Number (with all digits different and ordered in increasing order) multiplied by 9 has the sum of its digits equal to 9 
Bob tells his sister that if she thinks of a number with all its digits different and ordered in increasing order from left to right, and then multiplies the number she thought of by 9, he always knows how much the sum of the digits of the result of multiplication, although he does not know what number his sister thought.
Decide if Bob is lying or telling the truth and explain why.
Note: she thinks of a non-zero number.

The only useful thing that I found is that $9N=10N-N$, thus $9N=\overline{a_1(a_2-a_1)(a_3-a_2)\cdots (a_{n-1}-a_{n-2})(a_n-a_{n-1}-1)(10-a_n)}$, if $N=\overline{a_1a_2\cdots a_n}$.
Could anyone help me?
 A: Writing $9n=10n-n$ is a good start.  The idea then is to note that, since the digits are strictly increasing, it is easy to see how that subtraction must turn out.  Indeed, in most places you are subtracting a number from one which is strictly greater. More precisely, other than in the units and the tens place we are subtracting $a_i$ from $a_{i+1}$, thereby producing the gap between $a_{i+1}$ and $a_i$. (using your notation) In the units place we are, of course, subtracting the final digit from $0$ which will require a simple carry.
Consider the gap sequence, pretending that the number begins with a $0$.  Thus, if your starting number were $24578$ the gap sequence would be $\{2,2,1,2,1\}$.  We remark that the partial sums of the gaps return the original digits.  In particular, the sum of all the gaps is the final digit...let's call it $d$.
Then the digits in $9n=10n-n$ are precisely the digits in the gap sequence, save that the final one is lowered by $1$ and we append $10-d$.  All that remains is to remark that the sum of the digits in that string is precisely $d-1+10-d=9$ and we are done.
Example:  sticking with $n=24578$ we get $9n=221202$ as claimed.
Remark:  the lowering of the penultimate gap and the appending of $10-d$ are the result of the obvious carry, when you try to subtract $d$ from $0$.  Of course, if $n=0$ at the start, then there is no carry.  In all other cases, there is always a carry at that stage (and no other).
