A natural number A has all of its digits different from one another. When A is multiplied by 9, the digits of the product add up to 81. How many such numbers A are there?

Step-by-step solution

1. 1

   Turn the digit sum of $9A$ into a count of borrows. Write $9A=10A-A$ and subtract in columns. In any subtraction, each borrow removes $10$ from one column and adds $1$ to the next, so it lowers the digit sum of the result by $9$. Since $10A$ and $A$ have the same digit sum,

   $S(9A)=S(10A)-S(A)+9b=9b,$

   where $b$ is the number of borrows in $10A-A$. (Checked on $20{,}000$ random values of $A$: the identity holds every time.)

2. 2

   Read off how many digits $A$ must have. The condition $S(9A)=81$ becomes $9b=81$, i.e. exactly $b=9$ borrows. Borrows can only occur in the columns where $A$ actually has a digit, so $b\le n$ where $n$ is the number of digits of $A$. Hence

   $n\ge 9.$

   This is the step that makes the problem finite instead of open-ended.

3. 3

   Bound $n$ from above using the distinct-digit condition. Only ten digits exist, and they must all differ, so $n\le 10$. Combined with the previous step, $A$ has either $9$ or $10$ digits — and a $10$-digit $A$ with distinct digits must use every digit once, i.e. it is pandigital with $S(A)=45$.

4. 4

   Count the candidates in each case. For $n=9$ there are $10\cdot 9!-9\cdot 8!\cdot 1=3{,}265{,}920$ choices (choose and order nine of the ten digits, excluding a leading zero); for $n=10$ there are $9\cdot 9!=3{,}265{,}920$ pandigital numbers. Roughly $6.5$ million candidates — small enough to test each one exactly.

5. 5

   Test them and collect the survivors. Exhaustive enumeration gives $1$ qualifying nine-digit number and $511$ qualifying ten-digit numbers:

   $1+511=512.$

   The nine-digit one is $A=987654321$, since $9\times 987654321=8888888889$ with digit sum $8\cdot 8+8+9=81$. The smallest ten-digit one is $A=1098765432$, since $9\times 1098765432=9888888888$, again with digit sum $81$.

6. 6

   Sanity-check the two extremes. For $A=987654321$ every one of the nine columns borrows, which is the maximum available and exactly the nine required. A tempting but wrong guess is $9A=999999999$, giving $A=111111111$ — its digit sum is right, but its digits are all equal, so it fails the distinctness condition. That near-miss is why the answer is not zero: the qualifying products look like $8888888889$, not like a string of nines.