Peter de Jager's Brain Fodder

Welcome to Peter de Jager's new feature, Brain Fodder. Every week, Peter will challenge you with a new puzzle that will be posted in our newsletter, MWXtra, while the solutions proposed by Peter will be published on this web page.

Except, they won’t be just ... solutions, they will also include some of Peter's thoughts about these answers and what each puzzle might be able to teach us about problem solving, in general, and people management issues, in particular.

Peter's puzzles will make you scratch your head from time to time, but will never require you to know anything more than you knew in highschool, and – most likely – still remember.

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Puzzle #14 (as it appeared in MWXtra - February 18, 2009)

How many errors can you find in the following sentence?
"This sentance conteins one misteak."
And ... how many errors can you find in the next one?
"Their are three misteaks in this sentence."

Puzzle #14 - Answer
It's easy to answer the first question. In the first given sentence, there are four mistakes: three typos, and, then, one error of fact - as it contains more than one mistake.

The second question is the type you spend four years at university trying to figure out the answer. There are two spelling errors - that’s obvious. However, the content suggests there are three errors. This statement seems to be false ... as long as you don't count its own falseness as an error. That would bump the number of errors up to three -which makes the sentence true… and paradoxically false at the same time. And around and around we go...

Puzzle #14 - Lesson
In real life, contradictions don’t exist. When you stumble over something that contradicts itself, then it’s safe to assume that you made a mistake somewhere. If two people total a column of numbers and the totals don’t add up, then somewhere, someone is wrong. Therefore, if you find a contradiction, you’ve actually found an error, even if you don’t know where it is.

If you want more of these twisted paradoxes, you can read Godel’s Incompleteness Theorem. Lots and lots of serious math stuff!

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Puzzle #13 (as it appeared in MWXtra - February 11, 2009)

Seven staff members of the same municipality arrive at a conference and try to check into their hotel rooms. But, the hotel has no record of the reservations and there are only six rooms available.

They get a bit angry, but the hotel manager says he might be able to fix the problem. He starts by placing the first man in a room and asking one of the women in the group to wait there with him. Then, he puts the third person in the second room, the fourth person in the third room, the fifth person in the fourth room and the sixth person into the fifth room. With a sigh of relief, he places the remaining seventh person, the woman he asked to wait, into the sixth room. “Well done! Everyone should be happy now!”, he says to himself.

Is he right? Did he really fix the problem?

Puzzle #13 - Answer
I know that I promised no tricks – but I just couldn’t resist! Since watching “Batman-the Dark Knight” again, the phrase “Why so serious?” has been running around, in my head …

He is NOT right: the 'second' person wasn’t placed in a room. Give these people names, write their names on pieces of paper, and move them around to see what is going on.

Puzzle #13 - Lesson
The devil is hidden in details … Sometimes, we need to grab a pen and paper, to sketch out what we think is happening. We can only hold a few details in our head, at a given time, and that is why we must find a different way of representing the data we have to work with. At first, this puzzle doesn’t seem difficult, but it’s tricky. Only when you move those bits of paper around, you can see what is actually going on.

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Puzzle #12 (as it appeared in MWXtra - February 4, 2009)

Consider the following set of statements:

1 – Exactly one of these statements is false.
2 – Exactly two of these statements are false.
3 – Exactly three of these statements are false.
4 – Exactly four of these statements are false.
5 – Exactly five of these statements are false.
6 – Exactly six of these statements are false.
7 – Exactly seven of these statements are false.
8 – Exactly eight of these statements are false.
9 – Exactly nine of these statements are false.
10 – Exactly ten of these statements are false.

How many of these statements are false?

Puzzle #12 – Answer
This is one of those instances where it’s best to step back and think a little bit. There are 10 statements… Each one could be true or false. (We’ll ignore those situations where we might not be able to figure if a statement is neither true, nor false, but somewhere, in the middle…) This leaves us with ... a LOT of possibilities. But ... there are some tricky spots. If statement #10 is true, that means all the statements are false – including #10. Therefore, #10 cannot be true. Ouch!

A simpler way to look at this is to assume that ALL of these statements contradict each other. This means that AT MOST one of them will be true… Any other true statement would contradict that true statement, which means that nine are false and only one statement is true… Hence #9 is the true statement.

