MindBender Archive

Puzzles Archive
This is a list of the previous puzzles that have been sent out by E-mail.
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November 24, 1998 7. A man died and went to Heaven. There were thousands of other people there. They were all naked and all looked as they did at the age of 21. He looked around to see if there was anyone he recognized. He saw a couple and he knew immediately that they were Adam and Eve. How did he know?
8. A woman had two sons who were born on the same hour of the same day of the same year. But they were not twins. How could this be so?
9. A man walks into a bar and asks the barman for a glass of water. The barman pulls out a gun and points it at the man. The man says 'Thank you' and walks out.

...6. The poison in the punch came from the ice cubes. When the man drank the punch, the ice was fully frozen. Gradually it melted, poisoning the punch.
7. He recognized Adam and Eve as the only people without navels. Because they were not born of women, they had never had umbilical cords and therefore they never had navels. This one seems perfectly logical but it can sometimes spark fierce theological arguments.
8. They were two of a set of triplets (or quadruplets, etc.). This puzzle stumps many people. They try outlandish solutions involving test-tube babies or surrogate mothers. Why does the brain search for complex solutions when there is a much simpler one available?
9. The man had hiccups. The barman recognized this from his speech and drew the gun in order to give him a shock. It worked and cured the hiccups--so the man no longer needed the water. The is a simple puzzle to state but a difficult one to solve. It is a perfect example of a seemingly irrational and incongruous situation having a simple and complete explanation. Amazingly this classic puzzle seems to work in different cultures and languages.

November 30, 1998
I'm the beginning of eternity, the end of time and space. The beginning of every end, and the end of every place.
What am I?

...The letter "E".

A simple case of long division, where each of the letters A, B, C, D, E, F, G, H, I, and J uniquely represent one of the ten digits in the following cryptarithmetic problem in long division. The problem is best viewed in a mono-space font like Courier.
AAB quotient = AAB
_______
CDB | EEFGH divisor = CDB and dividend = EEFGH
IBC
____
JAJG
IBC
_____
DDHH
DGFE
____
DGB remainder = DGB
___________________________

...Answer to Cryptarithmetic Long Division
227 quotient = 227
_______
437 | 99506 divisor = 437 and dividend = 99506
874
____
1210
874
_____
3366
3059
____
307 remainder = 307
The above is the simple answer for this cryptarithmetic problem in long division. There are probably many methods of arriving at the final solution. The following are some of the steps in the thought processes I used to arrive at the solution:
AAB Line 1
_______
CDB | EEFGH Line 2
IBC Line 3
____
JAJG Line 4
IBC Line 5
_____
DDHH Line 6
DGFE Line 7
____
DGB Line 8
The following are mostly based on consideration of subtraction and any borrowing that is necessary or unnecessary.
Lines 6, 7, and 8 show G must be 0.
Lines 4, 5, and 6 show J must be 1.
Lines 2, 3, and 4 show I must be < E.
Lines 4, 5, and 6 show A must be < I.
Lines 4, 5, and 6 show A must be < D.
Lines 1, 2, and 3 show C must be < I since IBC is a multiple of CDB.
Lines 2, 3, and 4 and J=1 show F must be = C+1.
Lines 6, 7, and 8 and G=0 show H must be = F+1.
Lines 6, 7, and 8 show H must be < E.
Lines 6, 7, and 8 show H must be < B.
Lines 2 and 8 show D must be < C since the remainder must be less than the divisor.
Combining all these, we have:
G=0
J=1M
AA which means that:
A=2
D=3
C=4
F=5
and
H=6
Substituting those into the original problem and calculating, we find:
B=7
I=8
E=9
This then leads to the final solution.
I don't know any source for this Cryptarithmetic Long Division problem.

Mini-MindBender for Kids
This Mini-MindBender is suitable for even young kids. Appropriate age depends on ability and previous puzzle experience. Share this Mini-MindBender with a young person you know. It should be fun for both of you. Make sure you solve it yourself first though, because you know how kids can be --- they usually want an answer immediately after they have a solution.

