主题：【讨论】数学思考题 -- 伊粟

Agree with your opinion. Details as below:

Let the age of the older and younger people be A and B,

Also let x,y be the integer representing the tens digit and ones digit of the older's age.

Then, in the case of “exchange age” (对易数年龄), we have

A=10x+y, B=x+10y ; (A,B, x,y are all integers)

also, We know x<>y, A>B, 0<A,B<100, (x,y) are in the set of {0,1,2,3,...8,9}

Then we have

x>y .......(1) (if x <y then B>A) and

0<|x-y|< 9 ......(2) ; and

0<A-B<100......(3) ;

From (3), 0 <10x+y-x-10y <100,

or 0<9(x-y)<100,

or 0<x-y< 11

Combining (1), (2), (3) ,

The constrain is : 0< x-y < 9, and (x,y) are in the set of {0,1,2,3,...8,9}......(4)

From (4), there are only 9 cases meet the required situation:

x-y =1, or A-B=1*9=9;

x-y =2, or A-B=2*9=18;

x-y =3, or A-B=3*9=36;

........

x-y =8, or A-B=8*9=72;

x-y =9, or A-B=9*9=81;

Therefore,

1. The minimum age difference is 9 years, which the two people can have at less one chance to meet the "exchange age" requirement.

which is in the case of x-y =1, or A-B=9;

2. Also in the case of x-y =1, or A-B=9; the two persons can have maximum chance to meet the "exchange age" requirement.

i.e., {01,10},{12,23}, {23,32}, ....{89,98}; in total 9 pairs.

3. In the case of x-y =9, or A-B=81; or the age difference is 81 years, the two persons can have only one (最多出现一次) chance to meet the "exchange age" requirement.

which is {09,90}.