Can be inferred slowly; This number is divisible by 2 and 3, so it can be divisible by 6. According to the fact that only two numbers are not divisible, we can verify and exclude each one. If it is not divisible by 4, it cannot be divisible by 8, 12 (because 2 * 4 = 8, 3 * 4 = 12), which is obviously irrelevant, so it can be divisible by 4, 8, 12. There are 7, 9, 1 1, 13, 14 left. If it is not divisible by 7, it cannot be divisible by 14. If the other three are divisible, it will meet the meaning of the question. So one possibility is: No.7 and 14. If it is divisible by 7, it can be divisible by 14. So as to draw a conclusion

(1) There are four possible incorrect figures for two students: 7 and14,9 and1,9 and 13,1/kloc-0.

(2) Calculate the least common multiple of the above four possibilities and get: 2 * 3 * 4 * 5 * 3 *11*13 = 51480; 2*3*4*5*7* 13= 10920; 2*3*4*5*7* 1 1=9240; 2 * 3 * 4 * 5 * 7 * 3 = 2520, so: 1, when the numbers of two incorrect students are 7 and 14, this five-digit number is 51480,2, and when the numbers of two incorrect students are 9 and/kloc-0. This five-digit number is 10920 or 2 1840( 10920*2) or 43680( 10920*4) or 54600( 10920*5) or 770. 3. When the figures of two incorrect students are 9 and 13 respectively, the five digits are 18480, 36960, 46200, 64680, 73920 and 92400 respectively; 4. When two students with incorrect numbers are 1 1 and 13, The five digits are 10080, 15 120, 17640, 20 160, 22680, 25200, 30240, 35280 and 37800.