I counted the money and got some points.

Note: My algorithm uses a linear equation of one variable, but even in primary school, I learned how to solve the linear equation of one variable, so you can use it with confidence.

Analysis: 100 coin, * * 2 yuan is 200 cents, so the average face value of each coin is 2 cents.

So the scores of 1 and 5 should be grouped one by one, so the average face value is 2 points. Of course, you can calculate orally. Every time there are three 1 points, there will be a 5-point. At this time, the average face value of this group of four coins is (3 *1+1* 5)/(3+1) = 2 points.

Now we know that the total value of 2 points is more than 1 point 13 points, so there are 2 * x- 13 blocks 1 point and (2*x- 13)/3 blocks and 5 points. So the column:

x+(2 * x- 13)+(2 * x- 13)/3 = 100

X=32, so 1 has 5 1, and 5 has 17. The checking calculation shows that there is no problem.