PROBLEM COLLECTION

Here are a few math problems I am fond of. Solutions are provided. If you get a different solution to a problem, it may be because you made different assumptions about the nature of the problem. You may also have used a different method to arrive at a solution.

### HASTY CONCLUSIONS

Suppose you read the following items:

"Scientists have found that 90 percent of all people who acquire cancer are regular users of salt. Therefore, we can conclude that salt causes cancer."

"Recently released statistics from the police files reveal that 40 percent of all traffic accidents are caused by drunk drivers. This implies that sober drivers are responsible for most accidents. Therefore, based on the data, insurance companies should give discounts to heavy drinkers."

Are the conclusions reached valid ones? Why or why not?

solution

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### TRAINS AND A FLY

Two trains, 200 kilometres apart, are moving toward each other. Each one is moving at a constant speed of 50 kilometres per hour. A fly starting on the front of one of them flies back and forth between the trains at a rate of 75 kilometres per hour. It does this until the trains collide (with unfortunate results for the fly). What is the total distance the fly has flown?

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### INTEGERS & WIDGETS

A company sent out a large crate packed with widgets. When attempting to pack them with maximum efficiency, it was found that the widgets could not be arranged in groups of 2, 3, 4, 5 or 6, without at least one left over. They could, however, be arranged in groups of 7, with none left over.

How many widgets did the company send out?

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### GHASTLY CRIME

A ghastly crime is committed in a large city of 18 million people. The police arrest someone for the crime. A DNA test shows a close match. Experts indicate that approximately 3 people in the city would be expected to match this closely. The prosecutor argues that 3 out of 18 million shows that the probability of an innocent person having this close a match is 1 in 6 million and that, therefore, the accused must be guilty. Are there any flaws in the prosecutor's reasoning?

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### BASEBALL AVERAGES

Baseball batting averages are calculated by dividing the player's number of hits by the total number of times at bat. The result is given to 3 decimal places. Player A has a higher batting average (mean) than player B in the first half of the season. Player A also has a higher average in the second half of the season. However, for the season as a whole, player B has a higher average than player A. Is this possible?

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### FRACTAL CURVE

A Koch Curve is a type of fractal. It can be constructed by starting with an equilateral triangle. The first step is to divide each side of the triangle into 3 equal segments. Then the middle segment is removed and replaced by another equilateral triangle. This forms the next figure. For all subsequent steps, each line is divided into 3 parts and the procedure is repeated. This procedure is repeated infinitely many times. The result is a Koch Curve. If you examined a Koch Curve, you would notice that no matter how much you magnify any section of it, its appearance does not change. Consider a Koch Curve based on an original triangle with a side length of one unit.

What is the perimeter of the resulting Koch Curve?

What is the area of the curve?

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### AREA PROBLEM

The figure on the right is a square, containing four circular arcs, with a side dimension of two units.

Find the areas of the regions: x, y and z.

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### GAME SHOW

You are at a game show. The host shows you three doors. You know that behind one of the doors is a brand new car. Behind the other two doors are goats. You select a door. The host then goes to one of the doors you did not choose and opens it, revealing a goat. She then offers you the chance to change your mind and select the remaining door. The host knows which door has the car and at this stage of the game she always opens the door that has a goat behind it. Should you change your choice and pick the remaining door or should you stay with the one you originally picked? Does it make any difference in your probability of winning the car?

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### ROLLING DICE

Craps is a common casino game. The rules are straight forward. You roll two dice. If the total of the dice is 7 or 11, you win. If the total is 2, 3 or 12, you lose. If any other number comes up, you continue rolling. The number rolled on your first toss is called your point. You keep rolling until either your point is rolled again, in which case you win, or a 7 is rolled, in which case you lose. Numbers other than your point and 7 have no effect on the game after the first roll. What is the probability of winning a single game of craps?

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### HAT TURN A-BOAT

You are rowing a boat upstream. The river flows at 5 kilometres per hour. Your speed against the current is 7 kilometres per hour. You lose your hat on the water. Forty-five minutes later you realise it is missing and make an immediate turn and row back to get it. How long does it take to row back to your floating hat? What is the easiest way to solve this problem?

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### TEST RESULTS

Suppose that doctors have a test for a given medical condition. Assume that 0.5 percent of the population has this condition. The test is 99 percent effective. This means that 99 percent of those who have the condition will test positive and 99 percent of those who do not have the condition will test negative. The rest, in both cases, will receive false results. If someone tests positive for the condition, what is the probability that the person has it?

