There are a lot of such numbers:123112210111112120221210011111200210001101011020111011111111121120002201021000011011011100111111112000021000001100100110020011010101101110110121011100011110101111020111110111111111120000022001002100000011001100110100101101111011100001111011011111001111111111120000002100000001100010001100020001100101001100111001100121001101000101101010101101020101101101101101111101110000011110010011110020011110101011110111011111000111111010111111101111111111111...But it's hard to prove that there are no any palindromic number X, when X^2 is palindromic too
Aren't 10...01 numbers always palindromic?
This comment has been removed by the author.
I posted 12345654321, because the digits are consecutive
@BenVitale, ok.Squares of [1, 11, 111, 1111, 11111, 111111, 1111111, 11111111, 111111111] have consecutive digits at all
There are a lot of such numbers:
ReplyDelete1
2
3
11
22
101
111
121
202
212
1001
1111
2002
10001
10101
10201
11011
11111
11211
20002
20102
100001
101101
110011
111111
200002
1000001
1001001
1002001
1010101
1011101
1012101
1100011
1101011
1102011
1110111
1111111
2000002
2001002
10000001
10011001
10100101
10111101
11000011
11011011
11100111
11111111
20000002
100000001
100010001
100020001
100101001
100111001
100121001
101000101
101010101
101020101
101101101
101111101
110000011
110010011
110020011
110101011
110111011
111000111
111010111
111101111
111111111
...
But it's hard to prove that there are no any palindromic number X, when X^2 is palindromic too
Aren't 10...01 numbers always palindromic?
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteI posted 12345654321, because the digits are consecutive
ReplyDelete@BenVitale, ok.
ReplyDeleteSquares of [1, 11, 111, 1111, 11111, 111111, 1111111, 11111111, 111111111] have consecutive digits at all