Six digit 2003 can be divided by 99, and its last two digits are______ ．

Six digit 2003 can be divided by 99, and its last two digits are______ ．

Because 99 = 9 × 11, the number must be divided into 11 and 9 at the same time. According to the property that the sum of every digit is also a multiple of 9, then 2 + 0 + 0 + 3 = 5, 9-5 = 4, the last two digits sum is 4 or 18-5 = 13. Because the sum of odd digits is 2 + 0 = 2, and the sum of even digits is 0 + 3 = 3, if the last two digits sum is 4, the odd even digits can't be equal to 0, so the last two digits sum is 13 Let X be a number on the tens, 2 + x = 3 + (13-x), 2 + x = 16-x, & nbsp; 2x = 14, & nbsp; X = 7, 13-7 = 6, so the number is 200376. A: the last two digits of the number are 76. So the answer is: 76