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### CHESS PROBLEM

Place eight queens on a chess board is such a way that no queen can capture another queen in the next move.

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### CARD GAME

There are 1000 cards on a table in a row. Each card has two sides, one is red and one is blue. The first 500 cards are placed with the red side up. The last 500 cards are placed with the blue side up. First, you flip every card over. Then, you flip every second card over. Then, every third, then every fourth, and so on until you flip every thousandth and you are done. In the end, how many cards will end up with the red side up? Describe in words, and without a list, which cards will end up red.

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### PRIME PROBLEM

Prove that there are infinetly many prime numbers.

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### MEAN PROBLEM

Find an exact solution for X in the following equation, where X is a positive number. Name one other significant mathematical problem or formula that involves this number.

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### COVER BOARD

Given as many small rectangles, composed of squares, as you need (below left), is it possible to completely cover the larger figure (below right), composed of squares of the same size? The figures below are not to scale. None of the small rectangles can overlap or extend outside of the large figure. To get any value on this one, you must either show a way it can be done or prove that it cannot be done.

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### SAFE TRAVEL

A man who frequently travels by commercial airlines is afraid that someone will bring a bomb on board. Therefore, he always carries a bomb in his own suitcase. He reasons that, because the probability of there being a bomb on an aircraft is very low, the probability of there being two bombs on board must be virtually nil. Is his reasoning sound?

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### FAIR PLAY?

Imagine that a woman has three cards. One is black on both sides, one is red on both sides and one is black on one side and red on the other. She asks you to draw one of the cards out of a hat and to only look at one side, let's say it is red. She knows that it cannot be the black/black card and must therefore be the red/black card or the red/red card. She offers to bet you even money that it is the red/red card. Is the bet fair?

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### COIN ROLL

Two identical coins touch the side of a rectangle at the same point, one from the inside and one from the outside, as shown at right. The coins are rolled in the plane along the perimeter of the rectangle until they come back to their original positions. The height of the rectangle is twice the circumference of the coins and its width is twice its height. How many revolutions will each coin make.

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Tom and Betty applied for a marriage license. They were asked for their ages. Betty told the clerk that they were both in their twenties and that their birthdays fall on the same day. When the clerk told them that he would need more specific information, Tom told him that he was four times as old as Betty was when he was three times as old as Betty was when he was twice as old as Betty was. The clerk pondered for a few minutes and filled in their correct ages.

What ages did he fill in for Tom and Betty?

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### COMMON CENTS

Four men, whose last names were Conner, Morgan, Smith and Well, and whose first names were Al, Bill, Jack and Tom, though not necessarily respectively, amused themselves by playing a game. The winner of the first game would collect ten cents from each of the others. The winner of the second game would collect twenty cents from each of the others. The winner of the third game would collect thirty cents from each of the others. The winner of the fourth, and last, game would collect forty cents from each of the others. At the end, each man had won exactly one game. Jack won the first game, Morgan the second, Bill the third and Smith the last. Before they played, Tom had the most money and at the end, Wells had the most.

What was the full name of each of the players?

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### TWO MATHEMATICIANS

Two mathematicians met on the street one day. They had not seen each other for 20 years. The first asked the second what was new. The second said, "I have been blessed with three daughters since we last spoke." When the first asked how old they were, the second replied that the product of their ages was 36. The first then stated that this was not enough information.

The second mathematician then asked, "Would it help if I told you that the sum of their ages is the same as the number of that house?", pointing to one across the street.

"No", the first replied, "That is still not enough information."

The second then said, "Suppose I told you that my youngest daughter has blond hair?"

The first smiled and answered, "Yes, I now know the ages of your three daughters."

How old were the three daughters and how did the mathematician figure it out?

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### TRIPLE DUEL

You and two of your "friends" get into a dispute and decide to solve it with a three-way duel. Pistols are selected.

Friend A is a crack shot and never misses. Friend B hits the target 2/3 of the time. You are not quite as good as the others and hit your target 1/3 of the time. It is decided that you will take the fist shot, Friend B (2/3) will take the second shot (if alive) and Friend A (perfect shot) will go last (again if still alive). This rotation will continue until there is one person left alive. On each turn, one shot only is allowed. As stated, you are going first.

Which friend should you take your first shot at to have the greatest chance of winning the duel? What are your chances of winning by using this strategy?

