4 DIGIT PALINDROMIC & CUBE
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Prove that,there exists no palindromic number of 4 digits which is a perfect cube.
- Phlembac Adib Hasan
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Re: 4 DIGIT PALINDROMIC & CUBE
Pritom Vaia, please check your proof.Because $1331=11^3$.
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- Posts:190
- Joined:Sat Apr 23, 2011 8:55 am
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Re: 4 DIGIT PALINDROMIC & CUBE
Ya, Thanks. I made wrong because I was wrong in solving a equation.
I thought in this way.
Let,the palindrome is abba. Then, 1001a+110b=$k^3 $.
Now,11(91a+10b)=$k^3$.
so, 91a+10b=$ 11^2* n^3 $.
Let, n=1,then,91a+10b=121.
Then, the only positive integer solution is (1,3) & we got the case. But, When I made the problem I took there exits no solution... That was the wrong & so I wrote there exits no perfect cube..
& U can easily check n<2. coz, if n=2 then, R.H.S. is 968. But, the highest value of L.H.S. is 91*9+10*9=909.
I thought in this way.
Let,the palindrome is abba. Then, 1001a+110b=$k^3 $.
Now,11(91a+10b)=$k^3$.
so, 91a+10b=$ 11^2* n^3 $.
Let, n=1,then,91a+10b=121.
Then, the only positive integer solution is (1,3) & we got the case. But, When I made the problem I took there exits no solution... That was the wrong & so I wrote there exits no perfect cube..
& U can easily check n<2. coz, if n=2 then, R.H.S. is 968. But, the highest value of L.H.S. is 91*9+10*9=909.