There are some apples. If you put them in a pile of 6, you will get one less. If you put them in a pile of 8, you will get seven more. The number of apples is about 50. How many apples are there?

There are some apples. If you put them in a pile of 6, you will get one less. If you put them in a pile of 8, you will get seven more. The number of apples is about 50. How many apples are there?


The common multiple of 6 and 8 is 24, 48, 72
The number of apples is about 50
There are 48-1 = 47 apples



There is a pile of apples, if there are three, three, and the last two, if there are four, four, and the last three, if there are five, and the last four,
How many apples is there at least in this pile


Find the law
The multiple of 3 + the single digit of 2 may be 1 to 9, which is cycled every 30 digits
The multiple of 4 + the single digit of 3 may be 7 1 5 9 3, cycled every 20 digits
The number of multiples of 5 + 4 is only 9 or 4, which is cycled every 10 digits
From the multiple of 4 + 3, the single digit may be 7 1 5 9 3 and the multiple of 5 + 4, the single digit may only be 9 or 4 can only be 9
Find the final result according to the cycle
Multiple of 3 + 2: 29 59 89
Multiple of 4 + 3:19 39 59 79
Multiple of 5 + 4:9 19 29 39 49 59 69
The result was 59
The least common multiple minus one is the most scientific



1、 1. There is a pile of apples. If there are three, three and the last two, if there are five, five and the last four, if there are seven and the last seven,
The last six. How many apples are there in this pile?
2. Cut a rectangle 1.36M long and 0.8m wide into squares of the same size. If the area of the square is as large as possible and there is no surplus after cutting, how many squares can be cut?
3. If 16 pears and 19 apples are divided equally among several children, then there are two more pears and two less apples. How many children are there? (in the three sections above, please give reasons and list the formula.)


1. The common factor of 3,5,7 is 3 * 5 * 7 = 105, the remaining 2,4,6 are less than 3,5,7 by 1, so 105-1 = 104. There are at least 104 apples in this pile. 2,1.36 M = 136 cm, 0.8 m = 80 cm, the maximum common factor of 136,80 is 8136 / 8 × 80 / 8 = 17 × 10 = 170, 170 can be cut