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### TRI-AREA PROBLEM

The triangle shown is divided into four areas. Three of the areas are shown. Find the value of the area X.

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### SQUARE POINTS

Given that five points are randomly placed on or in a unit square, prove that at least two points are no further apart than:

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### CHESS CHANCE

Suppose that a rook and a bishop are randomly placed on an empty chessboard. What is the probability that one piece threatens the other?

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Thirty math students took a test. The average passing grade was 84. The average failing grade was 60. The overall average was 80. How many students passed the test?

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### DIGIT SUM

What is the sum of the digits of the decimal form of the product below?

2 1999 x 5 2001

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### SQUARES IN SQUARES

Consider the partitioning of squares into smaller squares. If a square can be partitioned into n squares, we will define n to be a "good number". The examples below show you five good numbers. Find all whole numbers that are not good numbers.

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### CORNERED SQUARE

Congruent isosceles triangles are cut from the corners of a square so that the remaining figure has an area equal to three-fourths of the area of the square. If the square has a side length of S units, how long are the legs of the triangles?

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### SQUARES AND RECTANGLES

The standard checlerboard below can be considered to contain a large number of rectangles. Any of which is composed of any number of small squares from one to sixty-four. How many distinct rectangles are contained within the checkerboard? Do not count. There is an easier way.

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In number theory, there is a curious relationship between the sum of consecutive cubes of the set of natural numbers and the square of the sum of the corresponding numbers themselves. It can be stated as:

Create an area representation of this relationship for the case of n = 5.

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Solutions

### HASTY CONCLUSIONS

It is also true that 100 percent of those who develop cancer regularly breath air but no one would decide that air causes cancer (polluted air is another question). Statistics often show a correlation between variables. It is important to note that correlations do not prove a cause and effect is present. It is possible that other variables are involved. To show a cause and effect, it is necessary to carry out more controlled studies that attempt to isolate the two variables. In addition, multiple controlled studies by different groups will lend additional support for a hypotheses of cause and effect. For example, originally a correlation was found between smoking and lung cancer. Since that time a vast number of independent controlled studies, in addition to medical evidence, have made the connection between smoking and cancer a virtual certainty.

Regarding the second part of the question, there are far more sober drivers than drunk drivers. To draw a conclusion, you would need to compare the percentage of drunk drivers involved in accidents with the percentage of sober drivers.

For example, suppose that, on average, 1 percent of those driving are impaired. Suppose that for a given daily sample, 100,000 people are driving. Then we have 1000 impaired drivers and 99,000 sober ones. Then let us assume that there are 100 accidents on this day. Because 40 percent of them are caused by impaired drivers, 40 were drinking and 60 were not not. But there are far more sober drivers than drunk ones. Forty of the the 1000 drunk drivers, or 4 percent of them caused an accident. Sixty of the 99,000 sober drivers, or 0.06 percent, caused an accident. These figures can be used to represent probabilities. It is, therefore, over 66 times more probable that a drunk driver will cause an accident that a sober one.

The above is an example and the figures are arbitrary. The point was to show that, mathematically, the question did not provide enough information to solve the problem.

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### TRAINS AND A FLY

The fly actually hits each train an infinite number of times before it gets crushed, and one could solve the problem the hard way with pencil and paper by summing an infinite series of distances. The easy way is as follows: Since the trains are 200 miles apart and each train is going 50 miles an hour, it takes 2 hours for the trains to collide. Therefore the fly was flying for two hours. Since the fly was flying at a rate of 75 miles per hour, the fly must have flown 150 miles. That's all there is to it.

This problem was posed to John von Neumann, the mathematician.

He immediately replied, "150 miles."

"Interesting," said the poser, "but nearly everyone tries to sum the infinite series."

"What do you mean?" asked Von Neumann. "That's how I did it!"

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### INTEGERS & WIDGETS

There are different approaches that can be used to solve this problem. Because the number of widgets cannot be divisible by 2, 3, 4, 5 or 6, the smallest number that fits this condition is the least common multiple of these numbers plus one. This number is 61. All multiples of 61 are consistent with the first condition. Next, the number of widgets must be divisible by 7. By examination, it can quickly be determined that the lowest possible solution is 301. The next solution must be greater than the first by the common factors (2*2*3*5*7 = 420).

