數學難題兩題~~~

1)Let the number be 10A + B ,
then 10A + B - 9 = 10B + A
9A - 9B - 9 = 0
A - B = 1.....(1)
sum of the digits is 11 :
A + B = 11.....(2)
(1)+(2):
2A = 1 + 11
A = 6
B = 5
The number is 65.
2)Let the number be 10a + b ,
10a + b - 18 = 10b + a
9a - 9b - 18 = 0
a - b = 2.....(1)
sum of the digits is 12 :
a + b = 12.....(2)
(1)+(2):
2a = 2+12
a = 7
b = 5
The number is 75.

Answer 001 is accurate, well-presented and helped the host earlier.

1.Let x be the tens digit and y be the units digit

x+y = 11 ----------- (1)
(10y+x) - (10x+y) = -9 ----------- (2)

(2)
(10y+x) - (10x+y) = -9
10y + x - 10x - y = -9
9y - 9x = -9
9(y-x) = -9
y-x = -1 ---------------(3)

(1)+(3)
(x+y) + (y-x) = 11 + (-1)
2y = 10
y= 5

Sub y = 5 into (1)
x+5 = 11
x = 6
So the number is 65.



2.Let x be the tens digit and y be the units digit

x+ y = 12 -------------(1)
(10y+x) - (10x+y) = -18 -------------(2)

(2)
(10y+x) - (10x+y) = -18
10y +x -10x -y = -18
9y -9x = -18
9(y-x) = -18
y-x = -2 ---------------------(3)

(1) + (3)
(x+y) +(y-x) = 12 + (-2)
2y = 10
y = 5

Sub y=5 into (1)

x+5 = 12
x = 7

So the number is 75.

It should be the correct answer if you didn't copy the question wrongly.

2010-01-14 22:40:25 補充：
I think my steps are more formal. Please trust me but not the one who answered you first.