### palindrome number

Posted:

**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.

The Official Online Forum of BdMO

https://matholympiad.org.bd/forum/

Page **1** of **1**

Posted: **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.

Posted: **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.

Posted: **Fri Feb 17, 2012 4:21 pm**

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

Posted: **Sat Feb 18, 2012 3:59 pm**

sorry,but i am a little confused what are you trying to say.qeemat wrote:number 23. Reverse and add that number to 23 to yield the palindrome 55.

the condition says to find a two digit palindrome square.

Posted: **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.

Posted: **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.

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.