Firstly, 1988 is decomposed into prime factors:

1988=2*2*7*7 1

There are three prime factors, 2,7,71.

Between 1 and 1987, any multiple of them is not the simplest fraction, and the problem is transformed into how many multiples of them are between 1 and 1987:

Multiplication of 2: 1988/2=994, but it should be noted that 1988 cannot be counted, because it is between 1 and 1988, so we have to subtract 1, that is, 994-/kloc.

Similarly, the multiple of 7: 1988/7- 1=283.

The multiple of 7 1:1988/71-1= 27.

That is 1987-(993+283+27)=684.

We haven't finished here, but we should pay attention to one problem: the common multiples between them are all reduced, such as 14, which is both a multiple of 2 and a multiple of 7, and has been reduced twice, so 1 should be added. These figures are as follows:

2 * 7 =/a multiple of kloc-0/4:1988/14-1=14/.

The multiple of 2 * 7 1 = 142:1988/142-1=13.

The multiple of 7 * 7 1 = 497: 1988/497- 1=3.

In addition, it also involves the common multiple of the three of them: 2*7*7 1=994. The last step was reduced three times, and the next step was added three times, so that he was not counted, and finally it was reduced, so the final result was:

684+( 14 1+ 13+3)- 1=840

In other words, the simplest score is 840.

2:

Xiaoming drinks milk 1 cup1/3+1/3+1/2 = 7/6 >1

So I drink more water.

3:

1 Class planted X trees, Class 2 planted Y trees and Class 3 planted Z trees.

Three classes always plant a tree.

Then A=X+Y+Z

A-X=40，A-Y=50，A-Z=60。

So there is 3a-(x+y+z) = 150.

3A-A= 150

A=75

Because A-X=40, A-Y=50 and A-Z=60.

So x = 35, y = 25 and z = 15.