Tennis Tournament
An elimination tennis tournament is to be held. This means that if you lose one match, you are eliminated from the rest of tournament play. 137 students are to be paired in matches in the first round of the tournament. Because 137 is an odd number, one randomly selected student will have a bye, meaning that student doesn't play in the first round and moves right on to the second round, as if she/he had won in that first round. Pairing continues on the second and later rounds, with a bye given to one randomly picked student in a round if an odd number of students is left for that round. The tournament schedule is planned so that a minimum number of matches is required to determine the tournament champion. How many matches must be played?

...Because all but one player (or 136) must be eliminated, and each match eliminates one player, exactly 136 tennis matches must be played.
This MindBender was modified from a puzzle in Martin Gardner's book, "The Unexpected Hanging and Other Mathematical Diversions."

December 3, 1998
Q. You have two strings of different lengths and thicknesses. Both are made such that when set aflame at one end, they take one hour to burn away completely. You have a box of matches, and no watch. How, using only the strings and the matches, (and no funny tricks like cutting the strings, measuring thier lengths, etc. ) would you measure a time of 45 minutes?

...A. Burn one string from both ends, and the other from one end only. Whenthe first string is completely burnt, light the other end of the second string. Viola!

December 7, 1998
Four Fours Part 1
This is the first part of a multi-part MindBender. It is an old problem but we will add a new twist to it later.
Using only simple mathematical symbols i.e., plus signs (+), minus signs(-), multiplication signs(*), division signs(/), left parentheses(() and right parentheses()) and the digit 4 four times, represent the numbers from 0 to 10. An example would be:
16 = 4 + 4 + 4 + 4 although 16 is not one of the desired numbers to represent today.
___________________________
Mini-MindBender for Kids
This Mini-MindBender is suitable for even young kids.
Appropriate age depends on ability and previous puzzle experience. Share this Mini-MindBender with a young person you know. It should be fun for both of you. Make sure you solve it yourself first though, because you know how kids can be --- they usually want an answer immediately after they have a solution.

Mini-MindBender for Kids
Water Glasses
This problem is best viewed in a mono-space font like Courier.
Six glasses are lined up in a row. The first three glasses are filled with water. The last three are empty. By moving one glass only, change the arrangement so that the glasses alternate full with empty. The original order would look like:
Glass Glass Glass Glass Glass Glass
1 2 3 4 5 6
|--| |--| |--| | | | | | |
|WW| |WW| |WW| | | | | | |
|WW| |WW| |WW| | | | | | |
---- ---- ---- ---- ---- ----
Full Full Full Empty Empty Empty
___________________________

...The following answers are just one of many possible representations of the numbers from 0 to 10
0 = 4 + 4 - 4 - 4
1 = 44 / 44
2 = (4 / 4) + (4 / 4)
3 = (4 + 4 + 4) / 4
4 = 4 * (4 - 4) + 4
5 = ((4 * 4) + 4) / 4
6 = ((4 + 4) / 4) + 4
7 = (44 / 4) - 4
8 = 4 + 4 + 4 - 4
9 = 4 + 4 + (4 / 4)
10 = (44 - 4) / 4

Answer to Mini-MindBender for Kids Water Glasses
Pick up Glass 2.
Pour its contents into Glass 5.
Replace Glass 2.
The result would look like:
Glass Glass Glass Glass Glass Glass
1 2 3 4 5 6
|--| | | |--| | | |--| | |
|WW| | | |WW| | | |WW| | |
|WW| | | |WW| | | |WW| | |
---- ---- ---- ---- ---- ----
Full Empty Full Empty Full Empty
This Mini-MindBender was modified from a puzzle in Martin Gardner's book, "The Unexpected Hanging and Other Mathematical Diversions."

