Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 124978 by rydasss last updated on 07/Dec/20

Answered by talminator2856791 last updated on 07/Dec/20

$$1000a+100b+10c+d+a+b+c+d+x=2013,$$$$x=S\left(S(1000a+100b+10c+d)\right)$$$$1001a+101b+11c+2d+x=2013$$$$a=1$$$$101b+11c+2d+x=1012$$$$b=9$$$$11c+2d+x=103$$$$x\mathrm{can}\mathrm{not}\mathrm{exceed}11$$$$c=8(c\ne 9.....)$$$$2d+x=15$$$$\mathrm{test}d=1.\mathrm{if}\mathrm{this}\mathrm{does}\mathrm{not}\mathrm{work}\mathrm{then}\mathrm{it}\mathrm{is}$$$$\mathrm{implied}\mathrm{that}\mathrm{the}\mathrm{leading}\mathrm{digit}\mathrm{of}x\mathrm{is}2$$$$2\left(1\right)+S(1+9+8+1)=2+1+9=12\ne 15$$$$2d\mathrm{is}\mathrm{always}\mathrm{even}\Rightarrow \mathrm{first}\mathrm{digit}\mathrm{of}x\mathrm{is}\mathrm{odd}$$$$2d+S(1+9+8+d)=15$$$$2d+2+(d-2)=15$$$$2d+(d-2)=13$$$$3d-2=13$$$$d=5$$$$n=1985$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com