What is the largest positive integer such that each digit is at least the sum of all the digits to its left? What is the largest positive integer such that each digit is at least the sum of all the digits to its left?
Can someone point me in the right direction with this problem? Don't give me the answer, please. I don't know where to start, the wording confuses me.
 A: The largest such number is 

 $11259$

(i) If the first digit were $\ge 2$, we could profitably replace it by $11$. So the first digit is $1$. 
(ii) If the second digit were $\ge 2$, then the next two digits would be at least $3$ and $6$, ruling out a 5-digit number.
(iii) If the third digit were $\ge 3$, then the next digit would be at least $5$, ruling out a 5-digit number.
Answer:

 A 6-digit number is clearly impossible: minimum digits are $11248$. So $11259$ is the best we can do.

A: Assume that all the digits have to be non-zero. Start working backwards. You know that the last digit has to be $9$, right? Now you want a number with as many digits as possible, that add up to at most $9$. So you start with a $1$. Then what is the smallest thing that the second digit can be? The third? and so on, until the sum of the digits is at least $9$. Then the number terminates. (Hint, consider the Fibonacci sequence).
A: The left most digit has to be 1,the smallest possible, because you want to have a number with maximum possible length, so the smallest possible for the next position has to be a 1 again because having a 2 you get a number with lesser digits.. Continue to play with digits to get a number whose slight modification will give you the result.. 
