Reality of Flatlanders

Last Problem:

At the chalkboard the teacher demonstrates the four factors of the number 6. That is to say, the whole numbers that can divide into 6 and leave no remainder. For the number six, the four factors are 1,2,3 and 6. (Remember, a number is always its own factor, as is 1.)

Between 1 and 100 there are five numbers that have exactly twelve factors. How quickly can you find all five?

Answer:

The five numbers between 1 and 100 that have twelve factors:

60: 1,2,3,4,5,6,10,12,15,20,30,60

72: 1,2,3,4,6,8,9,12,18,24,36,72

84: 1,2,3,4,6,7,12,14,21,28,42,84

90: 1,2,3,5,6,9,10,15,18,30,45,90

96: 1,2,3,4,6,8,12,16,24,32,48,96

Today’s Problem:

The sense of “Flatlanders” are limited to two dimensions (not three). If someone were to observe them from a point just “above” their world, the “Flatlanders” would have no way of seeing that particular observer.

What if you tossed a ball through the two-dimensional plane of Flatland? Would the Flatlander’s perceive the event as some sort of astronomical catastrophe? Describe exactly what the “Flatlanders” would see.

How Many Factors?

Yesterday’s Problem:

Three nickels and three dimes are distributed among three piggy banks such that each piggy bank holds two coins. One piggy bank has a label of 20 cents. One bank has a label of 15 cents. And, one bank has a label pf 10 cents.

There is a problem. Although each bank has a label showing how much money is supposed to be found in each piggy bank respectively, all three piggy banks are mislabeled.

Is it possible to determine how to correctly re-label the banks simply by shaking one of the banks until one of the coins drops out? If so, how could this be possible?

Answer:

If you shake one of the coins out of the bank labeled 15 cents, you can figure out how to correctly label all the banks.

Since you know that the bank is mislabeled, it cannot hold 15 cents- instead, the bank contains either two dimes or two nickels. The coin that drops out will tell you what the other coin is.

Say the answer is two dimes: that leaves you with three nickels and a dime between the two remaining piggy banks, one labeled 20 cents and one labeled 10 cents. Since the bank labeled 10 cents cannot have two nickels in it – because it is mislabeled – it must contain a nickel and a dime and the other bank must have the two nickels.

Today’s Problem:

At the chalkboard the teacher demonstrates the four factors of the number 6. That is to say, the whole numbers that can divide into 6 and leave no remainder. For the number six, the four factors are 1,2,3 and 6. (Remember, a number is always its own factor, as is 1.)

Between 1 and 100 there are five numbers that have exactly twelve factors. How quickly can you find all five?

Three Piggy Banks

Last Problem:

One grandfather, one grandmother, two fathers, two mothers, four children, three grandchildren, one brother, two sisters, two sons, two daughters, one father-in-law, one mother-in-law and one daughter-in-law attended a family reunion. If both halves of each relationship attended (i.e., the father and the son), how many people showed up?

Answer:

There were seven people at the reunion: a man and his wife, their three children (two girls and a boy) and the man’s mother and father.

Without stipulation that both halves of the relationships were present, there could be as few as four people. After all, one man can simultaneously be a father, a grandfather, a son, a brother and a father-in-law.

Today’s Problem:

Three nickels and three dimes are distributed among three piggy banks such that each piggy bank holds two coins. One piggy bank has a label of 20 cents. One bank has a label of 15 cents. And, one bank has a label pf 10 cents.

There is a problem. Although each bank has a label showing how much money is supposed to be found in each piggy bank respectively, all three piggy banks are mislabeled.

Is it possible to determine how to correctly re-label the banks simply by shaking one of the banks until one of the coins drops out? If so, how could this be possible?

The Family Reunion

Last Question:

A porter leads eight guests to their hotel rooms, roms 1 through 8. Unfortunately, the keys are not labeled and the porter has mixed up the order in which he originally held the eight keys. He dropped them while exiting the elevator.

Using trial and error, what is the maximum number of attempts the porter must make before he opens all the doors?

Answer:

The maximum number of attempts can be found by adding the following numbers:

8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36 attempts (maximum)

Today’s Problem:

One grandfather, one grandmother, two fathers, two mothers, four children, three grandchildren, one brother, two sisters, two sons, two daughters, one father-in-law, one mother-in-law and one daughter-in-law attended a family reunion. If both halves of each relationship attended (i.e., the father and the son), how many people showed up?