December 14, 1998
Four Fours Part 2
Using any mathematical symbols (this could include such symbols as ! for factorial and sqrt() for the square root function) and parentheses and the digit 4 four times, represent the numbers from 0 to 25 (or 50 depending on your interest level and dedication). An example would be:
58 = 4! + 4! + (4 / .4) although 58 is not one of the desired numbers to represent today.
___________________________

Mini-MindBender for Kids
JFK/de Gaulle
1963 must have been a very unusual year since, in that year:
A. John Kennedy's age was 46 and he had been in office 3 years. He was born in 1917. He became president in 1960. The sum of these four numbers is 3,926.
B. Charles de Gaulle's age was 73 and he had been in office 5 years. He was born in 1890. He became president of France in 1958. The sum of these four numbers is also 3,926.
Can you explain this remarkable coincidence?

...Answer to MindBender Four Fours Part 2
The following answers are just one of many possible representations of the numbers from 0 to 50.
0 = 4 + 4 - 4 - 4
1 = 44 / 44
2 = (4 / 4) + (4 / 4)
3 = (4 + 4 + 4) / 4
4 = 4 * (4 - 4) + 4
5 = ((4 * 4) + 4) / 4
6 = ((4 + 4) / 4) + 4
7 = (44 / 4) - 4
8 = 4 + 4 + 4 - 4
9 = 4 + 4 + (4 / 4)
10 = (44 - 4) / 4
11 = (44 / (sqrt(4 * 4)))
12 = 4 + 4 + sqrt(4) + sqrt(4)
13 = (4! + sqrt(4)) / sqrt(4)
14 = (4 * 4) - (4 / (sqrt(4))
15 = (4 * 4) - (4 / 4)
16 = 4 + 4 + 4 + 4
17 = (4 * 4) + (4 / 4)
18 = (4 * 4) + (4 / (sqrt(4))
19 = (4 + 4 - .4) / .4
20 = (4 + (4 / 4)) * 4
21 = 4! - 4 + (4 / 4)
22 = 4! -sqrt(4) + 4 - 4
23 = 4! - (4 / (sqrt(4 * 4))
24 = 4! - 4 + sqrt(4) + sqrt(4)
25 = 4! + (4 / (sqrt(4 * 4)))
26 = 4! + sqrt(4) + 4 - 4
27 = 4! + 4 - (4 / 4)
28 = 4! + 4 + 4 - 4
29 = 4! + 4 + (4 / 4)
30 = 4! + sqrt(4 * 4) + sqrt(4)
31 = 4! + ((4! + 4) / 4)
32 = 4 * (4 + sqrt(4 * 4))
33 = 4! + ((4 - .4) / .4)
34 = 4 * (4 + 4) + sqrt(4)
35 = 4! + (44 / 4)
36 = 4! + 4 + 4 + 4
37 = 4! + ((4! + sqrt(4)) / sqrt(4))
38 = 44 - 4 - sqrt(4)
39 = ((4 * 4) - .4) / .4
40 = 44 - sqrt(4 * 4)
41 = ((4 * 4) + .4) / .4
42 = 44 - (4 / sqrt(4))
43 = 44 - (4 / 4)
44 = 44 + 4 - 4
45 = 44 + (4 / 4)
46 = 44 + (4 / sqrt(4))
47 = 4! + 4! - (4 / 4)
48 = 44 + sqrt(4 * 4)
49 = 4! + 4! + (4 / 4)
50 = 44 + 4 + sqrt(4)
This can be extended farther if you are willing to try.
I don't know any source for this Four Fours Part 2 MindBender, but it has been around for many years.

Answer to Mini-MindBender for Kids
JFK/de Gaulle
Any date added to the number of years since that date will total the current year. Two such totals will be twice the current year.
This Mini-MindBender was modified from a puzzle in Martin Gardner's book, "The Unexpected Hanging and Other Mathematical Diversions."

