Can u find a number which added to itself one or several times will give a total having the same digits as that number but differently arranged and after the sixth edition will give a total of all nines?

142857 is the number

142857 which added for six times will give only the six digit answer and added for the seventh time will give the answer 999999 - all nines.

i. 142857+142857=285714
ii. 142857+142857+142857=428571
iii. 142857+142857+142857+142857=571428
iv. 142857+142857+142857+142857+142857=714285
v.142857+142857+142857+142857+142857+142857=857142
vi. 142857+142857+142857+142857+142857+142857+142857=999999

ya....
its correct

hii guys here is a doubt for me you are giving answers but can we remember as such
142857 so give us method to find the number tooo...............

heres my method for this

first we must get 9 in the last digit after six additions of same number meaning the last digit of the number multiplied by 7 must give 9 in the last position of resultant so it means only 7 is possible as last digit in the number....so we place 7 in last digit position...

number:_________7

now,we get carry 4 (7*7=49) to next digit so we need only 5 in next digit.......
to get 5 only 5 is possible(since 7*5=35)

number:_______57

now,we get carry 3(7*5=35) to next digit so we need only 6 in next digit...
to get 6 only 8 is possible(since 7*8=56)

number:______857

next,similarly carry:5 rdigit:4 nextdigit:2(7*2=14)

number:____2857

next,carry:1 rdigit:8 nextdigit:4(7*4=28)

number:__42857

next carry:2 rdigit:7 nextdigit:1(7*1=7)

number:142857