In 2006, the digits in Perry's age were the same as the digits in Laura's age, but in reverse order.

Question

In 2005, Perry was twice Laura's age.
What is the difference between Perry's and Laura's ages?

Answer 1

assume their ages were 2-digit numbers
in 2006:
Perry's age : 10x + y
Laura's age : 10y + x

in 2005:
Perry's age : 10x + y - 1
Laura's age : 10y + x- 1

10y+x-1 = 2(10x+y-1)
10y +x-1 = 20x+2y-2
19x -8y = 1

x = (8y+1)/19
or
y = (19x-1)/8

so 19x-1 has to be divisible by 8

of course we only have to try y's from 1 to 9 , possible values of (19y-1)/8 are:
1,18/8 --- 2, 37/8 ----- 3, 7 ---- ahhhh
sure enough , when y = 3, x = 7

So in 2006 Perry is 37 and Laura is 73
and the difference of their ages in 2006 is 36

Answer 2

By trial and error:
Perry's age = 73
Laura's age = 37
Difference of their ages = 36 years
Checking:
73-1 = 2(37-1)
72 = 2(36)
72 = 72

Answer 3

Great answer Reiny but line 8 is the wrong way round and should read
P=2*L
10x+y-1=2(10y+x-1)
but it doesn't change the answer