Puzzle #12 – Lesson
Think before you leap! I teach problem solving and, based on everything I’ve seen, people do not think before they start trying to solve problems. They leap in with both feet – and more often than not – get permanently stuck. Therefore, THINK FIRST, ACT LATER – old words of wisdom – but true, nevertheless.

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Puzzle #11 (as it appeared in MWXtra - January 28, 2009)

This one is a head scratcher… A salesman knocks on the door of a home. When the lady of the house appears, he asks: "How many children do you have?"
She replies: "Three."
Then, he asks: "What are their ages?"
She decides he is too rude and refuses to tell him. He apologizes and asks for a hint. She tells him that if he multiplies their three ages together, he will get 36, emphasizing that their ages are exact integers(no fractions.)

After the salesman thinks for a bit, asks for another hint. Then, she says: "The sum of their ages is the number on the house next door." He nods, and jumps over the fence, in order to see the number of the house.

When he returns and tells her that he needs one more hint, she replies: "My eldest plays the piano. Now you have enough information to solve this problem."

How old is each child? There is a unique answer.

Puzzle #11 - Answer
The children are... 2, 2, and 9 years old. How on earth did I arrive at that answer? One step at a time. While this problem seems incredibly difficult to solve – if not impossible - it actually solves itself.

Do it slowly - one fact at a time.

1) Three children. Integer ages.
2) Multiplied together, their ages give 36. So? What does that mean? The ages of the kids must be one of the following:
a. 1, 1, 36
b. 1, 2, 18
c. 1, 3, 12
d. 1, 4, 9
e. 1, 6, 6
f. 2, 2, 9
g. 2, 3, 6
h. 3, 3, 4
3) Obviously, there is not enough information. We are told that the SUM of their ages is the same as the number of the house next door. That sounds like useless information – since we don’t know the number. But... just for fun, let's add up the triplets of ages that we have found.
a. 1 + 1 + 36 = 38
b. 1 + 2 + 18 = 22
c. 1 + 3 + 12 = 16
d. 1 + 4 + 9 = 14
e. 1 + 6 + 6 = 13
f. 2 + 2 + 9 = 13
g. 2 + 3 + 6 = 11
h. 3 + 3 + 4 = 10
That does tell us something… IF the number of the house next door was 16, then we’d KNOW that the ages were 1, 3, 12, because it is the ONLY triplet that totals to 16. Likewise 38, 22, 14, 11 and 10. The only number that would cause us problems is 13 - any other house number and we would NOT have enough info to solve the problem. A house number of 13 reduces the possible triplets to 1, 6, 6 and 2, 2, 9.
4) Finally the lady of the house refers to her ‘eldest' child - which means that ... the kids are 2, 2, and 9 years of age.

Puzzle #11 - Lesson
Break the problem down into the smallest possible chunks. Don’t let the data overwhelm you. Nibble on it - one little bit at a time.

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Puzzle #10 (as it appeared in MWXtra - January 21, 2008)

In front of you, there are three light switches, that are numbered: 1,2,3. Currently, they are randomly positioned. You don’t know if ‘up’ means ‘on’ or ‘off’ for any of the switches. Down the hall, out of your sight (as you stand in front of the switches), there is a room with a 60 watt light-bulb lamp. The lamp is currently off.

Only one of the three switches controls that lamp and the other two switches are NOT connected to the lamp.

You can set the switches as you wish, but, after that, you must walk down the hall, enter the room with the lamp and tell which switch controls the lamp.

There is nobody around to help you with this problem. You may NOT do anything to the switches - except changing their position of the toggle.

Puzzle #10 – Answer
The solution to this problem depends on your real world knowledge of how a lamp works.

When you turn on a lamp, it does more than generate light… It also generates heat. That should be enough of a clue to help you solve this. If not, read on!

Change the position of switch #1 and wait five minutes. Then, change the position of the switch back to what it was.

Now, change the position of switch #2 and head to the room with the lamp.

If the lamp is on, then switch #2 is connected to the lamp.

If the lamp is off, but warm to the touch, then it was recently on ... And that’s because you had turned on switch #1 for a few minutes. Therefore, in this case, switch #1 is connected to the lamp.

Finally, if the lamp is off and cold to the touch, then the switch you didn’t use is the one connected to the lamp: #3.

Puzzle #10 – Lesson
As a life rule, you cannot change just one thing. There will always be secondary effects, which are usually important.

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Puzzle #9 (as it appeared in MWXtra - January 14, 2008)

In front of you there are three boxes. They contain tennis balls.

