Inspired to share a problem again.

Even digits must be at even positions, which leaves only odd numbers in odd positions.

0 must be last, and so 5 must be in position 5.

Last but one digit can be anything: a number made with digits from 1 to 9 (all but 0) is always a multiple of 9. We don't have to check anything.

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Numbers can only start with

123, 129, 147, 183, 189,

321, 324, 327, 369, 381, 387,

723, 729, 741, 783, 789,

921, 927, 963, 981, 987,

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Adding a 4th number (2 and 6 are the only possibilites) and the digit 5:

12365, 12965, 14725, 14765, 18325, 18365, 18925, 18965

32165, 32465, 32765, 36925, 38125, 38165, 38725, 38765,

72365, 72965, 74125, 74165, 78325, 78365, 78925, 78965,

92165, 92765, 96325, 98125, 98165, 98725, 98765.

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With 6th digit: if it was ending in 25, then it must be 258. If 65, it must be 654.

123654, 129654, 147258, 147654, 183258, 183654, 189258, 189654

321654, 324654, 327654, 369258, 381258, 381654, 387258, 387654,

723654, 729654, 741258, 741654, 783258, 783654, 789258, 789654,

921654, 927654, 963258, 981258, 981654, 987258, 987654.

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7th digit, only two odd digits left and result must be a multiple of 7:

1296547, 1472583, 3216549, 3816547, 7296541, 7836549, 9216543, 9632581.

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8th digit is the only even digit left and we check if the 3 last digits make a multiple of 8: 12965(478) - no, 14725(836) - no, 32165(498) - no, 38165(472) - yes ,

72965(418) - no, 78365(492) - no, 92165(438) - no, 96325(814) - no.

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Only 3'816'547'290 is valid.

Clearly e,j=5,0, and the remaining even digits are even and odd digits odd. Since 3|(d+e+f), d+f=1mod3, so {d,f} = {2,8} or {4,6}. Since f is even and 8|fgh it follows that 8|gh. Since g is odd and 8|gh, gh = 16, 32, 72, or 96. So h is 2 or 6.

Suppose, for reductio, that h is 6. Then gh = 96, so g=9. Since 3|ghi, i=3. So {a,c} = {1,7}, and b=1mod3 so b=4. Since 4|cd, and 10 and 70 are 2mod4, d=2mod4, so d=2. Since 3|(d+e+f), f=2mod3, so f=8. This leaves two options for a and c, and neither one yields 7|abcdefg. So the assumption that h is 6 is incorrect, and so h=2 (and we have only identified e,h, and j so far).

Since 3|def d+f=1mod3, {d,f} = {4,6} and b=8. Since h=2, gh is 32 or 72, so g = 3 or 7. Since 3|ghi, ghi is one of 321, 327, 723, or 729. Similarly a+c=1mod3, so {a,c} is one of {1,3}, {1,9}, and {7,1} ({3,7} is ruled out because g is one of 3 and 7). So cd is one of 16, 36, 76, and 96, and so d=6 and f=4. Putting in zeros for a, c, and g, we find that 7|0806540, and so 7|(a*10^6 + c*10^4 + g). Knowing that g is 3 or 7 and that either a or c is 1, the only way to achieve divisibility by 7 is for a,c, and g to be 3, 1, and 7, because 3*10^6=3mod7 and 1*10^4=4mod7.

That gives us 3816547290.