I can only use each digit once, 0-9 one will be spare.

xxx

xxx

xxx

= 1000
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You have to have 3 sets of 3 digits each.  The first set totals 10 (for the ones place).  The second and third sets total 9 (for the tens and hundreds place).  The 1 carries over from the total of 10 in the ones place, moving the 9 up to 10 in the tens place and the 9 up to 10 in the hundreds place.

Ways to make 10: 019, 028, 037, 046, 127, 136, 145, 235

Ways to make 9:  018, 027, 036, 045, 126, 135, 234

If 10 is 019, 9 leaves 234, not enough ways

If 10 is 028, 9 leaves 135, not enough ways

If 10 is 037, 9 leaves 126, not enough ways

If 10 is 046, 9 leaves 135, not enough ways

If 10 is 127, 9 leaves 036, 045, enough ways but 036 and 045 share 0

If 10 is 136, 9 leaves 027, 045, enough ways but 027 and 045 share 0

If 10 is 145, 9 leaves 027, 036, enough ways but 027 and 036 share 0

If 10 is 235, 9 leaves 018, not enough ways.

We checked all of the possible combinations of 3 digits totaling 10 and it looks like none of them leave any combinations to make two sets of 3 digits totaling 9 each.

Answer:  No solution.

by Level 13 User (103k points)

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