**Ron Barnette's Zeno's Coffeehouse
Challenge #49 Result**

**Dear Zeno's patrons: Here are the results of our previous challenge.
I have listed the original problem, with three winning results listed below. Thanks to all
you Zeno's patrons who took up the challenge! Over 200 entries were received, with over
175 winning solutions. Thanks! **

**Ron Barnette**

**One late night at the Coffeehouse, a customer, Helen, posed the following
challenge to Maggie and Charles. Helen removed four cards from her pocketbook and placed
them on the table. She revealed (truthfully) that each card has a letter on one side and a
number on the other. The cards below are what Maggie and Charles see. Helen then asked
them to consider this statement:**

**"Of the four cards shown, those with vowels on
one side have even numbers on the other"**

**Her challenge is this: what is the quickest way
(i.e. the least number of card-flips) to deductively establish the truth of the above
statement? And which card or cards would one flip over?**

**Must ALL be flipped? Fewer? Which ones? Obviously,
Charles and Maggie must flip at least one card to see what's on the other side.**

**Results from three sample solutions:**

**Congratulations to James, Simon, and
Gilles!**

**James Hess, Lafayette, LA, USA**

comments: If all the cards shown are supposed to have a vowel on

one side have an even number on the other side, the card

with E facing up and the card with 5 facing up need to be

flipped to decide if it is so.

The opposite side of the 'E' card must have an even number

on the other side, then the statement can be true,

otherwise: if there is an odd number on the other side of

the 'E' card, the statement must be false.

The card with 'F' facing up does not need to be flipped,

because the only vowel sounds are the letters 'A', 'E',

'I', 'O', 'U', and since it is established one side of each

card is a letter and the other a number, the other side

cannot be a vowel.

The card with '2' facing up does not need to be flipped,

because it is an even number, and the other side must

be a letter: whether a vowel or a non-vowel is found

on the other side, this card does not violate the statement.

The card with '5' facing up needs to be flipped.

If something other than a vowel appears on the other side,

then the statement can be either true or false, but

if a vowel is found on the other side, the statement must

be false.

Having considered all possible ways for the statement

to be violated: if there is an even number on the

opposite side of the 'E' card and there is also something

other than a vowel on the opposite side of the '5' card,

then the statement is true, otherwise it is false.

**Simon Lutterbie, St. Mary's City, MD, USA**

comments: There need only be 2 flips. First, you must flip the E, to make sure it
has an even number on the other side. You must also flip the 5, to make sure it
doesn't have a vowel on the other side.

This problem can be simplified as an p-->q (If p, then q) situation. To verify
this, one must confirm the antecedent, which is done by showing that if a vowel is
presented, an even number is on the flip side. Second, one must deny the consequent,
which means showing that, if an odd number is presented (~q), then a consonant will be on
the other side.

**Gilles Gour, Montreal, Quebec, Canada**

comments: You only have to flip the "E" card and the "5" card. Helen's
states only that every card with a vowell on one side has an even number on the other. She
does'nt say anything about cards with consonants on one side (the "F" card)and
she does'nt say that cards with an even number on one side have a vowell on the other. If
there waa an uneven number on the other side of the "E" card or a vowell on the
other side of the "5" card, her statement would be untrue. But there could be an
even or uneven number on the other side of the "F" card or a consonnant on the
other side of the "2" card and that would not falsify her statement.

I rest my case.