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

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