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Puzzle #56

You have a bank account which pays 5% interest. You make only 1 deposit and no withdrawals. After how many years will the amount of money double?
Basically, this is asking, if you deposit $100 into an account that pays 5% annual interest, how long will it take for this to become $200?

First, we will need the compoud interest equation:

Total = Principal * (1 + rate)years

and inserting the numbers for this problem, it becomes
200 = 100 * (1.05)years

which equals
(1.05)years = 2

Now, we have to solve for "years".
Notice that now we have to solve for an unknown (years) that is an exponent.
This is something that usually isn't taught in an introductory algebra course.
To do this we have to take logarithms of both sides.

log (1.05)years = log (2)

years * log (1.05) = log (2)

years = log (2) / log (1.05)

years = 0.30102999566 / 0.02118929907

years = 14.2066990827

Puzzle #55

In the United States Supreme Court, there are 9 justices. Each day, when they assemble, before going on the bench, they shake hands with each other.
How many handshakes occur?

First of all, we will assume that no justice shakes hands with himself.
To make things simpler, let's suppose there are only four justices A, B, C, D.
This would make a total of 6 handshakes:
(A B)   (A C)   (A D)   (B C)   (B D)   (C D)
Six is a triangular number and it is the third triangular number.
The equation for a triangular number is (n) • (n+1) ÷ 2

So, basically for any "handshake problem", we count the number of people ("n") and use the previous triangular number.
(By using the previous triangular number, this eliminates counting handshakes twice. 'A' shaking hands with 'B' and 'B' shaking hands with 'A' is counted as just one handshake.)
Since there are 9 Supreme Court Justices we will calculate triangular number eight.

(8) • (8+1) ÷ 2 =
72 ÷ 2 =
36 total handshakes

Another way to approach this problem is to think in terms of combinations.
We could reword this as "How many ways can 2 people be chosen from a group of 9?"
And again, the answer is 36

We could even take another approach to the solution.

Looking at the graphic, the judges have been numbered 1 through 9.
We see that judge 9 will shake hands with judge 8, judge 7, all the way to number 1, making a total of 8 handshakes.
Judge 8 will shake hands with judge 7 all the way to judge 1, making a total of 7 handshakes.
So, we can see the handshake total will equal 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 or 36.
Notice that this is how triangular number 8 is formed?

Puzzle #54       09/01/2003

This probability puzzle has been around a long time. Generally, it is referred to as the Monty Hall Problem because it is somewhat based on a show he hosted called "Let's Make A Deal". The problem is stated as follows.

You are a contestant on "Let's Make A Deal" and you have been shown three doors. You are told that behind one of the doors, is a new car, and behind each of the other two doors is a goat. Naturally, you want to win the car and you are told you can pick any one of the three doors. This would mean your chance of winning the car based on this condition is one in three or one third.

However, after you make your pick, Monty Hall opens a door you haven't chosen and shows that it's a goat. He then offers to let you change your selection. Would changing your selection decrease your odds of winning the car, increase your odds or make no difference whatsoever?

It might be a little easier to use an actual example.
You pick door number one.
Monty Hall shows that a goat is behind door number two.
He then offers you the choice of picking door number three or you can stay with your original choice of door number one.
Should you change your pick to door number three, stay with door number one or does it make any difference?

Conditions in effect when playing this game:
The placement of the car and the goats is done totally at random and you have no way of knowing what is behind which door.
Monty Hall knows exactly where all three are.
After you make your pick and Monty Hall shows you one of the other doors, it will always be a "goat" door and he will not show the "car" door nor will he open up the door you just chose.

Surprisingly, you will win two out of three times if you always change your choice.
The car is equally likely to be behind door 1, 2 or 3 (probability=1/3 for each), so let's examine each case.
To make it simple, let us say you always choose Door Number One in these 3 instances.

Car is behind Door Number One: You switch and always lose. Goats are behind the two doors you haven't chosen. You will be shown a goat and yet since you are always changing your choice, you always end up picking the other goat. This happens 1/3 of the time.