More formally, the solutions are given by the integer equations:

2*2*3*5*p+1 = 7*q

or

60*p + 1 = 7*q

The smallest solution is p = 5 and q = 43, giving 301.

Therefore the solutions are in the form:

301 + 420n where n = 0,1,2,3,...

Because of the nature of the problem, you may want to consider the possibility of an upper limit. This will result in a more subjective answer.

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### GHASTLY CRIME

Did the prosecutor consider conditional probability? The probability that a person, randomly selected from the city, has a DNA match is indeed 1 in 6 million. But is this the situation? The accused has a DNA match, and is not a random selection from the population. What is relevant is the probability that the accused committed the crime, given that there is a DNA match.

Conditional probability must be considered. For example, the probability that a student takes math, given that he or she is a student at CEC, is not the same as the probability that a student attends CEC, given that he or she takes math.

It is assumed that there are 3 people in the city with such a match. Therefore, the probability is 1 in 3 that the person is guilty because it is given that the accused has a DNA match. Of course, we are not considering the possible existence of collaborating evidence, only the prosecutor's mathematical statement. Additional evidence might change the probability of guilt.

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### BASEBALL AVERAGES

It is possible. For example, suppose player A hit .300 and player B hit .290 in the first half, with player A batting 200 times to player B's 100 times. During the second half, player A bats .400 and player B .390, but this time Player B comes to bat 200 times to player A's 100 times. Now, calculate how many hits each player had in each half of the season, add them, and then compare the result with their total number of times at bat for the year. Player A made 60 hits in the first half and 40 in the second (total=100 hits). Player B made 29 hits in the first half and 78 in the second (total=107). Both batters came to bat 300 times in the season. Therefore, player A had a season average of .333 and player B had a season average of .357. You cannot simply average averages.

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### FRACTAL CURVE

The initial perimeter is 3 units. At each iteration one third of the perimeter is removed and two thirds is added, for a net gain of one third of the previous value. Therefore, the perimeter of a Koch curve is infinite.

The area is a bit more challenging to find. The length of one side of the original triangle is one unit. Using the formula for the area of a triangle and the Pythagorean Theorem, we can find a formula for the area (A) of an equilateral triangle, given the length of a side (S):

A = S2*sqrt(3)/4

Therefore, the area of the initial triangle is sqrt(3)/4. The first iteration is found by adding 3 equilateral triangles, each with a side length of 1/3. Nothing is removed from the initial triangle. After this, each iteration is found by adding 4 times the number of triangles than were added in the previous step. For each iteration, the side length is reduced to 1/3 the previous value. When these values are used with the formula for the area of a triangle and the result is simplified, we get the series:

sqrt(3)/4 + sqrt(3)/12 + sqrt(3)/27 + 4*sqrt(3)/243 + ...

The terms of this series, except the first, are geometric with a first term of sqrt(3)/12 and a common ratio of 4/9. Using the formula for the sum of a converging geometric sequence, and simplifying, gives us:

sqrt(3)/4 + 3*sqrt(3)/20 = 2*sqrt(3)/5

It is interesting that an object of infinite perimeter encloses a finite area.

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### AREA PROBLEM

This problem can be solved by creating three equations in X,Y and Z. For simplicity, let P equal Pi and S(n) represent the square root of n. Remember that the side length of the square is 2. The first equation can be found using the area of the square (above).

equation one: 4X + 4Y + Z = 4

The second can be found using the formula for the area of a circle and using the quarter circle shown above and the side length as the radius.

equation two: 2X + 3Y + Z = P

The third equation is a bit more challenging to find. Consider the dome shaped area shown above. You can use calculus to find a formula for this area, but geometry will suffice. The angles of the highlighted triangle are all 60 degrees because of the reflective symmetry of the triangle in the vertical line through the vertex of the dome. Using the cosine function we can find the altitude of the triangle and thus its area. Next we can determine the area of the sector of the circle that is subtended by one of the lower 60 degree angles using the formula for a sector (area = Ar2/2, were A is the angle and r the radius). By subtracting the area of the triangle from the area of the sector we can find the area of each of the two identical highlighted areas. Putting these values together we have the third equation.

equation three: X + 2Y + Z = (4/3)P - S(3)

These three equations can now be solved.

X = -(2/3)P + 4 - S(3)

Y = P/3 - 4 + 2S(3)

Z = (4/3) + 4 - 4S(3)

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### GAME SHOW

When you first select a door the odds are 1 in 3 (1/3) that you have picked the one with the car and 2 in 3 (2/3) that you have not. If you do not change doors the odds will remain the same. The MC's action, in revealing the goat, is not relevant because she will always show you a goat and of the two remaining doors at least one must be a goat.

