Puzzles Archive
This is a list of the previous puzzles that have been sent out by
E-mail.
Don't forget to signup for "The MindBender" here.
To see the answer, click and hold your mouse button just to the right
of the red arrow 
and drag down. This will highlight
the answer and make it visible.
August 9, 1999
MindBender
Five Couples
Five couples spend an evening together. The women's first names are
Kris, Terri, Mary, Suzie, and Jeannie. The men's first names are Matt,
Pat, Dave, Scott, and Jeff. At a given time during the evening:
Matt's wife is not dancing with her husband, but with Kris's husband.
Scott and Jeannie are not dancing.
Suzie is not dancing either - she is eating ice cream.
Pat is playing the trumpet, with Mary at the piano, and Jeff is
lightly playing the drums. None of the musicians are married to each
other.
If Suzie's husband is not Pat, who is Jeff's wife?
___________________________
Mini-MindBender for Kids
Number Tree
56..88..79..96
.\../\...\../
..\/..\...\/
..27..16..31
...\...\../
....\...\/
.....9..11
.....\../
......\/
......??
What number belongs where the two question marks are?
___________________________
...Answer to MindBender
Five Couples
Kris's husband is not Matt. He cannot be Scott, Pat, or Jeff who are
not dancing. Therefore, Kris is married to Dave.
Likewise, Matt's wife is not Kris, Jeannie, Suzie, or Mary. Therefore,
Matt is married to Terri.
Mary's husband is not Pat or Jeff, the other musicians. From
above it can't be Dave or Matt. Therefore, Mary's husband is Scott.
Suzie's husband is not Pat. From above it can't be Dave, Matt, or
Scott. Therefore, Suzie is married to Jeff?
By elimination, Jeannie is married to Pat.
So, Jeff's wife is Suzie.
This MindBender was significantly modified from a puzzle in Pierre
Berloquin's book, "100 Games of Logic."
___________________________
Answer to Mini-MindBender for Kids
Number Tree
11. Each number is the sum of the digits in the number or numbers
linked to it from above.
This MindBender was modified from a puzzle in Pierre Berloquin's book,
"100 Games of Logic."
August 16, 1999
MindBender
Summer Houses
Five friends, Andrew, Bernard, Claude, Donald, and Eugene have summer houses
along the Atlantic coast. Each wanted to name his house after the daughter of
one of his friends that is, Anne, Belle, Cecelia, Donna, and Eve (but not
necessarily in that order). To be sure that their houses would have different
names, the friends met to make their choices together. Claude and Bernard
both wanted to name their house Donna. They drew lots and Bernard won. Claude
named his house Anne. Andrew named his house Belle. Eve's father had not come
to the meeting, and Eugene phoned him to tell him to name his house Cecelia.
Belle's father named his house Eve. What is the name of each friend's
daughter? What is the name of each friend's house?
___________________________
Mini-MindBender for Kids
Ties
Timothy's tie rack contains 17 blue ties, 11 yellow ties, 9 orange ties, 34
green ties, and 2 purple ties. The ties are not sorted by color. The light
bulb has burned out so Timothy cannot see what color the ties are. How many
ties does Timothy have to take to be sure he has at least two ties of the
same color?
___________________________
...Answer to MindBender
Summer Houses
We know that :
Andrew's house is Belle.
Bernard's house is Donna.
Claude's house is Anne.
Their daughters are not so named. Claude cannot be Donna's father and Eugene
cannot be Eve's father. Belle's father, who named his house Eve, can only be
Donald or Eugene. Similarly, Eve's father is Donald or Eugene. Since Eugene
phoned Eve's father, Eve's father is Donald. His house is Cecelia. Eugene is
Belle's father; Andrew is Donna's father; Bernard is Anne's father; and
Claude is Cecelia's father.
The answer summary is:
Father.....Daughter....House
-------....--------....-----
Andrew.....Donna.......Belle
Bernard....Anne........Donna
Claude.....Cecelia.....Anne
Donald.....Eve.........Cecelia
Eugene.....Belle.......Eve
My approach for this type of MindBender:
Create two matrices that match 1) the fathers' names with the houses'
names and 2) the fathers' names with the daughters' names. An entry in the
matrices contains X and a statement number for impossible or an ! and a
statement number for definite. Anytime an ! is entered in a matrix entry, all
other "same column" and all other "same row" entries in that matrix receive
X. Number the statements as follows:
0 No house is named after the owner's daughter.
1 Claude and Bernard both wanted to name their house Donna.
2 They drew lots and Bernard won.
3 Claude named his house Anne.
4 Andrew named his house Belle.
5 Eve's father had not come to the meeting, and Eugene phoned him to tell him
to name his house Cecelia.