Eight Hotel Keys

Last Problem:

A man and a woman with Parkinsons meet on the Parkinsons Recovery Cruise to Alaska. At the end of the cruise the woman tells the man her phone number. The man is sure he will remember the number so he does not write it down.

Upon returning home to Austin, Texas the man discovers that he can remember all the digits of her phone number, but he has completely forgotten the order of the numbers.

He decides to arrange the seven digits in various orders and begin calling his friend. what are the chances he will call the woman’s correct phone number if he calls various random numbers  for one hour?

Answer:

The total number of permutations in a seven digit phone number is seven factorial [7! or 7 x 6 x 5 x 4 x 3 x 2 x 1] which equals 5,040. The probability that any random number the man calls is 1 in 5,040 or about .02 percent. The chances he will land on the correct phone number after calling phone numbers for one hour are very slim indeed.

He can only hope that on the next Parkinsons Recovery cruise the woman will return. He will be careful the second time to write her number down on a piece of paper.

Today’s Question:

A porter leads eight guests to their hotel rooms, rooms 1 through 8. Unfortunately, the keys are not labeled and the porter has mixed up the order in which he originally held the eight keys. He dropped them while exiting the elevator.

Using trial and error, what is the maximum number of attempts the porter must make before he opens all the doors?

I Can Remember Her Phone Number

Last Problem:

Mister Ladybug meets Miss Ladybug on the petal of a flower.

I’m a boy says the one with red dots.

I’m a girl says the one with yellow dots.

Then they both laugh because at least one of them is lying. From this information can you tell which one has the red dots and which has the yellow dots?

Answer:

With two statements there are four possible combinations of truth or falsehood:

true/true
true/false
false/true
false/false

The first combination cannot be right because at least one of the statements is false. The second and third can not be right either because if one of the statements if false, it is impossible for the other to be true.

The only logically consistent possibility is that they both lied. This means that Mister Ladybug has the yellow dots and Miss Ladybug has the red dots.

Today’s Problem:

A man and a woman with Parkinsons meet on the Parkinsons Recovery Cruise to Alaska. At the end of the cruise the woman tells the man her phone number. The man is sure he will remember the number so he does not write it down.

Upon returning home to Austin, Texas the man discovers that he can remember all the digits of her phone number, but he has completely forgotten the order of the numbers.

He decides to arrange the seven digits in various orders and begin calling his friend. What are the chances he will call the woman’s phone number if he calls for one hour straight?

Mr and Ms Ladybug

Yesterday’s Problem:

A bookworm eats through 5 volumes of Robert Rodgers, Road to Recovery from Parkinsons Disease, that are sitting on his library shelf in Olympia, Washington. The bookworm finds itself on page 1 of volume 1 and begins eating through to the last page of volume 5.

If each book is 6 centimeters thick – Rodgers is wordy – including the front and back covers which are half a centimeter each, what is the distance the bookworm travels across the library shelf?

Answer:

Consider how the books are sitting on Rodgers’ library shelf.  The  bookworm eats through only the front cover of volume 1, all of the volumes 2, 3 and 4 and only the back cover of volume 5. The total distance traveled is thus 29 centimeters :

5.5+6+6+6+5.5=29

Today’s Problem:

Mister Ladybug meets Ms Ladybug on the petal of a flower.

I’m a boy says the one with red dots.

I’m a girl says the one with yellow dots.

Then they both laugh because at least one of them is lying. From this information can you tell which one has the red dots and which has the yellow dots?

The Bookworm

Last Problem:

If you draw the lucky ticket, you win the lottery jackpot. You are given the option to draw one ticket out of a box of 10 or draw ten times out of a box of 100, returning the drawn ticket each time before drawing again. Which choice gives you the best odds?

Answer:

The choices offer identical odds. But, in psychological experiments, about four in ten people preferred the singe draw and held to this decision even when the other choice was altered to provide fifty draws from the box of 100.

Today’s Problem:

A bookworm eats through 5 volumes of Robert Rodgers, Road to Recovery from Parkinsons Disease, that are sitting on his library shelf in Olympia, Washington. The bookworm finds itself on page 1 of volume 1 and begins eating through to the last page of volume 5.

If each book is 6 centimeters thick – Rodgers is wordy – including the front and back covers which are half a centimeter each, what is the distance the bookworm travels across the library shelf?