What?! Is that even possible?!
Let d ( n ) be the set of digits of n . Find the minimum of n that satisfies
- ( 1 ) : 1 0 5 ≤ n ≤ 1 0 6
- ( 2 ) : ∣ d ( n ) ∣ = ∣ d ( 2 n ) ∣ = ∣ d ( 3 n ) ∣ = ∣ d ( 4 n ) ∣ = ∣ d ( 5 n ) ∣ = ∣ d ( 6 n ) ∣ = 6
Details:
- If A is a set, then ∣ A ∣ is the number of elements in set A .
- This problem is from Thailand POSN, which is held today. XD
The answer is 126984.
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1 solution
What about 126984??
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Damn, I'm sorry for my stupid mistake. I've reported this to the moderator.
Thanks. I've updated the answer to 126984 because it is a smaller number. Any idea how to show that this is indeed the minimum?
Note: POSN is a computing olympiad. I've updated this problem into CS instead.
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Actually POSN has more than just CS. There're also math, physics, and stuffs. And no, this is from math POSN.
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@Samuraiwarm Tsunayoshi – Thanks for the clarification! Let me move it back to NT then.
The only approach that I can think of is tedious case checking, from like "123456" and so on, which is why I moved it to CS. Any idea on other approaches?
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@Calvin Lin – There's a special characteristics of 1 5 8 7 3 = 7 1 1 1 1 1 1 . If we multiply 1 5 8 7 3 by k such that k ≤ 6 2 and 7 ∣ k , we'll always get a number with different digits. I strict at k ≤ 6 2 because if 1 5 8 7 3 k ≥ 1 0 6 , it won't give you different digits any more.
EVERYTHING'S 142857!