Question

In a 3-digit number, unit's digit is one more than the hundred's digit and ten's digit is one less than the hundred's digit. If the sum of the original 3-sigit number and numbers obtained by changing the order of digits cyclically is 2664, find the number

Answer 1

the no is 879

let hundreds digit=x
thus tens digit=x-1
ones digit=x+1
the first form of our no. is x*100+(x-1)10+(x+1)
                                       =111x-9   (value no.1)
second form is
(x+1)100+x*10+(x-1)
= 111x+99       (value no 2)
third form is
(x-1)100+(x+1)10+(x)
=111x-90       (value no 3)

these three values add up to 2664
111x-9+111x+99+111x-90=2664
3(111x)=2664
111x=888
x=8
therefore our no is 8*100+(8-1)*10+(8+1)
                              =800+70+9
                              =879

Answer 2

This is the answer you want.