Either there are fewer numbers, fewer typos or fewer key sentences.

The first problem is generally factorized, such as a 98, a 2, a 1 1, and there are no numbers. . .

In the second question, I made up my mind that it should be the product of four different direct subordinates, and the minimum value is 2X3X5X7=2 10.

The third typo kills 39 = 3x1345 = 3x549 = 7x756 = 2x2x760 = 2x3x570 = 2x5x778 = 2x3x1384 = 2x3x791= 7x/kloc-0.

After decomposition, it is found that there are three 13, six 7 3, five 6 and three 9 2, and each group of numbers is 2x2x3x3x5x7x13.

There are 13 in the three numbers, so the three groups must be divided into three groups: 39 a b, 78 c d, 91e f.

More specifically, there are 3x3x5x7xx1378 left in group 56-56. If 39 doesn't have a number, it is 9 1 56 45, and the remaining 60 and 70 will be separated by 5, and 49 and 70 will be separated, so 60 49 78 will be a group of 39 70 84.

4 (a+b) x (a-b) =19961996 factorization 2X2X499 are two natural numbers, so a+b=998 a-b=2.