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a: b: c =1: 3: 5
a, b, c have 9 nonzero digits with each of them appearing exactly once. The smallest possible number: 123 The last digit of c should be a multiple of 5 ending in 5, so the first digit of a should end in an odd digit (3, 7, or 9). The first digit of a should be 1 because if it were larger, then c would be four digits. Process of trial and error:
a: b: c =1: 3: 5 = 123: 369 : 615 3 and 6 repeated
a: b : c = 127: 381: 635 1 and 3 repeated
a: b: c = 129: 387 645 This works!
So: a=129, b = 387, c = 645.
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In the name of love 12/03/2017 at 14:41a, b, c have 9 nonzero digits with each of them appearing exactly once. The smallest possible number: 123 The last digit of c should be a multiple of 5 ending in 5, so the first digit of a should end in an odd digit (3, 7, or 9). The first digit of a should be 1 because if it were larger, then c would be four digits. Process of trial and error:
a: b: c =1: 3: 5 = 123: 369 : 615 3 and 6 repeated
a: b : c = 127: 381: 635 1 and 3 repeated
a: b: c = 129: 387: 645 This works!
Answer : a=129, b = 387, c = 645.