If you do change doors the result is as follows:

If you had selected the car at first then you have lost. The odds of this are 1/3. If you had not chosen the car at first (odds 2/3) then you have definitely won because your door has a goat and the MC is showing you a goat. The door you are changing to is the last one and must have the car behind it. Thus the odds of winning the car by changing doors is 2/3.

You have the best chance of winning the car by changing doors. By showing you the goat, the MC is giving you information. She is giving you the knowledge of a door that does not have the car. You can use this information to reverse the odds.

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### ROLLING DICE

To calculate the probability of winning at craps, note that,

Where P(E) = the probability of event E then,

P(E) = number of favourable outcomes / total number of outcomes

Independent events are events, such that the outcome of one event does not affect the outcome of the other. For independent events E1 and E2,

P(E1 and E2) = P(E1) * P(E2)

P(E1 or E2) = P(E1) + P(E2)

Consider the total possible outcomes of rolling 2 dice. Use the table below to find the probability of each possible outcome. There are 36 possible outcomes.

```
DIE 1                    PROBABILITY

|   1   2   3   4   5   6       Number      Probability
-------|---------------------------
1  |   2   3   4   5   6   7         2                1/36
|                                 3         2/36 = 1/18
D   2  |   3   4   5   6   7   8         4         3/36 = 1/12
I      |                                 5         4/36 = 1/9
E   3  |   4   5   6   7   8   9         6                5/36
|                                 7         6/36 = 1/6
2   4  |   5   6   7   8   9  10         8                5/36
|                                 9         4/36 = 1/9
5  |   6   7   8   9  10  11        10         3/36 = 1/12
|                                11         2/36 = 1/18
6  |   7   8   9  10  11  12        12                1/36

```

To find the probability of winning the game, find the sum of the probabilities for each way the player can win. The player can win on the first roll or on one of the subsequent rolls. The probability of the player winning on the first roll is given as the sum of the probabilities of rolling a 7 or 11, and is,

1/6 + 1/18 = 4/18 = 2/9

Next consider the probability that the player will win on a subsequent roll. This is only possible if the player rolls a 4, 5, 6, 8, 9 or 10 on the first roll. Any other number (2, 3 or 12) will end the game.

There is no theoretical limit to the number of rolls that will take place. The player will roll until either the point is rolled again or a 7 is rolled (see rules).

Use a probability tree to represent the outcome of each roll. For each roll; W = win (making your point), L = lose (getting a 7) and N = neither event. In the latter case the player rolls again.

```
Roll 1       Roll 2       Roll 3       Etc.

|---- W      |---- W      |---- W      |---- W
|            |            |            |
----|---- N -----|---- N -----|---- N------|---- N ------- .....
|            |            |            |
|---- L      |---- L      |---- L      |---- L

```

To find the probability of an event, find the sum of the probabilities of all the branches that end with the event. Find the probability of each branch by multiplying all of the probabilities on that branch. For example, the probability of winning in the first or second attempt to make the point is,

P(W) + P(N)*P(W)

The probability of making the point "Pt" can be represented by the infinite series,

P(Pt) = P(W) + P(N)*P(W) + P(N)^2*P(W) + P(N)^3*P(W) + ...

Factoring the common term gives,

P(Pt) = P(W)*[1 + P(N) + P(N)^2 + P(N)^3 + ...]

The result is P(W) times an infinite geometric series. The formula for the sum of a converging infinite geometric series, where r is the geometric ratio and a is the first term of the series, is,

SUM = a/(1 - r)

The result of applying this formula to the one above, is,

P(Pt) = P(W)/[1 - P(N)]

Use the probability for making each point, on a single roll, to calculate the probability of not making the point (The sum of both must equal 1). Use these values, and the formula above, to find the total probability for making each point in the game, as shown in the table,

```
Point      P(W)      P(N)      P(Pt)

4        1/12       3/4       1/3
5        1/5       13/18      2/5
6        5/36      25/36      5/11
8        5/36      25/36      5/11
9        1/5       13/18      2/5
10        1/12       3/4       1/3

```

To find the total probability of winning at craps, find the sum of the products of the probability of winning each point with the probability of needing to make that point. To this, add the probability of winning in the first roll, calculated above. The final result is,

P(Winning at craps) = 2/9 + (1/12)(1/3) + (1/9)(2/5) +
(5/36)(5/11) + (5/36)(5/11) + (1/9)(2/5) + (1/12)(1/3) =
976/1980 = 244/495 = 0.49292929...