December 21, 1998
MindBender
Four Fours Part 3
(Although no fours are used in this part of this multi-part MindBender.)
This is the third part of a multi-part MindBender. It is an old problem but we expanded it before and now we will add a new twist to it.
Using only Pi and the mathematical symbols for addition, subtraction, multiplication, division, sqrt() for the square root function, rndwn() for the round-down function, and parentheses, represent the numbers from 0 to 20. The round-down function is defined such that rndwn(X) equals the greatest integer that is equal to or less than X. Each symbol and Pi may be repeated as often as necessary, but it is desired to use Pi as few times as possible. An example would be:
27 = rndwn(Pi) * rndwn(Pi) * rndwn(Pi) although 27 is not one of the desired numbers to represent.
___________________________
Mini-MindBender for Kids
A Simple Die Probability
Joe throws an ordinary six sided die, then Moe throws the same die. What is the probability that Joe will throw a higher number than Moe?
___________________________

...Answer to MindBender
Four Fours Part 3
The following answers are just one of many possible representations of the numbers from 0 to 20.
0 = Pi - Pi
1 = rndwn(sqrt(Pi))
2 = rndwn(sqrt(Pi * sqrt(Pi)))
3 = rndwn(Pi)
4 = rndwn(Pi + sqrt(Pi))
5 = rndwn(Pi * sqrt(Pi))
6 = rndwn(Pi + Pi)
7 = rndwn(Pi + Pi) + rndwn(sqrt(Pi))
8 = rndwn(Pi * Pi - sqrt(Pi))
9 = rndwn(Pi * Pi)
10 = rndwn(Pi * Pi) + rndwn(sqrt(Pi))
11 = rndwn(Pi * Pi + sqrt(Pi))
12 = rndwn(Pi * Pi) + rndwn(Pi)
13 = rndwn((Pi * Pi) + Pi)
14 = rndwn((Pi * Pi) + Pi + sqrt(Pi)) OR rndwn(rndwn(Pi) * (Pi + sqrt(Pi)))
15 = rndwn(Pi * Pi) + rndwn(Pi + Pi) OR rndwn(Pi) * rndwn(Pi * sqrt(Pi))
16 = rndwn((Pi * Pi) + Pi + Pi) OR rndwn(rndwn(Pi) * Pi * sqrt(Pi))
17 = rndwn(Pi * Pi * sqrt(Pi))
18 = rndwn(Pi * Pi) + rndwn(Pi * Pi) OR rndwn(Pi) * rndwn(Pi + Pi)
19 = rndwn((Pi * Pi) + (Pi * Pi)) OR rndwn(Pi * (Pi + Pi))
20 = rndwn(Pi * sqrt(Pi)) * rndwn(Pi + sqrt(Pi))
This MindBender was modified from a puzzle in Martin Gardner's book, "Wheels, Life and Other Mathematical Amusements."

Answer to Mini-MindBender for Kids
A Simple Die Probability
5/12.
The probability that both Joe and Moe will throw the same number is 1/6. Therefore, the probability that one will throw higher than the other is 1 - 1/6, or 5/6. Divide 5/6 (or 10/12) by two to get the probability that Joe will get a higher number than Moe.
This Mini-MindBender was modified from a puzzle in Martin Gardner's book, "The Unexpected Hanging and Other Mathematical Diversions."

December 23, 1998
Fourteen Questions
1. There's one sport in which neither the spectators nor the participants know the score or the leader until the contest ends. What is it?
2. What famous North American landmark is constantly moving backward?
3. Of all vegetables, only two can live to produce on their own for several growing seasons. All other vegetables must be replanted every year. What are the only two perennial vegetables?
4. At noon and midnight the hour and minute hands are exactly coincident with each other. How many other times between noon and midnight do the hour and minute hands cross?
5. What is the only sport in which the ball is always in the possession of the team on defense, and the offensive team can score without touching the ball?
6. What fruit has its seeds on the outside?
7. In many liquor stores, you can buy pear brandy, with a real pear inside the bottle. The pear is whole and ripe, and the bottle is genuine; it hasn't been cut in any way. How did the pear get inside the bottle?
8. Only three words in standard English begin with the letters "dw". They are all common. Name two of them.
9. There are fourteen punctuation marks in English grammar. Can you name half of them?
10. Where are the lakes that are referred to in the "Los Angeles Lakers"?
11. There are seven ways a baseball player can legally reach first base without getting a hit. Taking a base on balls -- a walk -- is one way. Name the other six.
12. It's the only vegetable or fruit that is never sold frozen, canned, processed, cooked, or in any other form but fresh. What is it?
13. How is it possible for a pitcher to make four or more strikeouts in one inning?
14. Can you name six or more things that you can wear on your feet, that begin with the letter "s"?
___________________________

