A Nice Car Talk Algebra Problem

I don't want to spoil the Car Talk puzzler - so I am releasing this today (the day they will share the solution). Here is the full version of the puzzle. In summary:

A person sees a sign that displays the temperature in both Celsius and Fahrenheit. On a fall day, the person see the temperature in the morning and the afternoon. The final temperature in Fahrenheit is 36 degrees cooler. The Celsius reading was the same as it was in the morning.

Read Car Talk's version of the puzzle, it is more entertaining.

And now for a solution. I will post this late enough that it won't help you with the show.

Fahrenheit vs. Celsius

The first thing to do is to determine a relationship between the two temperature scales. I know two things:

• 32° F is the same as 0° C (freezing point of water)
• 212° F is the same as 100° C (boiling point of water)

Let me call T_C the temperature in Celsius and T_F the temperature in Fahrenheit. If I assume a linear scale for both F and C, then I can write:

Where m and b are just two constants. Since I know two matching sets of variables, I can write the following two equations:

From the first equation, I get that b = 32° F. Plugging this into the second equation, I can solve for m:

Putting the values for m and b back in, I get:

What other info?

Now that we have that, what else is there? Really, there are 4 numbers in this problem. Let me call them:

• T_F1 - the temperature reading in Fahrenheit in the morning.
• T_F2 - the temperature reading in Fahrenheit for the afternoon.
• T_C1 and T_C2 - are the same readings (morning and afternoon) in Celsius.

And what do I know about these values? I know that:

Maybe you can see the problem. You can't really go forward with a solution unless you know what is wrong with the temperature sign. The puzzler indicated there was a malfunction, so what could it be? I suspect that the minus sign on the Celsius temperature doesn't turn on. Could this give us a reasonable answer? Well, that would change the above equations to:

Let me re-write them in terms of the change in temperature:

Since the change in temperature should be the same amount (but not the same value), I can say that:

Let me now convert the change in temperature of -36° F to Celsius:

And from before, the change in temperature in Celsius is just -2T_C1. This means:

So, if the initial temperature is 10° C, this would be 50° F. The final temperature would be -10° C and 14° F. That seems to work. Fits for a fall day (although one that ends quite cold).

But Wait!

Are there any other solutions? What if the "tens" spot doesn't work on the display board? This means you could start with a temperature of 22° C and end with a temperature of 12° C. However, the display would say "2 C" for both cases. If you had a temperature change of 10° C, this would be a temperature change of 18° F. Ok, that wouldn't work. But what if the temperature went from 22° C to 2° C? This would also display "2 C" for both cases and would produce a change of 36° F. Oh, and 22° C is the same as 72° F. So this starting temperature seems reasonable. I guess you could also do any change in temperature such that it was 20° C change.

Oh, here is another solution. What if the whole Celsius readout is stuck? Then it would just always read something like 17° C. Ok, that is a cheap solution.

In summary: I love Car Talk.