In other words, there is slightly less than a 50 percent chance of winning.

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### HAT TURN A-BOAT

The answer is 45 minutes, the same length of time that it took to row away from the hat. It is possible to solve this problem using the speed of the water, but it is not necessary. The hat is motionless relative to the water and the speed of the boat is constant relative to the water. The speed of the river relative to the land is not relevant to the problem. The easiest solution is to consider the hat motionless and, because the boat travels at a constant speed, note that it will take the same length of time to row away from it as toward it.

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### TEST RESULTS

Consider a population of 1,000,000. On average, 5000 people will have the condition and 995,000 will not. Ninety-nine percent of those who have the condition and 1 percent of those who do not will test positive. Thus, 99% of 5000 plus 1% of 995,000 equals 4950 plus 9950, which is 14,900. The percentage of people who test positive and have the condition is found by dividing 4950 by 14900. The result is that about 67 percent of those who test positive will not have the condition.

This may seem strange. Although the test is 99 percent accurate, the group of people who do not have the condition is so much larger than the ones who do that most of the people who test positive are in the group for which the results are incorrect. The reader is reminded that this question involves a hypothetical situation.

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### CHESS PROBLEM

Below is a possible solution for the queens problem.

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### CARD GAME

Consider how many times a given card is flipped over. The description of the problem tells us that a card is flipped each time we meet one of its divisors. For example, 12 will be flipped six times (1,2,3,4,6,12). Factors come in pairs (for 12 they are 1*12, 2*6, 3*4) and the result is an even number of factors. The only exceptions are numbers that are perfect squares of integers. For example there are five factors of 16 (1*16, 2*8, 4*4), an odd number. Therefore, Each number that is a perfect square will end up showing the opposite colour than it began with.

To calculate how many cards end up showing red, it is only nessary to find the number of perfect squares between 1 and 500, 500 and 1000 and make the nessessary calculation. The result is 487.

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### PRIME PROBLEM

Assume that there is a finite number of prime numbers. Let P be the largest prime. Let Q be a larger number equal to the product of all the numbers (integers) from 1 to P plus 1. Therefore, Q = P! + 1. No number from 2 to P divides evenly into Q because a remainder of 1 would be left in each case. If Q is not prime, it must be divisable by some prime number larger than P. If it is prime, then it is a prime number greater than P. In either case, P is not the largest prime number. Therefore, there is no largest prime and must be an infinite number of them.

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### MEAN PROBLEM

Because we are dealing with a recursive definition of X, X can be substituted into the right hand side. Solving for X gives us,

This number is called The Golden Mean, or The Golden Ratio. Its importance in mathematics is significant and it is usually symbolised by the Greek letter phi. It shows up in many seemingly unrelated areas. One involves the Fibonacci sequence. This sequence is defined by the recursive formula, f1 = 1, f2 = 1, fn = fn-1 + fn-2. An interesting property of the sequence is that,

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### COVER BOARD

Think of the figure as a chess board with two opposite corners missing, as shown below. Each rectangle must then be composed of a red half and a white half because the board has no two adjacent squares that are the same colour. Because we cannot cut a rectangle in half, there must be an equal number of red squares as white. However, the figure has 32 red squares and only 30 white squares. Therefore, it cannot be done. Although we have introduced a colour scheme, the colours represent specific squares. Proofs do not nessessarily need to be complex.

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### SAFE TRAVEL

Most people respond correctly to this question. The fact that he has a bomb with him cannot affect the chances that someone else does. However, the same error is often made in different contexts. For example. a lottery player might feel that because a given number has not been drawn in a long time, that number is somehow due and has a greater chance of coming up. What needs to be considered is not the probability that there are two bombs on board. It is the conditional probability that there is a second bomb, given that there is a first one. There is no advantage to bringing your own bomb.

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### FAIR PLAY?

There are two ways she can win and only one way you can win. The visible side of the card you picked could be the red side of the red/black card (she wins). It could also be the other side of the red/red card (she wins). Therefore, the probability that you will win is 1/3. It is true that the conditional probability of the card being red/red, given that it is not black/black, is 1/2. However, that is not the situation here. We not only know that the card is not black/black, we also know that a red side is showing.

