Find the total number of four-digit palindromes that are divisible by 3. (Recall that a palindrome is a nonnegative sequence of digits which reads the same forwards and backwards, such as 1331. Zero cannot be the first digit.)
1 - (1001 , 1111 , 1221 , 1331 , 1441 , 1551 , 1661 , 1771 , 1881 , 1991 , 2002 , 2112 , 2222 , 2332 , 2442 , 2552 , 2662 , 2772 , 2882 , 2992 , 3003 , 3113 , 3223 , 3333 , 3443 , 3553 , 3663 , 3773 , 3883 , 3993 , 4004 , 4114 , 4224 , 4334 , 4444 , 4554 , 4664 , 4774 , 4884 , 4994 , 5005 , 5115 , 5225 , 5335 , 5445 , 5555 , 5665 , 5775 , 5885 , 5995 , 6006 , 6116 , 6226 , 6336 , 6446 , 6556 , 6666 , 6776 , 6886 , 6996 , 7007 , 7117 , 7227 , 7337 , 7447 , 7557 , 7667 , 7777 , 7887 , 7997 , 8008 , 8118 , 8228 , 8338 , 8448 , 8558 , 8668 , 8778 , 8888 , 8998 , 9009 , 9119 , 9229 , 9339 , 9449 , 9559 , 9669 , 9779 , 9889 , 9999 )>>Total = 90 four-digit palindomes.
2 - (1221, 1551, 1881, 2112, 2442, 2772, 3003, 3333, 3663, 3993, 4224, 4554, 4884, 5115, 5445, 5775, 6006, 6336, 6666, 6996, 7227, 7557, 7887, 8118, 8448, 8778, 9009, 9339, 9669, 9999) >>Total = 30 palindromes that are divisible by 3
