Can we fill in the boxes with distinct digits so that the equation is satisfied?
Clarification:
The above equation represents a 4-digit number multiplied by 3 to produce a 5-digit number, so the leftmost box is non-zero on either side.
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There are 15 possible solutions:
4 6 0 9 ∗ 3 = 1 3 8 2 7
5 6 8 3 ∗ 3 = 1 7 0 4 9
5 6 9 4 ∗ 3 = 1 7 0 8 2
5 8 2 3 ∗ 3 = 1 7 4 6 9
5 8 3 2 ∗ 3 = 1 7 4 9 6
5 9 3 4 ∗ 3 = 1 7 8 0 2
6 3 5 8 ∗ 3 = 1 9 0 7 4
6 8 1 9 ∗ 3 = 2 0 4 5 7
6 8 3 9 ∗ 3 = 2 0 5 1 7
6 9 1 8 ∗ 3 = 2 0 7 5 4
8 1 6 9 ∗ 3 = 2 4 5 0 7
8 3 6 9 ∗ 3 = 2 5 1 0 7
9 0 4 6 ∗ 3 = 2 7 1 3 8
9 1 3 6 ∗ 3 = 2 7 4 0 8
9 1 6 8 ∗ 3 = 2 7 5 0 4