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## A Problem Begging to Be Extended

The following problem was published in Oregon Mathematics Teacher, a great source for problems that I have used regularly with students for more than 30 years. Though the solution is nice, the problem itself is begging to be extended...

Initial Problem: While driving on I-5 at a constant speed, Joe passes a milepost with a two-digit number. Exactly one hour later he passes a milepost with the same two digits, but in reverse order. Exactly one hour later he passes a third milepost with the same two inital digits, separated by a zero. what was the number on the first milepost?

Once you have solved the above problem, consider this extension (one of many that seem ready to appear!).

Extension: While driving on I-5 at a constant speed, Joe passes a milepost with a two-digit number. Exactly one hour later he passes a milepost with the same two digits, but in reverse order. Seeing that he is running late, Joe increases his speed by a factor of n (a positive integer < 11). Then, as before, exactly one hour later, he passes a third milepost with the same two inital digits, separated by a zero. what was the number on the first milepost?

Solve this new problem...and you should discover that for some cases, the answer is not unique. Also, ignore the fact that Joe should be arrested for speeding in most instances...

Source: Oregon Mathematics Teacher, Jan/Feb 2004, pp. 43, 45.

Hint: If the road sign has the digits AB, then it can be expressed as 10A+B. Also, given the problem situation, what do you know about the order relationship between A and B?

Now, express the full problem algebraically...and solve/reason.

Solution Commentary: For the first part, you should start with (10B+A) - (10A+B) = (100A+B) - (10B+A), which algebraically solves to B = 6A. Because A and B are single digits, the first road sign must be 16.

To shift to the extension, start with (10B+A) - (10A+B) = (1/N)[(100A+B) - (10B+A)] where Joe travels N times faster between the sign #2 and sign #3. By exploring this, you should find solutions for N = 1 (initial problem), 3, 4, 5, 7, and 10...with multiple solutions for some of these values (you find them!).

And, what happens if you remove the requirement that N < 11? Or, what happens if Joe slows down by a factor of 1/N? Or...and the extensions continue to come to mind....