The Jones Family

Last Problem:

Madame Malaise decided to buy a parrot to keep her company. But she wanted a parrot that talked. Madame Malaise asked the pet store clerk.

Does this parrot speak?

The clerk was unequivocal in her answer.

“This parrot,” she said, “repeats every word he hears. I guarantee it. “

That answer was sufficient to persuade Madame Malaise to purchase the parrot. After weeks of trying to teach her new parrot to speak, she never heard a peep out of him.

Did the pet store clerk lie? Or, is there something else going on here?

Answer:

The pet store clerk neglected to tell Madame Malaise that the parrot was deaf.

Today’s Problem:

Mr. and Mrs. Jones have two children. They tell you that at least one of them is a girl. Assuming that boys and girls are equally likely, what is the probability that their other child is a girl? (Hint: the answer is not as obvious as you might think).

The Parrot

Yesterday’s Problem:

There are three unmarked light switches that have been randomly set to “on” and “off” positions. Each is connected to a lamp in another room which will shine only when all three are in the “on” position.

If you were offered the chance to bet, for even money, that you could turn on the lamp with the flip of just one switch, should you take the bet?

Answer:

Such a bet is a losing proposition. Only three out of seven possible random settings allow for the light to be turned on with the flip of just one switch.

Today’s Problem:

Madame Malaise decided to buy a parrot to keep her company. But she wanted a parrot that talked. Madame Malaise asked the pet store clerk.

Does this parrot speak?

The clerk was unequivocal in her answer.

“This parrot,” she said, “repeats every word he hears. I guarantee it. “

That answer was sufficient to persuade Madame Malaise to purchase the parrot. After weeks of trying to teach her new parrot to speak, she never heard a peep out of him.

Did the pet store clerk lie? Or, is there something else going on here?

Would You Take the Bet?

Last Problem:

There are three light switches on the ground floor of a three story house. One of the switches turns on a light in the attic, but you do not know which one.

Your assignment is to find out which of the three light switches activates the lamp in the attic, but here is the rub: You are allowed only one trip to the attic to check on the light. Figure out how you can tell which of the three light switches turn on the light in the attic, allowing yourself to make only one trip to the attic.

Answer:

Many people who work on this problem assume that there is not sufficient information to solve it. This is because they are looking at the problem from the perspective or probabilities. The trick in solving the problem is to remember that when lights are turned on, they get hot. It remains warm many minutes after it is turned on.

With this in mind, here is the solution. First, turn on switch one and leave it on for several minutes so that the bulb in the attic will get good and hot (if in fact that is the switch that turns the attic bulb on). Next, turn off switch one and turn on switch two. Now, walk up to the attic.

If the light is on, then switch two obviously is the switch. If the bulb is dark but still warm, then switch one works the attic light. If the bulb is both dark and cold, then switch three – the one that has not yet been turned on – works the lamp.

Today’s Problem:

There are three unmarked light switches that have been randomly set to “on” and “off” positions. Each is connected to a lamp in another room which will shine only when all three are in the “on” position.

If you were offered the chance to bet, for even money, that you could turn on the lamp with the flip of just one switch, should you take the bet?

Which Switch Turns On the Attic Light?

Last Problem:

Place six glasses on a table – three upside down and three upright. Take any pair of glasses and invert them. If you continue to invert pairs for as long as needed, how long will it take to end up with all six glasses upright?

How long will it take for all six glasses to be upside down?

Answer:

The parity of the initial set up is odd. Three glasses and upside down and three are upright. An even number of moves each time will not change this.

This means that it will be impossible to have all six glasses upright or all six glasses upside down, regardless of the number of moves that are attempted.

Today’s Problem:

There are three light switches on the ground floor of a three story house. One of the switches turns on a light in the attic, but you do not know which one.

Your assignment is to find out which of the three light switches activates the lamp in the attic, but here is the rub: You are allowed only one trip to the attic to check on the light. Figure out how you can tell which of the three light switches turn on the light in the attic, allowing yourself to make only one trip to the attic.

How Long Will It Take?

Last Problem:

Perform the following trick for a friend:

Set up three glasses – two are face down and one is face up. Your goal is to bring all three glasses to the upright position in exactly three moves, turning over two glasses at a time. A quick test reveals that this is actually very easy to do. In fact, it can be done with any number of moves.

