The number, \(1406357289\), is a \(0\) to \(9\) pandigital number because it is made up of each of the digits \(0\) to \(9\) in some order, but it also has a rather interesting sub-string divisibility property.
Let \(d_1\) be the \(1\)st digit, \(d_2\) be the \(2\)nd digit, and so on. In this way, we note the following:
Find the sum of all \(0\) to \(9\) pandigital numbers with this property.
I did this one with a whiteboard and a calculator, purely because I could.
The first key insight is that \(d_6\) must be either \(0\) or \(5\). We can eliminate the \(0\) case because this would require \(0d_7d_8\) to be divisible by \(11\), which would imply \(d_7 = d_8\). Therefore, \(d_6 = 5\). Scanning through all multiples of \(11\) between \(500\) and \(599\), and excluding any with repeated digits, gives us
\[ d_6d_7d_8 \in \{506, 517, 528, 539, 561, 572, 583, 594\}. \]There are no multiples of \(13\) of the form \(06d_9\) or \(17d_9\), the only multiple of the form \(61d_9\) is \(611\), and the only multiple of the form \(94d_9\) is \(949\). Because \(286\), \(390\), \(728\), and \(832\) are multiples of 13, we have
\[ d_6d_7d_8d_9 \in \{5286, 5390, 5728, 5832\}. \]Likewise, the only multiple of \(17\) of the form \(32d_{10}\) is \(323\), but \(867\), \(901\), and \(289\) are multiples of \(17\). Thus
\[ d_6d_7d_8d_9d_{10} \in \{52867, 53901, 57289\}. \]Now we begin working from right to left. The only three-digit multiples of \(7\) ending in \(52\), \(53\), or \(57\) are \(252\), \(357\), \(553\), and \(952\). After eliminating the first and third for repeating digits, we are left with
\[ d_5d_6d_7d_8d_9d_{10} \in \{952867, 357289\}. \]In either case, \(d_5\) is divisible by \(3\), so the remaining rules simplify to: \(d_4\) is even, and \(d_3d_4\) is divisible by \(3\). In other words, \(d_3d_4\) is divisible by \(6\).
Consider first the case of \(d_5d_6d_7d_8d_9d_{10} = 952867\), for which the remaining four digits are \(0,1,3,4\). The only two-digit multiple of \(6\) that can be arranged from these four digits is \(30\). This case therefore gives us the pandigital numbers \(1430952867\) and \(4130952867\).
In the case of \(d_5d_6d_7d_8d_9d_{10} = 357289\), the remaining digits are \(0,1,4,6\). The two-digit multiples of \(6\) that can be formed here are \(06\) and \(60\). This case therefore gives us the pandigital numbers \(1406357289\), \(1460357289\), \(4106357289\), and \(4160357289\).
Summing the six pandigital numbers together yields the result, \(16695334890\).