Filling in blanks in a multiplication problem knowing only the set of digits in the product and that 9 divides each factor

Question

The 5774 Ulpaniada (part 2 of it) includes the following question:

The following multiplication exercises uses all $9$ digits $1,2,3,\ldots,9$. The digits are encoded by asterisks. We are told that the numbers in each of the three rows are divisible by $9$, and that the answer row consists of four consecutive numbers, not necessarily in order.

The sum of the last digits of the three numbers (denoted by red asterisks in the illustration) is:

$$\phantom{\times*}**\color{maroon}*\\\times\underline{\phantom{**}*\color{maroon}*}\\\phantom{\times}***\color{maroon}*$$

It gave five answer choices: $21$, $20$, $19$, $18$, and $17$.

I solved it, but I am unsatisfied with my method: it required far too much trial-and-error (guess-and-check) effort and far too little insight. I'm guessing there's a neater method I'm missing; does anyone have one?

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My method, for what it's worth:

Because the product has consecutive digits that sum to a multiple of $9$, it must have the digits $3,4,5,6$ in some order. (I assumed the question posers weren't counting $8,9,0,1$ as consecutive, since it had mentioned the "$9$ digits $1,2,3,\ldots,9$".) Since the factors are also divisible by $9$, the product is divisible by $81$; since there's only one $5$ in the multiplication problem, it can't be the last digit of the product. So we're looking for a product with the digits $3,4,5,6$ (not $5$ last), a multiple of $81$.

From there, I decided to guess and check mutliples of $81$. The first mutiple of $81$ I found was $4536$. It's $2^33^47$; choosing a $2$-digit product of some of those primes to serve as a possible $2$-digit factor in the multiplication problem shows that none actually works.

I then decided that that method takes too long, so tried to use the five given answer choices to eliminate possibilities, as follows. The product must end with $3$, $4$, or $6$. If it ends with $4$, then its factors must end with $2$ and $7$; if with $3$, then $7$ and $9$; if with $6$, then $2$ and $8$ or $7$ and $8$. The last three possibilities give last-digit sums among the five answer choices. I then went through the permutations of $3456$ ending in $3$ or $6$ (there are, of course, only $12$ such) to determine which are multiples of $81$, and found that (besides $4536$, already eliminated) only $5346$ is; thus the answer is $21$.

Answer 1

Write $a,b,c$ for the sums of each of the three rows. We have $a+b+c=1+\cdots+9=45$. The fact that each row is a multiple of 9 means that $a,b,c\in\{9,18,27,36\}$. So $a,b,c$ are either $27,9,9$ or $18,18,9$, in some order. As $1+2+3+4=10$, the 4-digit term cannot be $9$, so it is $18$ or $27$. But $27$ is not the sum of four consecutive integers, so the third row is $3,4,5,6$ in some order.

We are left with $1,2,7,8,9$ for the other two rows, one adding to $18$ and the other $9$. We cannot reach $18$ with two digits at all or with three digits not using the $9$, so the first row is necessarily one of the following two possibilities: $9,8,1$ in some order, and the middle one $2,7$; or $2,7,9$ in some order and $1,8$. This last case is easily discarded since the only 4-digit product is $279\times18=5022$, which produces the wrong last row.

Looking at the first row, the last digit cannot be $1$ because that would imply a repeated digit. $2$ and $9$ do not fit, because they would put an $8$ in the last row. Actually, if the first row starts with $8$ or $9$, as $819\times27$ already has five digits, we learn that the first number starts with $1$ and that the second one is $27$.

We have $189\times27=5103$, not good, so it has to be $198\times 27=5346$. So the answer to the question is $8+7+6=21$.