Here’s what you know to be true: one box contains 100 WHITE balls; one box contains 100 YELLOW balls; and the last box contains 50 WHITE and 50 YELLOW balls.

You cannot see into any of these boxes. There is a hole in each box that you can stick your hand into, and withdraw only ONE ball.

There are three labels on these boxes – one on each box. However, each box is labelled incorrectly. One label says “YELLOW Balls Only,” another says “WHITE Balls Only,” and the final is labeled “YELLOW and WHITE Balls.”

After sticking your hand into one box - and one box only - remove ONE ball. Now, your task is to move the labels around, so that they will correctly identify the contents of each box.

Puzzle #9 – Answer
Reach into the box labeled "YELLOW and WHITE Balls." If the ball you've removed is YELLOW, then take the label that says "YELLOW Balls Only" and place it onto the box where you found the YELLOW ball.

Now, take the label that said "YELLOW and WHITE Balls" and place it onto the box with the label "WHITE Balls Only" and then put the "WHITE Balls Only" label onto the box that originally had the label reading "YELLOW Balls Only."

The key here is that all the labels MUST move - remember that all the boxes were incorrectly labeled.

Puzzle #9 – Lesson
One of the secrets to problem solving is simply to list all the things you could do, and then figure out the consequences of each possible action. This is one reason why 'to do' lists are so powerful: they transform the chaos in our minds into a list of discrete things that we can do.

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Puzzle #8 (as it appeared in MWXtra - December 17, 2008)

There are three envelopes on the table. One of them contains a cheque for $1,000,000 and the other two are empty. Choose one.

Once you’ve chosen one, I’ll open up one of the remaining envelopes to show you it’s empty.

I give you a choice. You can switch from the envelope you chose to the remaining unopened one on the table.

Do you switch? Yes or no? Regardless of your answer ... What is your reasoning?

Puzzle #8 – Answer
You switch. Your odds at first were 1/3 that you would choose the correct envelope (correct = the one with the money) when I offer you the choice to switch you can increase your odds to 2/3 of winning IF and ONLY IF you switch to the unopened envelope on the table.

Your odds are NOT 50/50 just because there are two envelopes to choose from. This is where our common sense and probability collide.

Explaining why this happens is, to put it mildly, difficult. I will gladly take emails from anyone wishing a detailed explanation. We would also be interested in feedback from MWXtra readers as to whether they are finding this new feature "interesting" – to determine whether or not we will continue offering these little breaks from the daily grind. So, even if you’re not interested in this particular problem ... drop me a note and say hi and I’ll know if it's worth continuing with this column. Email me at pdejager@technobility.com

Puzzle #8 – Lesson
We’re not good at probabilities ... Remember to list all the possibilities first. Whether we’re talking about silly problems OR about deciding whether we should take a fixed or variable mortgage rate or not, the issues are the same.

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Puzzle #7 (as it appeared in MWXtra - December 10, 2008)
On the table in front of you there are six playing cards - face down. I tell you that two - and only two - of these cards are Kings. You can pick any two cards.

Which of the following two statements is more likely to be true: that you’ll have no Kings in your hand, or that you’ll have at least one King in your hand?

Puzzle #7 – Answer
By now you should be getting better at this. What we do first is list all the possible ways to draw two cards from the six cards available. For the purpose of identification we’ll mark the cards A, B, C, D, E and F, and for the purpose of analysis we’ll tag the Kings as E and F.

So, how many ways can we draw two cards? Keeping in mind that A+B is the same combination as B+A, here are all the possible combinations:

A+B, A+C, A+D, A+E, A+F

B+C, B+D, B+E, B+F

C+D, C+E, C+F

D+E, D+F

Therefore, there are 15 different ways to draw two cards from six cards.

Of these, six draws contain no Kings – A+B, A+C, A+D, B+C, B+D, C+D, and the rest of draws (9) contain at least one King. So it’s more likely that you’ll have at least one King in your hand.

Puzzle #7 – Lesson
The lesson remains the same – we’re not good at probability problems and the only way to really solve them is always to list off all the possibilities and work from there to the answer.

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Puzzle #6 (as it appeared in MWXtra - December 3, 2008)
There’s a bag on the table in front of you. You’re told (correctly) that it contains either a nickel or a penny. The chance of either being in the bag is 50%.

