **

a = 6q + r .................( A ) , where **0 ≤ r < 6.**

Obviously, r = 0 or **1** or **2** or **3** or **4** or **5.**

**Case-(i) **When r = 0 , then from ( A ) we have

*a = 6q + 0 *

=> **a = 6q = 2×3q = 2m _{ }**

**Case-(ii) When r = 1 , then **

**a = 6q + 1 = 2(3q) + 1 = 2m + 1 = odd number, where m = 3q.**

**Case-(iii) When r = 2 then **

**a = 6q + 2 = 2(3q + 1) = 2m _{1 }**

**Case-(iv) When r = 3 then **

**a = 6q + 3 = 2(3q + 1) + 1 = 2m _{1 }**

*Case-(v) When r= 4 then*

*a = 6q + 4 = 2(3q + 2) = 2m _{2 }*

**Case -(vi) When r = 5 then **

**a = 6q + 5 = 2(3q + 2) + 1 = 2m _{2 }+ 1 = odd number, where m_{2 = 3q + 2.}**

**Clearly from above cases ( ii ) , ( iv ) and ( vi ) we conclude that any +ve odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5.**

1

Comments

Let a and b are two integers, where a>b, then according to Euclid division lemma a=bq+r where 0<=r<b, Here let b=6, then a=6q+r so 0<=r<6 When r=0, a=6q, when r=1,a=6q+1, for r=2, a=6q+2, for r=3, a=6q+3, for r=4, a=6q+4, for r=5, a=6q+5.here 6q,6q+2 and 6q+4 are divisible by 2, so even. 6q+1,6q+3... read more

Let a and b are two integers, where a>b, then according to Euclid division lemma a=bq+r where 0<=r<b, Here let b=6, then a=6q+r so 0<=r<6

When r=0, a=6q, when r=1,a=6q+1, for r=2, a=6q+2, for r=3, a=6q+3, for r=4, a=6q+4, for r=5, a=6q+5.here 6q,6q+2 and 6q+4 are divisible by 2, so even.

6q+1,6q+3 and 6q+5 are not divisible by 2. so they are odd.

read less 0

Comments

View 1 more Answers

Now ask question in any of the 1000+ Categories, and get Answers from Tutors and Trainers on UrbanPro.com

Ask a QuestionRecommended Articles

Quest Academy - Institute of the month

Quest Academy is a professional Bangalore based NEET and JEE (Main + Advanced) training institute. The academy was incorporated in 2015 to cater to the needs of students, who aim to crack competitive exams by connecting with the best brains around. The institute helps students enhance their skills and capabilities through...

Meet Raghunandan.G.H, a B. Tech Tutor from...

Raghunandan is a passionate teacher with a decade of teaching experience. Being a skilled trainer with extensive knowledge, he provides high-quality BTech, Class 10 and Class 12 tuition classes. His methods of teaching with real-time examples makes difficult topics simple to understand. He explains every concept in-detail...

Meet Swati, a Hindi Tutor from Bangalore

Swati is a renowned Hindi tutor with 7 years of experience in teaching. She conducts classes for various students ranging from class 6- class 12 and also BA students. Having pursued her education at Madras University where she did her Masters in Hindi, Swati knows her way around students. She believes that each student...

Meet Sandhya R, a B.Sc tutor from Bangalore

Sandhya is a proactive educationalist. She conducts classes for CBSE, PUC, ICSE, I.B. and IGCSE. Having a 6-year experience in teaching, she connects with her students and provides tutoring as per their understanding. She mentors her students personally and strives them to achieve their goals with ease. Being an enthusiastic...

Looking for Class 10 Tuition ?

Find best Class 10 Tuition in your locality on UrbanPro.

Are you a Tutor or Training Institute?

Join UrbanPro Today to find students near you X ### Looking for Class 10 Tuition Classes?

Find best tutors for Class 10 Tuition Classes by posting a requirement.

- Post a learning requirement
- Get customized responses
- Compare and select the best

Find best Class 10 Tuition Classes in your locality on UrbanPro

Post your learning requirement