shubhamsrg
  • shubhamsrg
Fun question: Prove that a natural number formed by only 6's and 0's can never be a perfect square.
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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ganeshie8
  • ganeshie8
long time! good to see you again :)
shubhamsrg
  • shubhamsrg
Just came back to check if anything has changed. Woah ! the site became awesom-er ! Good to be back. ^_^ You have been a prominent member ever-since it seems :D
ganeshie8
  • ganeshie8
counter example \[0 = 0^2\]

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More answers

shubhamsrg
  • shubhamsrg
question edited.
ganeshie8
  • ganeshie8
Since \(n^2 \equiv 0,1 \pmod{4}\), we may try something like this : \[\sum\limits_{k} a_k*10^k\equiv \sum\limits_{k=0}^1 a_i2^k \pmod 4 \] because the terms are divisible by \(4\) for \(k\ge 2\)
ganeshie8
  • ganeshie8
Ahh that is just divisibility by 4 rule, its not going to help much...
shubhamsrg
  • shubhamsrg
This was a fantastic link that you added ! An interesting approach.
ganeshie8
  • ganeshie8
\[60060600 \\= 6*10010100 \\= 6*100101*10^2 \\= 3*100101*2^{\color{red}{3}}*5^{\color{red}{2}}\] I think it is easy to show that the exponents of \(2\) and \(5\) cannot be both even.
shubhamsrg
  • shubhamsrg
Spot on.
Kainui
  • Kainui
Here's my busted up excuse for a proof. It's complete :P If it's made up of only 6 and 0s then the leading digit must be a 6. Since it's square, some number's gotta square to it, the only way that's possible is if the leading digit of its square root is 8, since \(8^2=64\) is the only option. To get rid of that 4 we have to add 2 (since although adding 6 would turn this digit to 0, it would turn our 6 above into 7). In other words the first two digits are 6s: \[66... = (8X...)^2\] Here's where it breaks! It looks like puke if you write it out 'officially' so instead of wasting time typing up latex, we know X must be some value between 1 and 9, so quick trial and error gives that: X=1 \(\implies 65...\) X=2 \(\implies 67...\) And of course higher X just give 67 or higher, so we can't get 6 in that digit!
ganeshie8
  • ganeshie8
How do we know the leading digit of the root is 8 ?
Kainui
  • Kainui
No other numbers square to 6 in the leading digit. We can at most carry a 1 from lower digits, so if there was a digit that squared to a leading digit of 5 we might have had something to work with. But there isn't, so we don't.
Kainui
  • Kainui
Actually I think you're right I've overlooked somethin
Kainui
  • Kainui
Oh well, this is too trial-and-error based of a method for me to care to try to fix it up, but I think ultimately there might be a few more cases to check since it looks like there can be up to 2 carried over so 7 might be the leading digit. At any rate, this sorta idea will work as long as the statement to be proved is true I think, so I'm not too worried about fixing it, I just wanted to have fun coming up with an idea. :P
Kainui
  • Kainui
It would be interesting if this sorta algorithm never terminated and there were infinite-digit numbers made up of 6s and 0s that ARE perfect squares though...

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