6 Belle's father named his house Eve.
7 Anytime an ! is entered in a matrix entry, all other "same column" and all
other "same row" entries in that matrix receive X.
8 The only remaining choice for a row or column.
Work on the house matrix first.
..................House
..........A....B....C....D....E
.Father
...A......X7...!4...X7...X7...X7
...B......X7...X7...X7...!2...X7
...C......!3...X7...X7...X7...X7
...D......X7...X7...!5...X7...X7
...E......X7...X7...X7...X7...!8
.................Daughter
..........A....B....C....D....E
.Father
...A......X7...X0...X7...!8...X7
...B......!8...X7...X7...X0...X7
...C......X0...X7...!8...X1...X7
...D......X7...X7...X0...X7...!5
...E......X7...!6...X7...X7...X0
This results in the same answer summary as above.
This MindBender was modified from a puzzle in Pierre Berloquin's book,
"100 Games of Logic."
___________________________
Answer to Mini-MindBender for Kids
Ties
Six. There are five colors. The first five ties might be all different
colors, but the first six cannot be. Even if the first five are all
different, the sixth must match the color of one of the first five. Of course
a match could occur earlier, even on the second pick, but six are required to
guarantee at least one match.
This MindBender was modified from a puzzle in Pierre Berloquin's book,
"100 Games of Logic."
August 23, 1999
MindBender
Shopkeepers
Al is a butcher and president of the street shopkeepers' committee,
which also includes the grocer, the baker, and the candle maker. They
all sit around a table. Al sits on Bob's left. Cal sits at the
grocer's right. Dave, who faces Bob, is not the baker. What kind of
store does Cal have?
___________________________
Mini-MindBender for Kids
Multiplication
???,4?? X 7 = 6,743,?56
What are the missing digits (those represented by ?s)?
___________________________
...Answer to MindBender
Shopkeepers
Assigning Al the bottom seat, the four men can only sit this way:
......Cal
Dave.......Bob
......Al
Hence, Al is the butcher. Bob is the grocer. Dave is the candle maker.
Cal is the baker.
This MindBender was modified from a puzzle in Pierre Berloquin's book,
"100 Games of Logic."
___________________________
Answer to Mini-MindBender for Kids
Multiplication
Berloquin's Answer:
The ones digit of the first number multiplied by 7 must give a number
ending in 6. Only 8 will do that. So we have:
???,4?8 X 7 = 6,743,?56
8 times 7 is 56. Since the product already ends in 56, the tens digit
of the first number times 7 must end in 0. Therefore, the tens digit
is 0. We have:
???,408 X 7 = 6,743,?56
Now 408 times 7 is 2856. The missing digit in the product is 8:
???,408 X 7 = 6,743,856
The thousands digit times 7 must end in 1, since 1 plus "2 to carry
from 2856" is the 3 in the product. The only multiple of 7 ending in 1
is 21, so the thousands digit is 3:
??3,408 X 7 = 6,743,856
We repeat this procedure to obtain the remaining digits:
963,408 X 7 = 6,743,856
My Answer:
Change the multiplication problem into the equivalent division problem.
6,743,?56 / 7 = ???,4??
Then simply perform the division and use the 4 that is known in ???,4??
to produce the correct missing digits:
6,743,856 / 7 = 963,408
and equivalently
963,408 X 7 = 6,743,856
This MindBender was modified from a puzzle in Pierre Berloquin's book,
"100 Numerical Games."
August 30, 1999
MindBender
Two Dice
Timmy and Tammy play a game with two dice. But they do not use the
numbers. Some of the faces of the dice are painted red and the others
are painted blue. Each player throws the dice in turn. Timmy wins when
the two top faces are the same color. Tammy wins when the two top faces
are different colors. Their chances are even. The first die has five
red faces and one blue face. How many red faces and how many blue faces
are there on the second die?
___________________________
Mini-MindBender for Kids
Two Americans
There were two Americans waiting at the entrance to the British Museum.
One of them was the father of the other one's son. How could this be so?
___________________________
...Answer to MindBender
Two Dice
Berloquin's Answer:
Each die has six faces. When two dice are thrown, there are 36 equally
likely results. For chances to be even, there must be 18 ways of getting
the same color on top. Let X be the number of red faces on the second
die. We have:
18 = 5*X +1*(6-X)
X = 3
The second die must have three red faces and three blue faces.
