## 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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