... Answer to MindBender
Fourteen Questions
1. Boxing.
2. Niagara Falls. The rim is worn down about 2 and a half feet each year because of the millions of gallons of water that rush over it every minute.
3. Asparagus and rhubarb.
4. Ten times (not eleven, as most people seem to think).
5. Baseball.
6. Strawberry.
7. The pear grew inside the bottle. The bottles are placed over pear buds when they are small, and are wired in place on the tree. The bottle is left in place for the whole growing season. When the pears are ripe, they are snipped off at the stems.
8. Dwarf, dwell, and dwindle.
9. Period, comma, colon, semicolon, dash, hyphen, apostrophe, question mark, exclamation point, quotation marks, brackets, parenthesis, braces, and ellipses.
10. In Minnesota. The team was originally known as the Minneapolis Lakers, and kept the name when they moved west.
11. Batter hit by a pitch; passed ball; catcher interference; catcher drops third strike; fielder's choice; and being designated as a pinch runner.
12. Lettuce.
13. If the catcher drops a called third strike, and doesn't throw the batter out at first base, the runner is safe.
14. Shoes, socks, sandals, sneakers, slippers, skis, snowshoes, stockings, etc.

This MindBender came from Gary's Funny Email List.
Info available from:
Gary Noel Boone

December 27, 1998
The following problems involve tests conducted by a group of self-proclaimed psychics, who attempt to divine the identity of cards placed face down on a table.

FIRST PSYCHIC GUESS
Abe, Bea, and Cal begin by dealing three aces face down and guessing at each card in turn (as shown below). Each of the three aces is correctly identified by at least one person. Nobody gets exactly one right answer, however, and no two persons finish with the same number of correct answers. What are the three cards'!
...............1st Card.........2nd Card.........3rd Card
Abe.........Heart.............Spade..............Club
Bea..........Heart.............Diamond.........Club
Cal...........Diamond.........Spade............Heart

...The cards in order are Heart, Diamond, Club

December 29, 1998
SECOND PSYCHIC GUESS
Abe. Bea. and Cal repeat their test. This time everybody gets at least one right answer, but no two persons get the same number right. What are the three cards?
...............1st Card.........2nd Card.........3rd Card
Abe.........Heart.............Spade..............Diamond
Bea..........Club.............Diamond.........Heart
Cal...........Club.........Spade............Heart

...The cards in order are: Club, Spade, Heart

January 5, 1999
THIRD PSYCHIC GUESS
Abc, Bea, Cal, and Dee use four aces, Each ace is correctly identified by at least one person. When they check their results, they learn that each of them has the same number of correct guesses. What are the four cards?
...............1st Card.........2nd Card.........3rd Card..............4th Card
Abe.........Club.............Heart..............spade..............Diamond
Bea..........Heart.............Heart.........Diamond..............Diamond
Cal...........Diamond.........Heart............Club..............Heart
Deel...........Spade.........Diamond............Club..............Heart

...The cards in order were: Heart, Diamond, Spade Club

January 5, 1999
FOURTH PSYCHIC GUESS
The four repeat their test. Again, each ace is correctly identified by at least one person, and everybody makes the same number of correct guesses. What are the four cards?
...............1st Card.........2nd Card.........3rd Card..............4th Card
Abe.........Heart.............Club..............Diamond..............Spade
Bea..........Club.............Spade.........Diamond..............Heart
Cal...........Club.........diamond............Diamond..............Club
Deel...........Heart.........Heart............Club..............Spade

...The cards in order were: Club, Heart, Diamond Spade

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