20 Digits Problem?

Question

use the digits from 0 to 9 twice to fill in the 20 spaces in this multiplication problem.

_ _ _
x _ _ _
__________
_ _ _
_ _ _
_ _ _
__________
_ _ _ _ _

Answer 1

179
x224
-----------
716
358
358
-----------
40096

Answer 2

0, 1 and 5 not in the end.
0 must be in the middle.

100a + 10b + c => L1
100x + 10y + z => L2
-----------------------------------
[....][....][az][bz][cz] => L3
[....][ay][by][cy] => L4
[ax][bx][cx] => L5
------------------------------------
[p][q][r][s][t] => L6

1 < ax, ay, az < 10
a and x cant both be 1.
0 mist be in the middle (obviously) and cannot be in L2.
so x = y = z =/ 0
0 < x, y, z < 5
0 < a < 4
a = (0, 4)
c, z =/ 0, 1, 5

if x = 1, then L1 = L5
if y = 1, then L1 = L4

also if either one of x, y = 1,
then the other and any of L1 (a,b,c) will not equal 1.
a, b, and c must all be distinct.

from here UD(m) means unit digit of m

UD(cz) cannot equal either one of c or z, but it will equal t.
p will at least be 1 more than a.
since 0 < x, y, z < 5 and 0 < a < 4, b must be big enough so that (ay + bx) >= 10, meaning b cannot be 0 as well.
so L1 AND L2 do not contain any 0.

maybe if there's any repetition in L2, then there must also be a repetition in the answer (L6).

i give up.