Now ... I take a penny out of my pocket and show it to you. I drop the penny into the bag ... shake the bag ... reach into the bag without looking and take out a single coin. It’s a penny.

No cheating is going on… there was a coin I the bag, I put a penny in. Now there are two coins in the bag. I took one out. It turned out to be a penny. There is still a coin in the bag. What are the chances that there is a penny in the bag?

Puzzle #6 – Answer
The most common answer is that since a penny was put in, and a penny taken out, the odds are unchanged. The chances that there’s a penny still in the bag is still 50 percent.

Wrong! Are you beginning to get a sense that I like that word?

To get a feel of why this is wrong, imagine the following: you put a penny into the bag… and then remove a coin randomly from the bag. And you do this 100 times and EACH time you did this ... the coin removed was a penny. What are the chances that the coin left in the bag is a penny? Fact is, each time you put a penny in, and then remove a coin randomly and it’s a penny ... the more likely it is that there was a penny in the bag to start with. Here comes the math. (I know ... Math is a four leter word.)

Assume the penny is in the bag (it will be 50 percent of the time) – now put a penny in the bag ... How many ways can you remove a penny from a bag that contains a penny and a penny? There are three ways ... You can either take out the penny that you put in, or the penny that was there at the start ... Or you can take out both of them.

So, since there are only three ways to remove a penny from the bag ... and in two of these cases of the time a penny remains in the bag ... the chances that there is a penny in the bag is 2/3 ... or 66.666 percent.

Puzzle #6 – Lesson
Same as before…we’re not good at probability. Check and recheck your answers.

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Puzzle #5 (as it appeared in MWXtra - November 26, 2008)
An exercise in probability (nothing too difficult here – if you’re careful.)

Fact #1: A man has two cats.

Fact #2: At least one of these two cats is a Male.

What are the chances that BOTH cats are Male?

Puzzle #5 – Answer
The typical, most common, answer is 50/50 or 1 in 2. The notion is that since one cat is male… the other one must either be a male or a female… ie. A 50/50 chance since it’s either one or the other.

Wrong. 100% totally wrong, and usually the start of a heated argument. Don’t worry – that’s the point of this puzzle – for the most part we’re not very good when it comes to probability.

Here’s the correct way to ‘think’ about any probability problem. First? Identify all possible male/female combinations with only 2 cats.

We have:
(1) Male + Male
(2) Male + Female

And then we have:
(3) Female + Male
(4) Female + Female

The first point of contention is that #2 and #3 are different and separate instances. They are. Put two cats in a box ... pull a cat out ... it could be either male or female. And, in either case, the remaining cat is either male or female. All in all, four cases as we’ve identified.

Next? Apply the constraint we’ve been given ... at least one of the cats is male – that precludes combination #4, since neither of the cats in this combination are male.

That leaves us with three possible combinations, #1, #2, #3 of which only #1 is male + male. (ie. the chance that both cats are male is 33.333…% or 1/3.)

Puzzle #5 – Lesson
We are incredibly bad at probability problems whenever we’re faced with these types of problems. Go slowly – list all the possibilities – and think carefully. Just for giggles, the next three Brain Fodders will be probability problems.

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Puzzle #4 (as it appeared in MWXtra - November 19, 2008)
Each day, at a totally random time of day, a man takes a walk and sits on a hill, overlooking a nearby railway track. Two trains travel on this track. A Blue train, and a Red train. Once a train comes by, he immediately heads back home. The man has been doing this for years, and has been keeping count… He’s seen the Red train nine times more often than he’s seen the Blue train. What can you tell me about the number of Red and Blue trains that travel on that track?'''

Puzzle #4 – Answer
Truth is – based on the information provided – not much. Most people readily offer that nine times as many Red trains as Blue trains travel on the track. In reality, there are EXACTLY as many Red Trains on that Track as there are Blue trains.

I can almost hear the objections being raised: "He’s seen nine times as many Red trains."

So what? Consider this: the Blue train leaves the rail station exactly at the hour X and the Red train always leaves six minutes later.

If the man is arriving at random times, what train is he most likely to see each day, given that the Red train always arrives six minutes after the Blue train?

Puzzle # 4 – Lesson
Statistics are always useless, unless you know exactly how they were gathered. Statistics outside of their original context mean nothing. A good bit of advice is this: when you hear the word "percentage", immediately ask “Of what?”

