Question-130602 Tinku Tara June 4, 2023 Permutation and Combination 0 Comments FacebookTweetPin Question Number 130602 by EDWIN88 last updated on 27/Jan/21 Answered by mr W last updated on 27/Jan/21 =numberof4digitnumberswhicharedivisibleby3:1002,1005,…,9999⇒9999−10023+1=3000i.e.thereare3000fivedigitnumbersendingwith6anddivisibleby3. Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: by-using-diffintion-lim-z-1-1-z-2-1-help-me-sir-Next Next post: 0-0- Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.