Rohini Sector 16, Delhi, India - 110089
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1 Student
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 12 Tuition
5
Board
CBSE, ISC/ICSE, State
ISC/ICSE Subjects taught
Mathematics
CBSE Subjects taught
Mathematics
Taught in School or College
No
State Syllabus Subjects taught
Mathematics
1 Student
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 11 Tuition
5
Board
CBSE, State, ISC/ICSE
ISC/ICSE Subjects taught
Mathematics
CBSE Subjects taught
Mathematics
Experience in School or College
I've experience of more than 5 years of teaching mathematics of 11 and 12.
Taught in School or College
Yes
State Syllabus Subjects taught
Mathematics
1 Student
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 10 Tuition
5
Board
CBSE, ICSE, State
CBSE Subjects taught
Mathematics
ICSE Subjects taught
Mathematics
Taught in School or College
No
State Syllabus Subjects taught
Mathematics
1 Student
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 9 Tuition
5
Board
CBSE, ICSE, State
CBSE Subjects taught
Mathematics
ICSE Subjects taught
Mathematics
Taught in School or College
No
State Syllabus Subjects taught
Mathematics
5 out of 5 1 review
Nischal Kaushik
"Fantastic teacher and friendt is really lucky to have a teacher like Rahul sir.Excellent knowledge about the subject. "
Reply by Rahul
Thank you Nishchal.
Answered on 17/07/2019 Learn CBSE/Class 10/Mathematics/UNIT I: Number systems/Real numbers/NCERT Solutions/Exercise 1.1
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 = Even number , where m = 3q
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) = 2m1 = Even , where m1 = 3q + 1.
Case-(iv) When r = 3 then
a = 6q + 3 = 2(3q + 1) + 1 = 2m1 + 1= odd number, where m1 = 3q + 1
Case-(v) When r= 4 then
a = 6q + 4 = 2(3q + 2) = 2m2 = Even , where m2 = 3q +2 .
Case -(vi) When r = 5 then
a = 6q + 5 = 2(3q + 2) + 1 = 2m2 + 1 = odd number, where m2 = 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.
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Rohini Sector 16,
Delhi
