According to "if there are 9 classes on average, there are still 8", we can know that the multiple of the number of balls is 9, which is less1;

According to "the average number of classes is 12, then there is 1 1", we can know that the multiple of the number of balls is 12, and it is less1;

According to "the average number of classes is 15, then there is 14", we can know that the multiple of the number of balls is 15, and1is missing;

So the number of balls is 9 12, and the common multiple of 15 is less than 1.

Ask again, "How many balls are there at least in the school?"

So find the least common multiple of 9, 12 and 15 and subtract 1.

The least common multiple of (9, 12, 15) is 180.

So: the school has at least 180- 1 = 179 balls.

2、

By: "Cut rectangular paper with a length of 135 cm and a width of 105 cm into square paper with the same size. There is no redundancy."

It is known that this is the greatest common divisor of 135 and 105.

Simplify and get it

135=5*3*3*3

105=5*3*7

Therefore, the greatest common divisor of (135, 105) is 3 * 5 = 15.

In other words, the maximum side length of these squares is 15cm.

3、

From: "The decimal point moves one place to the right, which is equal to B"

It is known that the number b is 10 times of the number a.

So: "The sum of A and B is 4.62"

It can be understood as: (1+10 =)11times a number is 4.62.

So: the number A is 4.62/ 1 1 = 0.42.

Then: the number B is 0.42 * 10 = 4.2.