Car is behind Door Number Two: You switch and always win. The car is behind a door you haven't chosen and Monty has no choice but to show you a "goat door" and the car is behind the remaining door. This happens 1/3 of the time.

Car is behind Door Number Three: You switch and always win. (see above) (Happens 1/3 of the time).

Adding up the probabilities, we see that by always switching, you win 2/3 of the time and lose only 1/3 of the time.

Granted, this was based on your choosing door number one each time but if you look at the above 3 explanations, you will see that choosing door number 2 or door number 3 will also result in your winning the car two out of three times.

Another Explanation: When you make your first pick, you will be wrong 2 out of 3 times or putting this another way, the odds are always 2 out of 3 that the car is definitely behind one of the two doors you did not choose. Since Monty Hall always eliminates one of the unchosen doors, two out of three times the car will be behind one of them and Monty has no choice but to show a "goat" door.

And Another Explanation: Suppose there are 1,000 doors, 999 goats and 1 car. For your first choice, you select Door Number One. Then, (just as in the previous example), Monty Hall opens all the unchosen doors except one. So, Monty Hall opens 998 doors to reveal 998 goats and only door number one and door number 557 remain unopened. Do you think it would be worth taking a risk and switching from your original choice?

Sorry No August Puzzle

Please come back 09/01/2003

Puzzle #53       07/01/2003

The above graphic shows a chessboard that has had its upper left square and its lower right square "sawed off". Underneath that is a domino that is exactly the size of 2 of the chessboard squares. If you had 31 of these dominoes, would you be able to arrange them so that ALL 62 squares were covered ?

31 dominoes could NOT cover the chessboard completely and the proof is quite simple. Each domino must cover a white square and a black square. However, the chessbord has had 2 corner squares removed - and they are both white. Therefore, it cannot be done.

Puzzle #52       06/01/2003

Let's assume that a $20 gold piece weighs twice as much as a $10 gold piece. So, which would you rather have, a half pound of $20 gold pieces or a pound of $10 gold pieces?

At first glance this seems to be similar to that old puzzle - "what's heavier - a pound of feathers or a pound of lead?" And the answer is they are the same thing.
However, this is a twist on that old puzzle. Let's suppose (to keep things simple) that a $10 gold piece weighs an ounce and a $20 gold piece weighs 2 ounces. So a half pound (8 ounces) of $20 gold pieces is 4 coins equaling $80. A pound (16 ounces) of $10 gold pieces is 16 coins totaling $160.
Basically, this puzzle is asking, "would you rather have a half pound of gold or a pound of gold?"

Puzzle #51       05/01/2003

An airplane pilot overheard a woman say that she had missed her flight. The pilot said "I can give you a ride and I'll charge you nothing because you won't even be taking me out of my way." The woman was very suspicious of this offer especially when she said "But you don't even know where I'm going !!" The pilot said "It doesn't matter. It will not be out of my way." The pilot was not lying so just where the heck was he going?


He was going to the opposite side of the Earth. For example, if he were in Honolulu, Hawaii he was headed for Mozambique, Africa.
On the surface of the Earth, the shortest distance between 2 points is a "great circle route". So, when you are traveling to the exact opposite side of the Earth, there are an infinite number of great circle routes you can choose, any one of which would take you over the destination where the woman wanted to go.
Granted, in reality it would be difficult to fly ANY great circle route you chose. You have to consider places for refueling. There are many places that have restricted airspace and some countries would not allow you to have ANY air rights whatsoever. Then the weather of a particular region might make things difficult if not dangerous. (Flying over Antarctica in the dead of winter, for example).
STILL, it does make an interesting puzzle doesn't it?

Sorry, NO April Puzzle - Come Back In May

Puzzle #50       3/1/2003

How about an easy puzzle this month? A class consists of 58% girls and 42% boys. There are 6 more girls than boys in the class. How many TOTAL students are there?

