Rajesh06 wrote…

Tom and Ray said that there was no algebraic solution to this puzzler.

To be fair, what they said was that they weren’t sure whether an algebraic solution existed. That’s silly, of course, because even a Course XV slacker like me knows this can be solved algebraically.

You can’t blame them for staffing out the puzzler responses. I wouldn’t want to read all of the mail from losers like us. Still, you’d think their staff would call attention to the fact that some of us provided an algebraic explanation.

My version:

Aside from cheating and just looking up the answer on a conversion table, we can solve the problem algebraically.

If we express the temperatures as combinations of digits A and B, then we can write the expression, (10A + B - 32) / 1.8 = 10B + A. (This takes into account the conversion from degF to degC, which comprises subtracting 32 and dividing by 1.8.) The expression, when solved for A, boils down to A = (17B + 32) / 8.2.

We can now plug in values for B to arrive at A, and, therefore, the temperatures that satisfy the conditions of the puzzler. However, we can constrain our choices because we know certain things – e.g., A and B can only represent single-digit integer values, the lower of the two temperature values must be above freezing because it was raining, neither digit can be zero because the display would not show a leading zero, etc.

So, we plug in ‘1’ for B and arrive at ‘6’ for A. In fact, 61 degF = 16 degC (allowing for roundoff error).

Next, we plug in ‘2’ for B and arrive at ‘8’ for A. In fact, 82 degF = 28 degC.

By the time we plug in ‘3’ for ‘B’, we are out of range for A – it is no longer a single digit.

So, the two temps are 61 and 82 degF (corresponding to 16 and 28 degC).