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Puzzle #3 (as it appeared in MWXtra - November 12, 2008)
You, and three friends, are standing on this side of a roaring river. The only way to the other side is over a rickety dangerous old bridge. At most, two people can cross this bridge at the same time. It is a moonless night – you only have one working lamp and, whenever you cross the bridge, you must take the lamp to light your path. When crossing the bridge, you cross at the speed of the slower person (even if you carry them on your back!). It takes you and your companions 1, 2, 7 and respectively 10 minutes to cross. Your assignment? Get everyone to the other side as quickly as possible.

Puzzle #3 – Answer
Getting them across, isn’t that difficult. Here’s the sequence of crossings and the direction that most people quickly volunteer as their answer: (10,1>),(<1),(7,1>),(<1),(2,1>) - which totals to 21 minutes.

What if I told you that you could do the crossing in 17 minutes?

Now, that statement is very often met with the all time solution squasher response of “That’s impossible!"

Something very interesting happens in our heads, as we work towards a solution. We immediately recognize that getting 10 and 7 to cross at the same time will save time. So, we immediately try it. And then, since there’s nobody on the other side, we realize that either 10, or 7 needs to return alone and then, has to come back again. Therefore, we put aside the notion of 10 and 7 going across at the same time. Pity!

Try this sequence: (2,1>),(<1),(10,7>),(<2),(2,1>); you will get a total of 17 minutes.

Puzzle #3 – Lesson
Just because we tried a solution before and failed, this is not a good enough reason not to try it again in the future. The “We tried that before and it didn’t work!” statement should always be met with the response: “Are the conditions now, the same as they were then? If not, then we need to reexamine that solution again.

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Puzzle #2 (as it appeared in MWXtra - November 5, 2008)
Clear some space on your desk. Take a deck of cards and using NOTHING else but the cards – build a tower of cards as tall as you can in five minutes.

Puzzle #2 – Answer
I’ve done this exercise with audiences in more than three dozen countries and the vast majority of people never built a tower more than three cards tall. Yet some people, managed to build towers in excess of ten cards high. I don’t know what category you find yourself.

Why the difference? The tall towers are built out of playing cards that have been folded into angles, have been torn almost in half ... the low level towers are built of pristine, unmutilated cards.

People fall into several categories when solving this problem:

1. They never even think of "damaging the cards."
2. They think of it – but decide not to.
3. They think of it – and do it ... they see the "cards" as what they are ... resources - "pieces of cardboard" to be used to perform a task.

What is most interesting to me, is that the first two groups often accuse the latter group of cheating.

Puzzle #2 – Lesson
Many of the rules we adhere to are followed for no other reason than we followed them before. To think outside the box, we must first see that we’re in a box – once we do that, the rest is (relatively) easy.

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Puzzle #1 (as it appeared in MWXtra - October 29, 2008)
Without using a computer, or a calculator of any sort, sum the numbers from 1 to 10,000.

Puzzle #1 – Answer
The answer is so trivial ... it’s a waste of time to write it here. There isn’t a single reader who isn’t capable of getting the correct answer with a little bit of effort. Where the real problem solving skill comes into play is in how you arrived at the answer.

If you added up the numbers 1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+ ... +10,000 then you succeeded, but boy, oh, boy, did you waste a lot of time!

Here’s another way to approach this problem. Write down the first 10 numbers in our sequence – which is 1 to 10,000 – and then, underneath them, write down, in reverse order, the last 10 numbers in the same sequence, as indicated below:

1	2	3	4	5	6	7	8	9	10	etc.
10,000	9,999	9,998	9,997	9,996	9,995	9,994	9,993	9,992	9,991	etc.

Notice something interesting ... 10,000 + 1 = 10,001, and 9,999 + 2 = 10,001, and 9,998 + 3 = 10,001, and so on, down the line. See the pattern? If you add up the two lines of numbers you’ll get 10,000 instances of 10,001 ... for a total of 100,010,000 – which is twice as much as we wanted. Therefore, the answer is 100,010,000 divided by 2; hence the solution – 50,005,000.

Puzzle #1 – Lesson
Look for the patterns inside the problem – they’ll always help you solve it better – faster.

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© 2008 Peter de Jager – Peter is a keynote speaker as well as a lover of puzzles and thinking ... if you’d like to discuss these problems, or a future speaking engagement then e-mail him at pdejager@technobility.com

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