Ten Problems

Here are some of your (and our) favorite problems. Enjoy — trick or treat!

***********************************************

Problem One: Gadsby

If youth, throughout all history, had had a champion to stand up for it; to show a doubting world that a child can think; and, possibly, do it practically; you wouldn’t constantly run across folks today who claim that “a child don’t know anything.” A child’s brain starts functioning at birth; and has, amongst its many infant convolutions, thousands of dormant atoms, into which God has put a mystic possibility for noticing an adult’s act, and figuring out its purport.

What’s significant about this quotation from a book (and this conundrum)?

Source: Gadsby by E. V. Wright

Solution

Problem Two: Thrifty Thugs

Five thugs steal 100 diamonds. They come up with a way to distribute them. They get in line and the first thug suggests a sharing method. Then, a vote is taken. If more than 50% agree, the method is used. Otherwise, the suggester is killed and the next person in line suggests another solution. Assuming all five people are intelligent and self-serving, what are the end results?

Solution

Problem Three: Estmodus Inrebus

Since this is quite a tough problem, I’m not going to post nine more problems today, and I hope everyone try this because it is actually really interesting.Replace the letters and asterisks with digits to satisfy the multiplication:

Image

Don’t ask me what “Estmodus Inrebus” means. I have no clue. FYI this puzzle comes from “The Master Book of Mathematical Recreations” created by a mathematician by the name of Fred Schuh. I’ll try to post a full solution to this problem in a few days.

Edit: I just found out what the phrase means. It’s actually “Est modus in rebus”, a quote from Horace (Ancient Roman poet), meaning “There is a measure in things”.

Solution

Problem Four: The Four Doors

A man wakes up in a room. He sees no windows, a table with a note on it, and four doors ahead of him. Each door has a different symbol on its face: rectangle, kite, rhombus, and trapezoid.He then walks to the table and reads the message:
“Three of these doors hold behind them a deadly booby trap, and one of them leads to the exit of this house. You can only open one door. The door that leads to the exit of this house can be deduced by looking at a string of numbers hidden somewhere in the house. Figure it out quickly, however, because as soon as the clock strikes noon, all the doors in the house will become permanently locked.”He then searched everywhere in the room, until he found a string of numbers carved under the table:1 - 4 - 8 - 11 - 1He thought long and hard as the clock ticked away, getting closer and closer to his potential doom.Suddenly he realized what the numbers meant, and promptly went over to the correct door, which he opened and managed to escape the house unhindered.

Which door leads to the exit?

Solution

Problem Five: Mayhen

If a hen and a half lays an egg and a half in a day and a half, how many and a half who lay better by half will lay half a score and a half in a week and a half?

Source: 536 Puzzles and Curious Problems by Dudeney (p. 50)

Solution

Problem Six: The Blind Mathematician

Here’s is a fun problem I found on the Internet today while browsing around:

Once upon a time a king wanted to hire the best Mathematician in his kingdom to work in his palace. His servants brought to him the best two mathematicians, one of them was blind. The king told both mathematicians that he can’t hire both so he will ask a question and whoever answers gets the job.

The king said: “I have three sons, whoever guesses their ages will be hired.” The king told them that if they multiply the ages of his sons the result will be 36. Both mathematicians told the king that this information is not enough. The king then said: “The number of windows in the building across the street is equal to the sum of the ages of my sons.” The first mathematician (who can see) counted the windows, and told the king that he still could not figure it out. The Blind mathematician (who could not count the windows) told the king that he does not have an answer. The king then said: “My oldest son has red hair.” Right away the blind mathematician gave the correct answer and got hired.

The question is: How did the blind guy know the answer and what is the answer?

There are no ‘tricks’ to this problem. An you can assume the age are whole numbers and if their ages differ by less than a year they count as having the same ages (i.e. two six year olds have the same ages regardless of when they are born).

Source

Solution

Problem Seven: 16 Coins

Johnny was given 16 coins by his older, somewhat meaner brother, Mark. He told him that he could keep them all if he could place all 16 on the table in such a way that they formed 15 rows with 4 coins in each row.After 10 minutes, Johnny walked away with the coins and Mark, after complaining futilely to his mother, left with nothing.How did Johnny place the coins?

Solution

Problem Eight: Connect the Dots II

Image

Go through all 49 dots in twelve straight contiguous lines. Begin and end on the red dot.

Solution

Problem Nine: Eating for Free?

