Ten divided by three equals three plus one, which solves the problem.

Solution: let everyone be assigned to X and leave Y.

x*3+y= 10

y= 10-3x

Everyone gets at least 1, x>= 1, x: n *, and the minimum value of n * is 1, so it contains x >；; = 1, so X >；; = 1 This condition is merged with N* and then merged into x:N*.

The remaining number is 0.

1 & lt； = 10-3x & lt； =9

-9 & lt； =-3x & lt； =- 1

1/3 & lt； = x & lt=3

x:N*

0.33 & lt= x & lt=3

Draw the number axis, x= 1, 2, 3.

max{ 1，2，3}=3

x=3，y= 10-9= 1

A: Each person will be allocated 3 at most, and the rest will be 1.