The penultimate son will be assigned to: N- 1+N/6 horses.
The penultimate son will be assigned to: the sum of the number of horses assigned to the last two sons after N-2+/6.
= N-2+(2N- 1+N/6)/6 = N-2+( 13N-6)/36
And so on, the fourth son from the bottom. ...
At least ensure that the penultimate son is assigned an integer, then N/6 is an integer, and n must be an integer multiple of 6. Let N = 6M, where M= 1, 2,3.
At least make sure that the penultimate son is assigned an integer, then (13N-6)/36 is an integer.
Let (13N-6)/36 = k, where K = 1, 2,3. ...
13N-6 = 36K
13N = 6(6K+ 1)
13M = 6K+ 1
The integer solution of this equation is
M= 1、K=2
M=7、K= 15
M= 13、K=28
……
When M=7, N=6M=42, that is, the elderly have 42 sons, which is unreasonable in real life. At the same time, there is no guarantee that the number of horses allocated by the penultimate son is an integer.
When M= 13 N=78, there is no guarantee that the penultimate son will get an integer number of horses.
In short, after incomplete induction, only M= 1 and N=6 are reasonable.
That is, * * * has six sons. The number of horses is:
Liuzi: 6+0/7 = 6.
Fifth son: 5+6/6 = 6.
Four sons: 4+(6+6)/6 = 6.
Third son: 3+(6+6+6)/6 = 6.
Second son: 2+6*4/6 = 6.
1 sub: 1+6 * 5/6 = 6.
There are 36 horses above.
From front to back:
Son 1: 1+35/7 = 6.
Second son: 2+28/7 = 6
The third son: 3+2 1/7 = 6.
Fourth son: 4+ 14/7 = 6.
Fifth son: 5+7/7 = 6
Liuzi: 6+0/7 = 6.
===================================
Modifications and supplements:
The son of N- 1 took the horse of N- 1 first, and then took the 1/7 of the rest of the horses. This means that 6/7 of the remaining horses will be left to the nth son. 1/7 is 6/7 of 1/6, and the nth son got n horses, so it is N- 1+N/6.