My Answer:
Since the chances are even, the probability of matching colors must equal
the probability of different colors. Let R be the number of red faces on
the second die and B be the number of blue faces on the second die.
We have:
(5/6 * R/6) + (1/6 * B/6) = (5/6 * B/6) + (1/6 * R/6)
5R/36 + B/36 = 5B/36 + R/36
5R + B = 5B + R
4R = 4B
R = B
Since R+B = 6,
R = B = 3
This answer somehow just doesn't seem intuitive to me.
This MindBender was modified from a puzzle in Pierre Berloquin's book,
"100 Numerical Games."
___________________________
Answer to Mini-MindBender for Kids
Two Americans
They were husband and wife.
This MindBender was modified from a puzzle in Paul Sloane's book,
"Lateral Thinking Puzzlers."
September 7, 1999
MindBender
Word
Find a common English three-letter word, given that:
LEG has no common letter with it.
ERG has one common letter, not at the correct place.
SIR has one common letter, at the correct place.
SIC has one common letter, not at the correct place.
AIL has one common letter, not at the correct place.
Note: The first two givens are not really required.
___________________________
Mini-MindBender for Kids
Walking
Tim loves to walk. His constant speed is 6 kilometers per hour. Every day at
noon, he meets a friend at an inn about halfway between their homes. The
friend walks a little slower than Tim, 5.5 kilometers per hour. If both
walkers reach the inn exactly at noon, after about three hours of walking
each, how far are they from each other at 11:00?
___________________________
...Answer to MindBender
Word
Since the word contains no E or G (as in LEG), the only good letter in ERG
is R. Then the good letter of SIR is R, which is the third letter of the
desired word. I and L are not in the desired word, so the good letter of AIL is A.
The A is not the first letter of the desired word, so it must be the second.
The good letter of SIC is C, which must begin the desired word. The desired
word is CAR.
Without the first two givens: SIR has one common letter, at the correct
place. SIC has one common letter, not at the correct place.
The S and the I cancel each other (since there is only 1 letter in each
word and it can't be both in the correct and incorrect place at the
same time) meaning that the correct letters are R and C respectively.
The R is in the correct place so we have either C_R or _CR.
AIL has one common letter, not at the correct place.
It can't be the I due to SIR/SIC so it must be the A or the L. A word
always has a vowel so it's the A. The A can't be in the first spot so
it's CAR (besides even if the 'always has a vowel' weren't true in
this case neither CLR or LCR is a word).
This MindBender was modified from a puzzle in Pierre Berloquin's book,
"100 Games of Logic."
___________________________
Answer to Mini-MindBender for Kids
Walking
They are 6 + 5.5 or 11.5 kilometers from each other at 11:00.
This MindBender was modified from a puzzle in Pierre Berloquin's book,
"100 Numerical Games."
September 13, 1999
MindBender
Bikes/Hats
Three friends, Andrew, Bernard, and Claude, are bicycling. Each one
is riding the bicycle of one friend and wearing the hat of another.
The one who wears Claude's hat is riding Bernard's bicycle. Who is
riding each bicycle and who is wearing each hat?
___________________________
Mini-MindBender for Kids
Sequence
What is the next letter in the following sequence:
O T T F F S S E ?
___________________________
...Answer to MindBender
Bikes/Hats
The person riding Bernard's bicycle and wearing Claude's hat can't
be Bernard or Claude. Therefore, he is Andrew. Claude can't be riding
his own bicycle or Bernard's (Andrew is), so he is riding Andrew's
bicycle. Therefore, Bernard is riding Claude's bicycle. Similarly,
Bernard can't be wearing his own hat or Claude's (Andrew is), so he
is wearing Andrew's hat. Therefore, Claude is wearing Bernard's hat.
This MindBender was modified from a puzzle in Pierre Berloquin's book,
"100 Games of Logic."
___________________________
Answer to Mini-MindBender for Kids
Sequence
N. N is the first letter in the number Nine. The first eight letters
in the sequence are the first letters in the numbers One to Eight.
The MindBender moderator is the source for this Mini-MindBender.
September 20, 1999
MindBender
Three Questions
A history test had three questions on presidents of the United States.
Here are the answers of six students:
#1. Nixon, Nixon, Reagan.
#2. Reagan, Reagan, Nixon.
#3. Carter, Carter, Nixon.
#4. Reagan, Nixon, Carter.
#5. Carter, Reagan, Reagan.
#6. Reagan, Carter, Carter.
Every student answered at least one question correctly. What are the
correct answers?
