Can you find the smallest positive number such that if you shuffle the digits of the number in a particular order, the shuffled number becomes twice the original number.
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125874 => 251748
251748 is twice the 125874 and have same digits 1,2,4,5,7 & 8
How can we show that this is indeed the smallest such number?
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I cant see any method to prove that it is smallest positive number .
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I think that would be the interesting part. It is not clear to me how to even arrive at this solution in the first place (other than through brute force computation).
For example, another solution (ignoring smallest) would be 1 4 2 8 5 7 × 2 = 2 8 5 7 1 4 , which would be the answer that I expect most people to give.
The Mathematica function
Test[n_] := Sort[IntegerDigits[2n]] == Sort[IntegerDigits[n]]
identifies if a number n can have its digits reordered to form its double. The command
Select[Range[130000],Test[#] &]
gives the answer
125874, 128574
which seems to settle the matter, if in an unsatisfying manner!
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Yea, I don't think there is a nice mathematical proof to this.
This is particularly interesting since these are the recurring digits from division by 7
Another person posted this problem, and gave the answer 142857. But, clearly, your answer is smaller. 1/7 vs. 18/143?
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If you look at my reply to Calvin, there are two 6-digit answers possible less than 130000, and the one here is the smaller of the two. There are doubtless even larger ones.
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Yes, I know. I saw it. I just commented to get the 18/143 in there somewhere.
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@James Wilson – And to let you know someone else posted this question with the incorrect answer.
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Expecting it to be a 2 digit no. taking 2 (10a + b) = 10b + a gives 19a=ut 8b now d taking L.C.M of 19 and 8 ............ to find later that condition is not satisfied and continuing to 3 digit and goin on ...............................................
but is a very long process .... does any1 hav a shorter way by d way ididn't continued i googled