A natural number ends in 2. If we move this digit 2 to the beginning of the number, then the number will be doubled. Find the smallest number with this property.
@Numb3r1
ab...2=2ab.../2 more specifically, whatever number ab... is (multiple digits, clearly, as 2d cannot equal d2*2 because that would be (2d)4, but 42 is not twice 24 or vice versa) ab... times ten plus two equals 2*10^(?) plus ab....
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That means 2*10^(?)=9(ab...) plus 2. ab... must end in a 2, because 9*ab.... ends in an 8 to make 9*ab...+2 end in a zero.
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Wait, that has so many issues... ignore it.
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Here's a fixed version:
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That means 2*10^(?)+ab...=20(ab...) plus 2. The twenty accounts for the doubling of the original number.
Therefore, 2*10^(?)=19(ab...) plus 2. 2*10^(?)-2 looks like 19....98. Divide by 19 and we have 1...?
Nope, that makes no sense.
ab2=2ab/2... let's see... ab must start in a 1, so I was right anyway.
Unless we carry... that's it!
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2ab2=ab22*2... nope. I should just go to sleep. Good luck!
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Thank you sir
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