Algebra and Logic Twist!

First of all, consider the digit 0. If 0 were to go in anyone of the boxes on the left hand side, the right hand side would equal 00, hence repeating a digit. Therefore, the digit 0 must be part of one of the results.

If 0 were the first digit of a result, there would have to be 3 digits which multiplied to make a 4th. The only digit which has 3 factors (other than itself) is 8, using 1,2, and 4. This gives us: 1 x 2 x 4 = 08 and ? x ? x ? = ?? (the digits 3,5,6,7,9 remaining) However, the only 2-digit result from multiplying 3 of the remaining numbers, is 3 x 5 x 6 = 90, but this is impossible. Therefore, we have shown that the 0 cannot be the first digit of a result and must be the second digit of one of the two results.

Using the fact that a number ending in a 0, must be a multiple of 5, this equation must have 5 as one of its digits. Therefore: ? x ? x 5 = ?0 and ? x ? x ? = ?? (the digits 1,2,3,4,6,7,8,9 remaining) We can try all these digits in the blank before the 0.

_ x _ x 5 = 10

_ x _ x _ = _ _ (2,3,4,6,7,8,9) None of these digits can make the first sum correct

1 x 4 x 5 = 20

_ x _ x _ = _ _ (3,6,7,8,9) Only 1 and 4 can fit in the first sum. However, multiplying any 3 of the remaining digits gives a 3-digit result ,

1 x 6 x 5 = 30

_ x _ x _ = _ _ (2,4,7,8,9) Only 1 and 6 can fit the first sum. However, using at least two of the digits (7,8,9) on the LHS gives a 3-digit result. We can check 2x4x7, 2x4x8 and 2x4x9, none of which leave digits to finish the sum. ,

1 x 8 x 5 = 40

_ x _ x _ = _ _ (2,3,6,7,9) Only 1 and 8 can fit the first sum. However, all of the possible combinations don't fill all of the boxes with a single digit. ,

3 x 4 x 5 = 60

_ x _ x _ = _ _ (1,2,7,8,9) Only 3 and 4 can fit the first sum. ,

_ x _ x 5 = 70

_ x _ x _ = _ _ (1,2,3,4,6,8,9) None of the remaining digits can finish the first sum. ,

_ x _ x 5 = 80

_ x _ x _ = _ _ (1,2,3,4,6,7,9) None of the remaining digits can finish the first sum without repeating a digit. ,

3 x 6 x 5 = 90

_ x _ x _ = _ _ (1,2,4,7,8) Only 3 and 6 can fit the first sum.

Of these, only 60 and 90 leave possible results. We can check all of the possibilities and we find that there is one possible second sum for each of these.

3 x 4 x 5 = 60 and 1 x 8 x 9 = 72

3 x 6 x 5 = 90 and 1 x 4 x 7 = 28

As using the sum containing a 0 as the top one, without loss of generality, the two equations can be flipped around, making the desired two boxes any of the results. Therefore the answer is 60 + 72 + 90 + 28 = $\boxed{250}$ Note: The numbers multiplied together in each equation can be rotated, without changing the answer

5 and 0 has a relationship that if the two are not together in one of the equations, then both would have to be the tens of the results each. This is only possible for 06, 08, 54 and 56, but the first two would require both 1 & 2, while the last two would require at least one of them (either 1 or 2 but not both), so it's obvious that the 5 and 0 has to be in the same equation.

Furthermore, for an equation of

we can see that 2 cannot be on the LHS, else the other number would have to be used twice. Also, because 7 is a prime that has no other multiple under 10, it can't be in this equation at all.

For that other equation,

since the result is less than 100, we cannot have big value digits to be all together on the LHS. How big is considered big? I'd consider any multiplication of two different digits over 50 as big, since it would have to force a 1 to be the third digit on the LHS. Since 6 × 9 = 54 > 50, so 6, 7, 8 and 9 are all big value digits.

Returning back to the equation with 5 & 0, after the elimination of 7 as prime solo, let's consider 3 and 2, too. If there is / are factor(s) of 3 in it, then it could be either 3, 6 or 9 on both the LHS & RHS. On the other hand, if there wasn't, then only 1, 4 & 8 are possible on its LHS with an addition of 2 for the RHS.

Anyway, since the resulting number is a 2-digit one, then the multiplier on LHS must be even and < 100/5, that's less than 20. Without a factor of 3, the possibilities are 4 and 8 for 20 and 40 each, but with 3, we have 6, 12 and 18 for 30, 60 and 90 respectively.

20 = 1 × 4 × 5
Other numbers to fill in the other equation are 3, 6, 7, 8, 9, but even the least three of them have a resulting multiplication of more than 100 with 3 × 6 × 7 = 126 > 99.

40 = 1 × 8 × 5
Other numbers to fill in the other equation are 2, 3, 6, 7, 9. Since there are 3 digits with factor(s) of 3 and only 2 space for the resulting multiplication (2-digit number), there would have to be at least one of them on LHS. Also, since there's only 3 odd numbers, it's impossible to have an odd result because that would require three odds on LHS and one odd on RHS, so it's the case of not enough odds to go around. Our result should be an even multiple of 3 (can use the fact that its sum of digits will also be a multiple of 3), with the tens an odd number. Therefore, it's either 36, 72 or 96, but we know the first and the last don't have a factor of 7 in them while the middle doesn't have enough factor of 3 packed in.

30 = 1 × 6 × 5
Other numbers to fill in the other equation are 2, 4, 7, 8, 9. If 9 is on LHS, then RHS must only have 72 as its result, but it won't have enough factor of 2 (about another 2 of them). Other than that, 9 must be the tens itself with an even ones, but neither 92 or 94 is a multiple of 7 and 98 would need two factors of 7 ( 98 = 2 × 7² ). If we recall back about big values mentioned above, it's easy to see that neither two of the three among 7, 8 & 9 should be together on LHS without the company of the digit 1, so it should be enough to check whether 78 divides 9 and 98 divides 7 with no other results than factors of 2.

60 = 3 × 4 × 5
Other numbers to fill in the other equation are 1, 2, 7, 8, 9. Again, 9 on LHS and 18 or 72 on RHS, but 18 does not have a factor of 7 but this time 72 works, since 1 × 8 × 9 = 72. For completeness, 92 does not divide 7 and 98 needs another factor of 7.

90 = 3 × 6 × 5
Other numbers to fill in the other equation are 1, 2, 4, 7, 8. With 3 evens, it's easy to see that we are far from having enough odds to be shared by both LHS and RHS, and LHS would have been forced to accept at least an even digit either way. Since all the factors of 3 are utilised in the first equation, our result can't have a digit sum of multiple of 3. Recalling big value rules, 7 and 8 must also be on different sides of the equation. It suffice to check whether 28 and 82 divides 7, and 74 divides 8. Only the first one is true, since 1 × 4 × 7 = 28.

So, a list of working equations :
3 × 4 × 5 = 60
1 × 8 × 9 = 72
3 × 6 × 5 = 90
1 × 4 × 7 = 28

Answer
= 60 + 72 + 90 + 28
= 250