Show-that-the-number-122-n-102-n-21-n-is-always-one-less-than-a-multiple-of-2020-For-every-positive-integer-n- Tinku Tara June 4, 2023 Number Theory 0 Comments FacebookTweetPin Question Number 63674 by Tawa1 last updated on 07/Jul/19 Showthatthenumber122n−102n−21nisalwaysonelessthanamultipleof2020.Foreverypositiveintegern. Commented by Prithwish sen last updated on 07/Jul/19 (101+21)n−(101+1)n−21n={101n+n101n−121+……+n101.(21)n−1+21n}−{101n+n101n−1+…..+n101+1)−21n=n101n−1(21−1)+……..+n101(21n−1−1)−1=101×20{101n−2+…….n(21n−2+21n−2+…+1)}−1=2020×m−1m=integerwhichisalways1lessthanmultipleof2020.Proved Commented by Tawa1 last updated on 07/Jul/19 Godblessyousir Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: 1-calculate-0-2pi-dt-cost-x-sint-wih-x-from-R-2-calculate-0-2pi-sint-cost-xsint-2-dt-3-find-the-value-of-0-2pi-dt-cos-2t-2sin-2t-Next Next post: Given-f-x-2x-1-2x-1-and-f-1-x-2-1-f-1-x-2-1-1-x-2-then-x- Leave a Reply Cancel replyYour email address will not be published.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.