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Mathematical reasoning problem
If there are three people at each table, there will be two more people, and the total number of people is a multiple of three, less 1.

If there are five people sitting at each table, there are four more people, and the total number of people is a multiple of five, and 1 is missing.

If there are 7 people at each table, there will be 6 more people, and the total number of people is a multiple of 7, less 1.

If there are nine people at each table, there will be eight more. The total number of people is a multiple of 9, minus 1.

Let the total number of people be n, then n = 5× 7× 9× k-1= 315k-1,where k is a natural number.

When each table is seated at 1 1 person, there are not many people, and n is a multiple of 1 1, that is, 3 15k- 1 is a multiple of 1 1.

3 15K- 1 =(308+7)K- 1 =( 1 1×28+7)K- 1 = 1 1×28K+7K- 1

So 7k- 1 is a multiple of 1 1, and k can be 8,19,30, ..., 1 1m-3, (where m is a natural number).

When k = 8, the total number of people n = 25 19.

25 19÷3=839……2

25 19÷5=503……4

25 19÷7=359……6

25 19÷9=279……8

25 19÷ 1 1=229