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1 to 9 concatenation (Posted on 2016-04-12) Difficulty: 3 of 5
What is the largest number n so that n, 2n, and 3n together contain every digit from 1-9 exactly once?

Analytical solution preferred.

No Solution Yet Submitted by Ady TZIDON    
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A start | Comment 3 of 7 |
A few observations

a) n must be three digits.  Two digits is too little and 4 are too many.

b) n must be greater than 123.

c) n must be less than 330.  Otherwise, either digits are repeated or 3n has 3 digits.

d) if n, 2n and 3n contain all digits, then their sum of digits = 45, so the sum is divisible by 9.
   but the sum = 6n, so n must be divisible by 3.
d) So, that leaves less than 70 that might work   

  Posted by Steve Herman on 2016-04-12 12:19:59
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