A family of eight were eating at a nice hometown diner. After all of them had their fill of grub, they received the bill. The bill ended up being exactly $69.69, but the family left the diner without paying a dime. They were not arrested for not paying the bill, in fact, their waitress wished them a good night before they left. How did they get away with a seemingly free meal?

Solution

Problem Ten: Critical Moment

This is probably one of the most famous puzzles of all time, but I’ll post it here anyways in case someone hasn’t seen it before.In a game show you can pick one of three doors. A,B or C. One of the doors holds the car you got your eyes on. The others contain goats. You guess door A. The host (knowing which door holds the prize) opens one of the 2 other doors that he knows holds no prize. He opens door B. He tells you to try again, and pick one of the 2 doors that are left, A or C. Which door should you pick?

Solution

If you have any comments or problems, please contact us!

Answer One: Spot the Differences VIII

Show ▼

Answer Two: Sudoku VI

Show ▼

Answer Three: Coin in the Bottle

Show ▼

Answer Four: All Five Vowels

Show ▼

Answer Five: Detention Essay

Show ▼

Answer Six: Fifty

Show ▼

Answer One: The Blood of the Earth
Show ▼

Answer Two: The Scrapbookers
Show ▼

Answer Three: Royalty?
Show ▼

Answer Four: Self-Esteem Builder
Show ▼

Answer Five: Speeding Blues
Show ▼

Answer Six: Tinman’s Pick
Show ▼

Answer Seven: Screaming and Soaring
Show ▼

Answer Eight: 1 to 12
Show ▼

Answer Nine: Big Splash
Show ▼

Answer Ten: Blessing or Bane
Show ▼

This is going to be quite a long post, since the solution is quite involved. Here is the original problem.

This is the diagram in question:

First, we note that there are both alphabets and asterisks in the equation. We see that there are 12 different letters in total, so while the same letters always represent the same digit, each digit may be represented by more than one letter. For example, while “E”s may always represent “3″, other letters, such as “M”, may also represent “3″.

To simplify our analysis, let’s call the eight digit multiplier (ESTMODUS) and the other (INREBUS) . Since there are seven partial products (the seven lines of asterisks in the middle), we can deduce that there are no “0″ in . We also note that the last two digits of , , and their product are the same, namely “US”. Thus, the square of “US” must also end in “US”; for example, and . It turns out that these are the only two possibilities (among two digit numbers, which we are looking for here), and so “US” must be or . To prove this point, let the two digit number “US” be , then the last few digits of the product must be . But the last two digits of is just , therefore must be divisible by 100. From this, we can see that either is divisible by 4 and is divisible by 25, or is divisible by 25 and is divisible by 4. In the first possibility, ; in the other possibility, .

(more…)

Here’s a fun game for two. The first person says a number between 1 and 10. Then the other will say a number that is at least 1 higher and at most 10 higher than the previous number. They go back and forth until one says 50 and wins. What’s the winning strategy for the person who starts?

An annoyed detention teacher told his misbehaving student, “You have a minute to produce a 1000-word essay. Go!”

What should the student do?

Which 6-letter English word contains all five vowels?

A bottle containing a coin is sealed with a cork at the bottleneck. How can you take out the coin without taking out the cork or breaking the bottle?

Link

Solution

Source: puzzles.about.com

Link

Solution

© Bonnie J. Malcolm

Answer One: Spot the Differences VII

Show ▼

Answer Two: Sudoku V

Show ▼

Answer Three: Missing Square Puzzle

Show ▼

Answer Four: Crazy Driver

Show ▼

Answer Five: 1=2

Show ▼

Since Percy posted the 1=2 thing a few days ago, I’m tempted to make up something of my own. So here it is…

To proof:

proof:

multiply both sides by , we have

From property of logarithms, we have

Therefore, , and

What went wrong?

Update: Solutions are posted!

Which two whole numbers (not fractions or decimals) multiply together to give you the luck number 13?

Update: Solutions are posted!

How can you distribute 10 sugar cubes in 3 glasses such that each glass contain an odd number of such cubes? No consumption of anything whatsoever!

Update: Solutions are posted!

Tony and Janet decide to play a game to settle their differences. The game begins with N coins. They alternate turns with Tony going first. Every turn, they must remove one, three or four coins from the table. The player who takes the last coin wins the game. For which values of N between 31 and 35 inclusive does Janet have a guaranteed winning strategy?

Update: Solutions are posted!