No products in the cart.

Exercise 1.1 q 2

CBSE, JEE, NEET, CUET

CBSE, JEE, NEET, CUET

Question Bank, Mock Tests, Exam Papers

NCERT Solutions, Sample Papers, Notes, Videos

Exercise 1.1 q 2
  • 1 answers

_Jass_ Mahey_ 1 year ago

Let 'a' be a postive odd integer also. Let 'q' be quotient and 'r' be the remainder after dividing by six. Then a=6q+r, where 0<r<6. Postive integer are 1,2,3,4,5 a=6q+0, 6q+1, 6q+2, 6q+3, 6q+4, 6q+5. But a=6q, 6q+2 and 6q+4 are even. When a is add it is the form 6q+1, 6q+3, 6q+5 for some integer q. .°. Hence proved...
http://mycbseguide.com/examin8/

Related Questions

3.11stques
  • 0 answers
5025 prime factorisation
  • 1 answers
a+b ka hool square
  • 1 answers
Ch 3 ex 3.3
  • 2 answers
7+7+√7
  • 1 answers

myCBSEguide App

myCBSEguide

Trusted by 1 Crore+ Students

Test Generator

Test Generator

Create papers online. It's FREE.

CUET Mock Tests

CUET Mock Tests

75,000+ questions to practice only on myCBSEguide app

Download myCBSEguide App