Once you have succeeded in demonstrating this feat to your friend, turn all three glasses over to the inverted position such that the tops of all three glasses are face down on the table. Now, challenge your friend to duplicate your feat.

Why does your friend become so frustrated because they cannot do what you just did so effortlessly?

Answer:

Try as your friend might, they will always fail. This is because turning over two glasses at a time changes the number of upright glasses by two or by zero. And, although the number of upright glasses in the first set up was one, so that adding two gave you a total of three, the number of upright glasses in the second set up is zero. Changing two at a time will allow your friend to fluctuate between zero glasses and two glasses but they will never get to three glasses.

The first set up has an odd parity while the second set up has an even parity. In both instances, turning over two glasses at a time will not change that parity. The trick lies in the fact that most people do not see that the original set up (where you demonstrated you ability to accomplish this feat) is different from the second set up.

Today’s Problem:

Place six glasses on a table – three upside down and three upright. Take any pair of glasses and invert them. If you continue to invert pairs for as long as needed, how long will it take to end up with all six glasses upright?

How long will it take for all six glasses to be upside down?

Secret Behind the Magic Trick with Three Glasses

Last Problem:

One of the most eloquent coin tricks is often explained as a feat of extrasensory perception when in fact if is simply an example of parity.

Here is the trick: Ask someone to toss a handful of coins on a table. After a quick peek at the result, turn your back and ask the person to turn over pairs of coins at random – as many pairs as he or she would like to turn over. Then ask the person to cover up one coin.

When you turn around, you can tell immediately whether the covered coin is showing heads or tails.

Can you figure out the mathematical secret which lies at the heart of this trick which is bound to impress your friends, especially when you can do it over and over again?

Answer:

Before you turn your back, check to see how many coins are showing heads. You know that the number of heads will increase by two, decrease by two or stay the same for every pair of coins that is turned over. Therefore, if the initial number of heads is odd, the number has to remain odd, no matter how many coins are turned over.

When you turn back around, count the number of heads that are now showing. If the number is odd, as at the start (or even, as at the start), the covered coin must be a tail. If the number of heads is even for an odd start (or odd for an even start) the covered coin must be a head.

The odd-even parity of this system is always preserved as long as pairs of coins (not individual coins) are turn over.

Today’s Problem:

Perform the following trick for a friend:

Set up three glasses – two are face down and one is face up. Your goal is to bring all three glasses to the upright position in exactly three moves, turning over two glasses at a time. A quick test reveals that this is actually very easy to do. In fact, it can be done with any number of moves.

Once you have succeeded in demonstrating this feat to your friend, turn all three glasses over to the inverted position such that the tops of all three glasses are face down on the table. Now, challenge your friend to duplicate your feat.

Why does your friend become so frustrated because they cannot do what you just did so effortlessly?

What is the Secret of this Coin Trick?

Yesterday’s Problem:

What three numbers have a sum that is equal to the product of those same three numbers?

Answer:

1+2+3=1 x 2 x 3 = 6

Today’s Problem:

One of the most eloquent coin tricks is often explained as a feat of extrasensory perception when in fact if is simply an example of parity.

Here is the trick: Ask someone to toss a handful of coins on a table. After a quick peek at the result, turn your back and ask the person to turn over pairs of coins at random – as many pairs as he or she would like to turn over. Then ask the person to cover up one coin.

When you turn around, you can tell immediately whether the covered coin is showing heads or tails.

Can you figure out the mathematical secret which lies at the heart of this trick which is bound to impress your friends, especially when you can do it over and over again?

Name the Three Numbers

Last Problem:

Anne and her friend are dancing in a circle. The circle is set up so that every dancer is next to two people who are both the same gender.

How many girls are there if there are twelve boys in the circle?

Answer:

Anne’s neighbors can be either two boys or two girls. If they are girls, then each of them must be neighbored by another girl since they are both next to Anne. So, in the instance where Anne’s neighbors are girls, the entire circle must be girls.

Since there are boys in the circle, the circle is obviously not all girls. That means Anne’s neighbors must be both boys, each of whom is neighbored by Anne and another girl. The alternating pattern continues around the circle, so the circle contains twelve boys and twelve girls.

Today’s Problem:

What three numbers have a sum that is equal to the product of those same three numbers?