George leaves Crawford, driving at a constant speed. After a while he passes a mile marker displaying a two digit number. An hour later he passes a mile marker displaying the same two digit, but in reversed order. In another hour he passes a third mile marker with the same two digits(backward or forward) separated by zero. What is the speed of George's car?

45 miles per hour?

16

61

106

 

very fast. It was a easy one.

Suppose the first sign had the two-digit number "AB".
   The value is actually:  10A + B

An hour later, the second sign had the number "BA".
   The value is actually:  10B + A

The difference is the distance he drove in an hour.
   (10B + A) - (10A + B)  =  9B - 9A

This is also his speed:  (9B - 9A) miles per hour.


Another hour passes and the sign says "A0B" or "B0A".
   The values are:  100A + B  or  100B + A, respectively.

The difference of the third number and the second number is also his speed.


Suppose the third sign said "B0A" or 100B + A.
Then his speed is:  (100B + A) - (10B + A) =  90B

We've expressed her speed in two ways; they must be equal.
   9B - 9A  =  90B   ÂÂ¨   -9A  =  81B   ÂÂ¨   A  =  -9B

But both A and B are positive digits.
So this case is impossible.


The third sign must have said "A0B":  100A + B
   The difference is:  (100A + B) - (10B + A)  =  99A - 9B

This equals his speed:  9B - 9A  =  99A - 9B
   then:  18B =  108A   ÂÂ¨   B = 6A


Since A and B are digits:  A = 1, B = 6

The first sign had "16", the second had "61", the third had "106"

his speed is 45 miles per hour.