Finding a 6 digit number The sum of digits of a six digit number is $41$.
At first we omit its leftmost digit and subtract it from the original number. Then we omit the two leftmost digits and add it with the result. Next we omit the three leftmost digits and subtract it from the current result. This process of successive subtraction-addition continues until all the digits of the six digits numbers have been omitted. The final is $706010$. How can we find the original number?
Any hint will be very much appreciated. And it is perhaps best to add that I have not been able to find anything yet.So I am stuck now.
 A: Some hints (it is a rather old question...) : 


*

*the net effect of the first subtraction is to return the upper digit followed by five $0$ (you may try this with any actual $6$ digits number). 

*the net effect of the following addition and subtraction is to give the most significative digit (as previously), $0$ and the third most significative digit followed by three $0$.

*so what can you deduce from the final $706010$ ?

*the sum of digits is $41$ so what should the three $0$ really be ?
A: We know that we can write any six digit number in the form 100000a+10000b+1000c+100d+10e+f,where a,b,c,d,e,f are integers.Now removing the leftmost digit means that we are going to remove a,which in turn means we are going to subtract 100000a from the original number,thus getting 1000b+100c+. . . +f.Hence subtracting it from the original would just give us 100000a.By repeating the successive subtraction-addition thing,we have
100000a+1000c+10e=706010
and the only values that a,c and e can take are 7,6 and 1 respectively.Adding them gives us 14.Therefore,
f+b+d=27
But since f,b and d are digits,the only value that all of them can take is 9.Thus our required number is 796919.
