What is the largest number n so that n, 2n, and 3n together contain every digit from 1-9 exactly once?

Analytical solution preferred.

987/3 = 329 is the upper bound for n.

Assume n starts with 3:

Then n, 2n, 3n are 3xx, 6xx, and 9xx. There can be no carry from the tens to the hundreds in this case. This implies the tens digit in 9xx is at least 6. To get that high the tens digit in 3xx must be 2, which leaves 7 as the only possible tens digit in 9xx. Then the numbers are 32x, 6xx, 97x. This leaves only n=324, 325, or 326 as possible candidates. None of those satisfy the problem.

Then n starts with 2. Assume n is just below 300, so in this case n, 2n, 3n are 2xx, 5xx, 8xx.

The smallest value for 8xx is 813, so the smallest 2xx is 813/3 = 271. Then the tens digit of 2xx must be 7 or 9. If n is 29x then at least one of the tens digits of 5xx and 8xx is another 8 or 9, contradicting the problem. Then n must be at most 279.

From this it is easy to find **n=273, 2n=546, 3n=819**