Given-that-4362-12-9495-21-3816-18-Then-9649- Tinku Tara June 3, 2023 Arithmetic 0 Comments FacebookTweetPin Question Number 6687 by Tawakalitu. last updated on 13/Jul/16 Giventhat4362=129495=213816=18Then=……..9649=? Answered by Rasheed Soomro last updated on 14/Jul/16 4362=129495=213816=189649=?−−−−−−−−−−−−−−−−−−−−−Ihavediscoveredonelogic.Otherlogicsmaybepossible.Iamwritingmylogicinaformofformulahere.(Sumofdigits)−[(unitdigit+1)(mod7)](4+3+6+2)−[(2+1)(mod7)]=15−3=12(9+4+9+5)−[(5+1)(mod7)]=27−6=21(3+8+1+6)−[(6+1)(mod7)]=18−0=18So,(9+6+4+9)−[(9+1)(mod7)]=28−3=∥25∥== Commented by Tawakalitu. last updated on 15/Jul/16 Wowthanks. Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Question-72218Next Next post: limit-1-1-1-x-x- Leave a Reply Cancel replyYour email address will not be published.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.
![4362 = 12 9495 = 21 3816 =18 9649 = ? −−−−−−−−−−−−−−−−−−−−− I have discovered one logic. Other logics may be possible. I am writing my logic in a form of formula here. (Sum of digits)−[ (unit digit+1)(mod 7) ] (4+3+6+2)−[(2+1)(mod 7)]=15−3=12 (9+4+9+5)−[(5+1)(mod 7)]=27−6=21 (3+8+1+6)−[(6+1)(mod 7)]=18−0=18 So, (9+6+4+9)−[(9+1)(mod 7)]=28−3=∥25∥_(=) ^(=)](https://www.tinkutara.com/question/Q6697.png)
