7 out of 12 × 3 out of 28 × 16 out of 21!

7 out of 12 × 3 out of 28 × 16 out of 21!

7 / 12 × 3 / 28 × 16 / 21 divide 7 into 28, divide 3 into 12, and then divide the remaining two 4 into 16
＝1/21

（1983+1985+…… +1999)/（1984+1986+…… +2000）

（1983+1985+…… +1999)/（1984+1986+…… +2000）
=【（1983+1999）×9÷2】/【（1984+2000）×9÷2】
=（1983+1999）/（1984+2000）
=3982/3884
=1991/1992

Among the 1987 scores of 1 / 1988, 2 / 1988, 3 / 1988,..., 1986 / 1988 and 1987 / 1988, how many are the simplest scores?
The answer given at the back of my homework is 841; the answer given on the third floor is 840, only one short. Can you check it for me again?

First, we decompose 1988 into mass factor
1988=2*2*7*71
There are three prime factors, 2,7,71
The problem is to find out how many multiples there are between 1 and 1987
The multiple of 2: 1988 / 2 = 994, but it should be noted that 1988 can not be included, because it is between 1 and 1988, so we need to subtract 1, that is, 994-1 = 993
Similarly, the multiple of 7: 1988 / 7-1 = 283
Multiple of 71: 1988 / 71-1 = 27
That is, 1987 - (993 + 283 + 27) = 684
There is still one problem to pay attention to: the common multiple between them has been subtracted. For example, 14, which is a multiple of both 2 and 7, has been subtracted twice, so 1 has to be added
Multiple of 2 * 7 = 14: 1988 / 14-1 = 141
Multiple of 2 * 71 = 142: 1988 / 142-1 = 13
The multiple of 7 * 71 = 497: 1988 / 497-1 = 3
In addition, we also need to deal with the problem of their three common multiples: 2 * 7 * 71 = 994. In the previous step, it is subtracted three times, and in the next step, it is added three times. In this way, it is not taken into account. In the end, we have to subtract it, so the final result is:
684+(141+13+3)-1=840
That is, there are 840 minimalist scores