A meeting of number family

lets assume the number to be 'N'. We want to find the smallest such number, and for this we will start with the smallest possible numbers starting from the leftmost digits of N and 2N. Let us start with 1 as the first digit of N. This makes the first digit of 2N 2 or 3 and 2 is the smaller of these. 0 cannot be a digit of N because then its corresponding digit in 2N would either be 0 or 1, both of which are already taken. The next smallest number we can use for the second digit of N is 3. The next smallest number that we can use for the third digit of N is 4. Now that we have no more numbers less than 5 that we can use for N, there must be a carryover onto the third digit of 2N, making it be 9 instead of 8. The next smallest number we can use for the fourth digit of N is 5. However, there will be a carryover because there are no more available digits less than five, and the fourth digit of 2N would be 1, which we have already used. The next smallest number we can use for the fourth digit of N is 7. The fourth digit of 2N would then be 5. However, this leaves us with the numbers 8 and 0 for the last digits of N and 2N, and 28 = 16, whose last digit is not 0. So, our next smallest number which we can use for the fourth digit of N is 8. This would mean that the fourth digit of 2N is 7. This leaves only 5 for the last digit of N. 52 = 10. 0 would be the last digit of 2N, and everything fits together perfectly. So, the smallest numbers N and 2N such that the digits 0 through 9 are used exactly once are:

N = 13485

2N = 26970

Let $N = \overline{abcde}$ and $2N = \overline{fghij}$ . Clearly $a \ge 1$ . Suppose $a = 1, b < 5$ , then $f = 2$ .

Now, if $b = 0$ , then $g \in \{0,1\}$ , impossible. So $b \ge 3$ . Suppose $b = 3$ . If we also assume $c < 5$ , then we have $g = 6$ .

Similar reasoning can be used to prove that $c = 0$ is impossible, so there is only one remaining number: $c = 4$ . This means $h = 8$ or $h = 9$ .

If $h = 8$ , then $d < 5$ . Since we have used $1,2,3,4$ , then we need to have $d = 0$ , but then $i \in \{0,1\}$ , contradiction. So $h = 9$ .

Since $2N$ is even, $j \in \{0,8\}$ . But if $j = 8$ , then $e \in \{4,9\}$ , contradiction. So $j = 0, e = 5$ , and the rest can be easily tested: $d = 8, i = 7$ . This gives $N = \boxed{13485}, 2N = 26970$ . Since we have been taking the smallest values, and it's clear that greedily taking the smallest digits for the leftmost places work, this is the smallest possible value.