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### COIN ROLL

When a coin rolls the distance of its circumference, it makes one full revolution. The perimeter of the rectangle is 12 circumferences. At each vertex the outside coin also makes an additional quarter turn. Therefore, it makes a total of 13 revolutions. To find the distance traveled by the inside coin, we need to subtract 8 times its radius (the four corners) from the perimeter. The radius is the circumference divided by 2 times pi. Therefore, the inside coin makes 10.7 revolutions.

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Consider, and translate their ages into three time intervals. Let A be the present and B and C be the times referred to in the question. Let Betty's age when he was twice as old as her, be "X". Tom's age is therefor "2X". The difference between their ages is "X" and this value remains constant. We can use the constant difference in ages and the information provided to work the table up to the top as shown.

 Time Interval TOM BETTY A - now 8X 7X B - when 3 times older 3X 2X C - when twice as old 2X X

We know that they are both in their 20's. Substitute integer values for X and check the values of 8X and 7X. Only with X equal to 3 does the solution fit this condition. Therefore, Tom is 24 and Betty is 21.

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### COMMON CENTS

You could start by setting up a table with first names for the rows and last names for the columns. Because each man won exactly one game some of the possibilities can be eliminated because they would mean that a player won more than one game. Other possibilities can be eliminated because it would mean that a man had not won any games. The possibilities remaining are shown.

 Al Bill Jack Tom Conner XXX Bill Conner Jack Conner XXX Morgan Al Morgan XXX XXX Tom Morgan Smith Al Smith XXX XXX Tom Smith Wells XXX Bill Wells Jack Wells XXX

We can now list the final monetary change for each possible man based on the information in the problem (1st 10 cents, 2nd 20 cents, 3rd 30 cents and 4th 40 cents, from each man). The outcomes are shown below using the men's initials.

 BC JC AM TM AS TS BW JW 20 -60 -10 -10 60 60 20 -60

There is only one combination of names where Tom starts with the most money and Wells ends with the most.

• Jack Conner
• Tom Morgan
• Al Smith
• Bill Wells

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### TWO MATHEMATICIANS

If we list all possible sets of three numbers that have a product of 36, we get the following possibilities:

 Option A B C D E F G 1st age 1 1 1 1 1 2 3 2nd age 6 3 4 1 2 2 3 3rd age 6 12 9 36 18 9 4 Sum 13 16 14 38 21 13 10

Because they have not seen each other in 20 years, we can eliminate D, although this step is not needed to solve the problem. Any combination that has a sum different from any of the others can be eliminated because if it was correct, the mathematician would have been able to solve the problem as soon as the sum was revealed. This leaves A and F. A can be eliminated because the mathematician is then told the colour of the oldest girl's hair (she is not a twin). The daughters are 2, 2 and 9.

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### TRIPLE DUEL

This is a kind of problem where the best strategy can be found logically, without actually making the calculations. First, let's assume that the other two use an optimal strategy, therefore,

• If it is B's shot and all three are still alive, B will shoot at A because the result is certain, the remaining opponent will get just one shot and A is a better shot than you.

• If it is A's shot and all three are still alive, A will shot at B because A knows what B's strategy will be, as stated above.

Regardless of which opponent you shoot at, there is a 2/3 chance that you will miss, this leaves you in the second situation above. Regardless, if A makes the shot, you are in a stand-off with A, firing back and forth till one dies, and if A misses, you have one chance to shot at B. Therefore, regardless who you shot at first, if you miss, you have the same chance of living through the ordeal.

If you shot at A first and hit A then B will kill you next shot. If you shoot at B first, and hit B, then you are in a stand-off with A, with some chance of winning. Therefore, you have a better chance of winning if you take your first shot at B. In short, if you shoot at A, you better miss or you've lost.

This is an interesting problem. The exact probabilities require the use of stochastic processes, using Bayer's theorem, combined with methods for working with an infinite geometric series. They are:

• First shot at A – Chance of winning = 26.5%

• First shot at B – Chance of winning = 31.2%

But is this the best you can do? What if you waste your shot by shooting in the air? Consider that your situation now is exactly the same as when A shoots at B (above). But now the probability of a win is not reduced by 2/3. An exact calculation shows a chance of winning of 39.7% in this case. Interestingly, the best overall stratagy is to waste your shot.