We can set up 2 equations:

.58*g = .42*b + 6
g = b + 6
Substituting the second equation into the first we get:
.58b + 3.48 = .42b + 6
.16b = 2.52
b = 15.75
Since we can't have decimal portions of people, we can truncate it to 15.
So, there are 15 boys and 21 girls for a total of 36 students.
Due to rounding of the percentages there are two more answers.
16 boys + 22 girls = 38 students and 17 boys + 23 girls = 40 students
So, all three answers are correct !!!

Puzzle #49       2/1/2003

You are offered an even-money bet involving the flipping of two coins.
You have 2 chances to flip 2 coins and if you get two heads in either flip, you win.
For each flip of the two coins there are 4 possible outcomes:

"Head, Head"; "Head,Tail"; "Tail,Head"; "Tail,Tail"

You reason that since 2 Heads is 1 of the 4 combinations, then flipping 2 coins twice will give you a "50/50" chance of getting 2 heads at least once.

Is your reasoning correct ?

This is not a good bet either. There are 4 possible outcomes of one flip of two coins and so there are 16 possible outcomes of 2 flips:

  1st Outcome    2nd Outcome     Successful ?

1)   H H             H H             Y
2)   H H             H T             Y
3)   H H             T H             Y
4)   H H             T T             Y
5)   H T             H H             Y
6)   T H             H H             Y
7)   T T             H H             Y
8)   H T             H T             N
9)   H T             T H             N
10)  T H             T H             N
11)  T H             H T             N
12)  T T             H T             N
13)  T T             T H             N
14)  T H             T T             N
15)  H T             T T             N
16)  T T             T T             N

Notice that the successful outcomes are only 7 out of 16 and not 8 out of 16. (Note: Case #1 is considered only 1 success - not two). Therefore, you will only be successful 43.75% of the time and not 50%.
If you still doubt the reasoning of this, try flipping 2 coins yourself OR write a computer program to simulate this. Even the RAND() function in MS Excel™ will allow you to run such a simulation.

Puzzle #48       1/1/2003

When rolling 2 six-sided dice, there are 36 possible combinations. Out of these 36 combinations, there is only one way that a twelve (2 sixes) can appear.

You are offered an even-money bet (let's say for $10) that if you can throw a twelve in 21 throws, you win $10. If not, you lose $10. Also, you may continue to make bets for as long as you'd like.

You reason that a twelve is 1 out of the 36 dice combinations, therefore your chances are 50-50 after 18 rolls and with 21 tries you are going to come out a winner more often than not. Is this reasoning correct?

For any 'n' throws of the dice , all possible outcomes equal 36n. So, for example, after 10 throws of the dice, the number of all possible outcomes = 3610 = 3,656,158,440,062,980. Then you would have to calculate the successes which is quite a complicated procedure and involves the binomial distribution theorem.
Fortunately, here's an easier method. Instead of calculating the successes, look at it from a "non-success" point of view. For example, on the first roll, there is a 35 in 36 chance (.97222222) that you won't roll a twelve. For the second roll, the probability of not succeeding is (35/36)2 or (.97222222)2 or about .945216.
As we proceed to calculate (35/36)n for higher values of 'n' we will reach a point where the failure probability falls below .5 which is the point at which the odds are better than 50-50 that you will roll a twelve.
Continuing on to 20 we see:
(.97222222)20      =   0.5692602
(.97222222)21       =   0.553447
(.97222222)22       =   0.538073
(.97222222)23       =   0.523127
(.97222222)24       =   0.508596
(.97222222)25      =   0.494468
So after 25 throws, we see that the "chance for failure" has fallen below 50% or the chance for winning is now better than 50 per cent. Therefore, taking a bet involving rolling a "12" in 21 throws would not be a good idea.
In fact in 1952, a gambler named "Fat the Butch", agreed to this bet (except at $1,000 for 21 rolls). After losing $49,000 he realized that something had to be wrong with his reasoning.
To read more about this, click here.


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