## Take the number $192$ and multiply it by each of $1$, $2$, and $3$:

\begin{align}

192 \times 1 &= 192\\

192 \times 2 &= 384\\

192 \times 3 &= 576

\end{align}

By concatenating each product we get the $1$ to $9$ pandigital, $192384576$. We will call $192384576$ the concatenated product of $192$ and $(1,2,3)$.

The same can be achieved by starting with $9$ and multiplying by $1$, $2$, $3$, $4$, and $5$, giving the pandigital, $918273645$, which is the concatenated product of $9$ and $(1,2,3,4,5)$.

What is the largest $1$ to $9$ pandigital $9$-digit number that can be formed as the concatenated product of an integer with $(1,2, \dots, n)$ where $n \gt 1$?

### The largest 1 to 9 pandigital number surely starts with 9, and since n>1, we want as high digits as possible following it. Therefore, the integer we use for multiplication should start with 9 and be as small as possible, so that we could involve up to “n=5”, hence higher digits – 3, 4, 5 in the pandigital. We are restricted to the 1 to 9 range, so we may attempt with a 3-digit number starting with “9” that is further maximized by arranging the digits in reverse order, i.e. we try with “987”.

If we multiply 987 by 1, 2, 3 and then concatenate the products, unfortunately, we run into a repetition of “9” in the third product, so the 3-digit starting number is too large for this process. The next obvious option to try would be a 2-digit integer starting with “9”.

Let’s take the highest 2-digit integer starting with “9”, which is “95”.

If also this is not small enough, we can reduce the last digit. We go in reverse order so we get “94”, then “93” etc until “91”. It’s reasonable to start with 95, then go down to 91, as we are trying to get the largest pandigital.

After some calculations, we find that 93 and 94 give an 8-digit end result, 92 yields a pandigital but with only 18 at the end (not optimal), and reaching 91, we find:

91 × 1 = 91,

91 × 2 = 182.

Concatenating these products results in the 9-digit number 91182, which is short by 4 digits of being a 9-digit number.

Let’s take the next multiplying number, 3, and calculate:

91 × 3 = 273.

Concatenating this number with the previous, we get 91182273. Still short by 1 digit. We continue with the next, 4:

91 × 4 = 364.

The concatenation of all our products results in the 9-digit number 9118227364, but this is unfortunately a 10-digit number. Hence, we need to backtrack and try with 90.

The next option is therefore “90”, however, 0 is not a valid digit in a pandigital number. Thus, no 1 to 9 pandigital 9-digit number can be formed as the concatenated product of an integer with (1,2,…,n) for n > 1 using a two-digit starting integer.

The only option left is then to use a one-digit integer which must be 9 (to ensure we get the largest pandigital). And indeed:

9 × 1 = 9,

9 × 2 = 18,

9 × 3 = 27,

9 × 4 = 36,

9 × 5 = 45.

When we concatenate these products, we get a 1-9 pandigital number: 918273645.

Hence 918273645 is the largest 1-9 pandigital 9-digit number that can be formed as the concatenated product of an integer with (1,2,…,n) where n > 1.

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