**Puzzle 71**

This puzzle was designed for high school Algebra 1 students. An exercise that is frequently assigned in such a class is to identify what is done in each step in the solution of an equation (or the derivation of a formula or the proof of a theorem) and, assuming the truth of the given, if each step is justified, the final conclusion must be true. In this puzzle, we are given the justifications, but the conclusion is troublesome.

We’ll start by letting *a* and *b* represent the same quantity. Since any quantity is equal to itself, *a* = *b*

Multiplying both sides of the equation by the same quantity, *a*, we get *a* ² = *ab*

Subtracting the same quantity, *b* ², from both sides of the equation, we get *a* ² – *b* ² = *ab – b *²

Factoring both sides of the equation, we get (*a* + *b* )(*a* – *b* ) = *b* (*a* – *b* )

Dividing both sides of the equation by the same quantity, (*a* – *b* ), we get *a* + *b* = *b*

Substituting *b* for *a*, its equal, we get *b* + *b* = *b*

Substituting 2*b* for *b* + *b*, its equal, we get 2*b* = *b*

Finally, dividing both sides of the equation by the same quantity, *b*, we get 2 = 1

This is obviously not true. The puzzle is to find the error in reasoning that led us astray, despite the justification we provided for each step of the way.

Show Solution*a*+

*b*=

*b*, we divided both sides of the previous equation by the same quantity, (

*a*–

*b*), a seemingly permissible operation. Ater all, it is exactly what we do to correctly solve the equation, 2

*x*= 6. However, since we are given that

*a*=

*b*, it follows that

*a*–

*b*= 0, and it is not permissible to divide by 0. (See current Math Stuff posting on division by zero.)

**Puzzle 72**

How many distinct positions of the hands of an accurately-working analog clock are there that have both hands in exactly the same position and how long a time is it between two chronologically successive such positions?

Show SolutionThere is only one time when both hands are pointing exactly at one of the twelve numbers on the clock and that is at 12:00, when both hands are pointing exactly at the “12”. The next time that both hands are in the same position is some time between 1:05 and 1:10 when both hands are pointing at the same point between the “1” and the “2”. We’ll get to exactly what that time is in a moment. The next time this happens again is between 2:10 and 2:15, when both hands are pointing at the same point between the “2” and the “3”. It happens again between 3″15 and 3:20, between 4:20 and 4:25, etc.,… until both hands are between 10:50 and 10:55. Each time this happens, the time is closer to the second of each pair of times until the minute hand overtakes the hour hand not between 11:55 and 12:00 (the equivalent of 11:60), but at exactly 12:00, when the hands are in the same position that we started from, so we don’t count it again. That means that there are 11 times that the two hands are in exactly the same position.

Since each hand is moving at a constant rate, the times when they are in the same position are equally spaced. Since there are 11 positions in 12 hours, the time between two such successive positions is hours or minutes, which comes to minutes or hour and 5 minutes. To find out what all the exact times are, just add hour and 5 minutes to the previous exact time. The first three times after 12:00 are , and .

**Puzzle**** 73**

Find two positive integers such that the product of 11 and the first integer plus the product of 13 and the second integer is 300.

Show Solution*x*and

*y*be the two positive integers. Then the puzzle states

(1) . Solving for *x* in terms of *y*, we get

(2) .

(3) Let . Solving for *y* in terms of *A*, we get

(4) .

(5) Let . Solving for *A* in terms of *B*, we get

(6) . Substituting for *A* [Eq. (6)] and *B* for [Eq. (5)] in Eq. (4), we get

(7) , so

(8) . Substituting for *y* [Eq. (8)] and for [By Eq’s. (3) and (6), both are equal to *A*] in Eq. (2), we get

(9) , so

(10) .

It is clear from [Eq. (8)] that *B* must be positive for *y* to be positive, and it is clear from [Eq.(10)] that for *x* to be positive, *B* < 3, so or . and , so the two positive integers are 19 and 7 or 7 and 18.

This puzzle is an example of a Diophantine equation, an equation requiring an integral solution, the most famous one being the one discussed in Fermat’s Last Theorem, which states that there is no integral solution for *x*, *y* and *z* to the equation if *n* is a positive integer greater than 2. The general methodology for a two-variable linear equation such as the one presented here is to find parametric equations for the two variables, using the algorithm illustrated in the solving of the current puzzle, and then apply any other conditions given in the problem to find the set of allowed solutions.

**Puzzle 74**

On that legendary island of the Trutels and the Falstels (see puzzles 69 and 70), there is another indigenous population called the Thirders, members of which tell the truth exactly once in every three utterances that they make in any individual conversation. (Strangely, in any conversation, they always make a number of utterances that is divisible by three. They must get a lot of puzzle fanciers among their tourist population.) One day, I overheard the following six-utterance conversation:

A: Today is Saturday. B: It’s Sunday. A: But yesterday was Thursday. B: It was Friday. A: Well, tomorrow is Wednesday. B: No, it’s Tuesday.

On what day of the week did I hear this conversation?

Show Solution

**Puzzle 75**