palindrome number

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MATHPRITOM
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palindrome number

Unread post by MATHPRITOM » Sat Feb 04, 2012 7:58 pm

prove that, there exits no palindrome number of 2 digit which is a perfect square.no trial & error,plz.try to prove with logic.

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nafistiham
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Re: palindrome number

Unread post by nafistiham » Sun Feb 05, 2012 4:13 pm

all the two digit palindromes are $10k+k=11k$ type. where $k$ can be $1,2,3,4,5,6,7,8,9$ but if $11k=p^2$ $11$ must divide $k$. which it does not.so it isn't possible.
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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qeemat
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Re: palindrome number

Unread post by qeemat » Fri Feb 17, 2012 4:21 pm

number 23. Reverse and add that number to 23 to yield the palindrome 55.

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nafistiham
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Re: palindrome number

Unread post by nafistiham » Sat Feb 18, 2012 3:59 pm

qeemat wrote:number 23. Reverse and add that number to 23 to yield the palindrome 55.
sorry,but i am a little confused what are you trying to say. :?
the condition says to find a two digit palindrome square.
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
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Prosenjit Basak
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Re: palindrome number

Unread post by Prosenjit Basak » Fri Jan 25, 2013 10:19 pm

The palindromes of two digits always have the same digit .So there are $9$ palindromes of two digit number. If there remains a perfect square palindrome,its first digit must be $1,4,5,6,9$ .So the palindromes must be $11,44,55,66,99$ .But neither of them is perfect square.So there doesn't exist any palindrome perfect square. :idea:
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zadid xcalibured
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Re: palindrome number

Unread post by zadid xcalibured » Sat Jan 26, 2013 12:16 am

Prosenjit,
Your solution is close to brute force.you have to show that 11 must divide them.Eventually in order to be a perfect square 121 must divide them.Which is not possible.

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