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### TRI-AREA PROBLEM

Consider the diagram shown. A line has been constructed between C and F. Let X1 and X2 denote the areas of CDF and CEF respectively. Therefore X = X1 + X2.

Because triangles ABF and BEF have same area, then because the two triangles have a common height, the bases are the same. Therefore, AF = EF.

The triangles ACF and CEF also have a common height. Therefore, they have equal bases, AF and EF. Thus they have equal areas. We will state as equation one:

X1 + 3 = X2

Because triangles ADF and ABF have a common height and their areas are 3 and 7 respectively, we can state the proportion DF:BF = 3:7.

The triangles CDF and CBF also have a common height and they have bases DF and BF. Therefore, their area ratio is 3:7. Thus we can state as equation two:

7 * X1 = 3 * (X2+7)

By solving the simple system of two equations, we have:

X1 = 7.5
X2 = 10.5
X = X1 + X2 = 18

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### SQUARE POINTS

Construct two lines that bisect the square vertically and horizontally, as shown at right. No matter where the points are placed, at least two of them will be in the same small square, as shown. These points can be no more distant than the diagonal of the square, which can easily be calculated to be:

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### CHESS CHANCE

We can work this out by considering just one of the chess pieces. The following figure shows four possibilities for the rook. We count the number of squares it threatens and the number of squares that threaten it (and considering how a rook and bishop move). If the rook is in one of the middle four squares (this is 27 squares out of a possible 63), we multiply this by the four squares in the inner square out of a total 64 possible squares. To this we add similar calculations for the next three larger squares as shown below.

The final calculation gives us the solution:

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Let X = the number of students who passed the test.
Let P = the sum of the passing grades.
Let F = the sum of the failing grades.
Then 30 - X = the number of students who failed the test.

Using the formula for averages, we can use the three facts given in the problem to state:

P / X = 84
F / (30 - X) = 60
(P + F) / 30 = 80

Rearranging these equations into standard form, we have:

-84 X + P = 0
60 X + F = 1800
P + F = 2400

These are three equations in three unknowns. Solving the systems gives us 25 as the solution for X. Therefore, 25 students passed the test.

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### DIGIT SUM

The following solution is arrived at through inductive reasoning (This is not the same as proof by induction). Therefore, this solution is not proved here, just offered. The exponents have a difference of 2. Consider the general case:

2 n x 5 n + 2

The table below shows the first few calculations.

 n 2 n 5 n + 2 2 n x 5 n + 2 Digit Sum 1 2 125 250 7 2 4 625 2500 7 3 8 3125 25000 7 14 16 15625 250000 7 1999 ... ... 250 ... 7

The pattern shown by the first few calculations leads to the conclusion that the product will be 7.

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### SQUARES IN SQUARES

To solve this we can first consider sequences of good numbers. The series below starts with one and then each partitioning of the lower right into four squares adds three more (see the shaded squares. Continuing this pattern gives us all numbers that start with one and add a multiple of three (shown below).

1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, ...

The next sequence follows a similar pattern but now we start with nine and add multiples of three.

9, 12, 15, 18, 21, 24, 27, 30, 33, ...

The last sequence starts with seventeen and adds multiples of three in the same way.

17, 20, 22, 25, 28, 31, 34, ...

Adding these three sequences gives us the following.

1, 4, 7, 9, 10, 12, 13, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, ...

All whole numbers that are greater than fourteen are good numbers. The numbers that remain are:

2, 3, 5, 6, 8, 11, 14

Because the minimum number of new squares that can be added is never less than three, these numbers cannot be formed by partitioning a square into squares, and are not good numbers (you can check this by starting at each square number and adding different combinations of three or eight).

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### CORNERED SQUARE

Consider the figure below. Let L equal the length of the legs of the triangles. Using the formula for the area of a right angle triangle and the information given we can get the first equation below. Solving this for L gives us the solution to the problem, shown in the second equation.

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### SQUARES AND RECTANGLES

Consider that any rectangle can be uniquely defined by two horizontal lines and two vertical lines. The example below shows a three by two rectangle. There are nine possible vertical lines and nine possible horizontal lines.

In each case we can cover all possibilities by choosing any two out of the nine lines. Using the language of mathematics the answer is given by

Suppose you were asked to find the number of squares within the checkerboard. This problem cannot be solved directly as above. Perhaps a solution can be found without counting. Try starting with smaller checkerboards and look for a pattern.