Harder question: Binary perfection. Can a number consisting of 300 ones and some number of zeros be a perfect square? Explain your answer.
moneybird taught me the trick on these...
Explain pls.
you mean the answer? No, the problem hasn't been up long enough.
Pumas campeón!
Thinking....
I'm gonna see if Jyqft has something doable this time...
No-idea
It was destined for meta-math anyway, but I know moneybird will solve it if I put it there...
This number is divisible by 3 (because of the divisibility rule) but not by 9. Therefore it cannot be a perfect square. :)
Yeah TracysaurusRex, that's it! I didn't know about that rule until I came on this site.
wow that is very nice rule,
moneybird referenced the rule used above as a special case of this theorem, which I had never heard of either: http://answers.yahoo.com/question/index?qid=20080716092006AAWNZNu
Not sure I see the connection though, but I'm sure it's insightful.
i think that rule is simply because of the fact that a perfect square must have only 2 factors 1 and the number itself, and if a number is a multiple of threee, 3 must come in its prime factorization, and it must come even number of times, hence it hasto be a multiple of nine to be a perfect square, as far as that theorem is concerned, it is elegant
Stom, you were probably thinking of primes! A perfect square has to have an ODD number of factors, like 100 has 1,2,5,10,20,50,100.
what i meant was a perfect squares must have only 2 factore 1 and the number itself, the number itself can have its own factors as well, thats not a n issue at all,
Just in case it's relevant, is the number in base 10 or base 2?
It sounds like it should be binary now that you mention it, but the answer is what TracySaurus gave, so it must be base 10 I suppose.
By "binary" they must just mean the fact they are using two numbers, not as a base.
Oh great, so that was the official answer, and it makes sense too.
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