Pandigital Products

We shall say that an $n$-digit number is pandigital if it makes use of all the digits $1$ to $n$ exactly once; for example, the $5$-digit number, $15234$, is $1$ through $5$ pandigital.
The product $7254$ is unusual, as the identity, $39 \times 186 = 7254$, containing multiplicand, multiplier, and product is $1$ through $9$ pandigital.
Find the sum of all products whose multiplicand/multiplier/product identity can be written as a $1$ through $9$ pandigital.
HINT: Some products can be obtained in more than one way so be sure to only include it once in your sum.

Let’s try to solve this problem systematically.

The total number of digits from the multiplicand, multiplier, and the product is $9$, which matches the pandigital requirements of $1$ through $9$ exactly once. Thus, we should consider the possible ways this could happen:

1) A $1$-digit number multiplied by a $4$-digit number to give a $4$-digit number
2) A $2$-digit number multiplied by a $3$-digit number to give a $4$-digit number

These are the only ways to have $2$ numbers multiply to give a $4$-digit product, since a $5$ digit product would need one multiplicand to be at least $3$ digits – meaning in total there would be more than $9$ digits, which isn’t allowed.

Having established this, we should test each scenario:

1) A $1$-digit number multiplied by a $4$-digit number:

If we think about it, the one-digit number could only be $1$, $2$, $3$, $4$ or $5$. Why? Because if we used $6$, $7$, $8$ or $9$ one of the resulting product’s digits would of necessity be repeated in the four digits of the multipliying number. And that’s not allowed because we are only allowed to use $1$ through $9$ exactly once. Also, using $1$ as our single digit is useless because it doesn’t modify the $4$ digit number.

If we used $2$ as our single digit then the $4$ digit multiplied number should be in $4000$ range but this would imply that the number is either $4356$ or $4378$, which will result in a product greater than $9000$, so it is invalid.

The same reasoning is applicable for the single-digit numbers $3$, $4$ and $5$, the resulting product will be greater than $9000$.

This shows that the first scenario is impossible. Now, examine the second scenario:

2) A $2$-digit number multiplied by a $3$-digit number:

For these conditions to be met, the $2$-digit number must be at least $12$ and the $3$-digit number $123$. As a result, the smallest possible product is bigger than $1000$, which suits us perfectly.

Now, considering there are only $9$ items and $5$ digits have already been accounted for in both the two and three-digit numbers, there are only $4! = 24$ permutations that allow us to rearrange the remaining $4$ numbers. We can easily manually test these.

After examining all the cases, we find there are only two pandigital products corresponding to the situation: $12 \times 483 = 5796$ and $18 \times 297 = 5346$.

Now both of the products are different, hence the sum of all products whose multiplicand/multiplier/product identity can be written as a $1$ through $9$ pandigital is $5796 + 5346 = 11142$. So, the answer is $11142$.

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