___________________________
Mini-MindBender for Kids
Next Number
.. 02 .. 04 .. 07
06 ~~ 07 ~~ 03 ~~ 05
.. 11 ~~ 06 ~~ 01
..... 10 ~~ 04
??
What number belongs where the ?? is written?
___________________________
...Answer to MindBender
Three Questions
Note that the three correct answers need not all be different.
Can the first answer be Nixon? If so, only student #1 answered the
question correctly. Then student #3 and student #4 must each be right
on one of the last two questions, where the answers must be either
Carter or Nixon. But student #5 must be right on one of the last two
questions also, although he answered Reagan to both. This is
impossible, so the first answer is not Nixon.
Can the first answer be Carter? If so, by similar reasoning on the
answers of student #1 and student #2, the last two answers must be
either Nixon or Reagan. But student #6 answered Carter to both. Still
another impossibility. By elimination, the correct first answer is
Reagan. Then the second answer is Carter and the third answer is
Reagan again. This is to make students #1, #3, and #5 have at least
one correct answer.
Another solution approach is to list the 27 possibilities for the
correct answers to the three questions. Then eliminate any of the
27 possibilities that result in any student having no correct
answers. The one possible answer set that remains is:
Reagan, Carter, Reagan.
This MindBender was modified from a puzzle in Pierre Berloquin's book,
"100 Games of Logic."
___________________________
Answer to Mini-MindBender for Kids
Next Number
08. Each ~~ produces below itself the sum of the two horizontal numbers
next to the ~~ minus the number above the ~~.
This MindBender was modified from a puzzle in Pierre Berloquin's book,
"100 Games of Logic."
September 29, 1999
MindBender
4 Letter Word
Find a common English four-letter word, knowing that each of the
following four words have two letters in common with the desired
word: EGIS, PLUG, LOAM, and ANEW. In each case, the two letters in
common with the desired word are not at the correct places.
___________________________
Mini-MindBender for Kids
Pets
Heidi and Cathleen visited a pet store to look at animals. They
looked at 3 more rabbits than birds. They saw one half as many
birds as kittens. There were one third as many kittens as puppies.
They looked at 36 puppies. How many animals did they look at
altogether?
___________________________
...Answer to MindBender
4 Letter Word
The eight letters of EGIS and LOAM are all different. The four
letters of the desired word are among them. Then in PLUG, L and G
are the good letters; and in ANEW, A and E are the good letters. Or
alternatively, since each of four words have two letters of the
desired word, there must be some duplicates. The letters of the
desired word must occur in the four given words a combined total of
8 times. The four given words have four letters that are duplicated
twice each and 8 non-duplicated letters. Therefore, the four
duplicated letters must be the letters of the desired word. These
letters are: E, G, L, and A. The first letters of the four given
words include A, E, and L. So G must be the first letter in the
desired word. The third letters of the given words include A and E.
So L is the third letter. GELA is not a "common English word."
Therefore, the answer is GALE.
This MindBender was modified from a puzzle in Pierre Berloquin's book,
"100 Games of Logic."
___________________________
Answer to Mini-MindBender for Kids
Pets
Start at the back. 36 puppies. One third of 36 is 12, therefore 12
kittens. One half of 12 is 6, so 6 birds. 6 plus 3 is 9, so 9 rabbits.
Then 9 + 6 + 12 + 36 = 63 animals altogether.
This MindBender was modified from a puzzle in "The Problem Solver 4"
by Judy Goodnow and Shirley Hoogeboom.
October 13, 1999
MindBender
Brothers/Sisters
If LEAH is LOUIS's sister and
CLARISSE is BRUNO's sister and
MAUD is CHRISTOPHER's sister,
then who is HAMILTON's sister?
Is it IRENE, CLAIRE, SUE, or PEGGY?
___________________________
Mini-MindBender for Kids
Soccer Party
Several soccer teams are having a party. The team captains are
putting square tables tightly together in one long row for the party.
They can put two chairs on each side of a table. The tables are all
the same size. If they put together ten tables in a row, how many
people can sit down?
___________________________
...Answer to MindBender
Brothers/Sisters
SUE. The names of each brother/sister pair contain each of the five
vowels once.
This MindBender was modified from a puzzle in Pierre Berloquin's book,
"100 Games of Logic."
___________________________
Answer to Mini-MindBender for Kids
Soccer Party
44. Two on the opposite sides of each of the ten tables and two more
at the open end of the two end tables.
This MindBender was modified from a puzzle in "The Problem Solver 4"
by Judy Goodnow and Shirley Hoogeboom.
October 18, 1999
MindBender
Votes
In a small town hall, the five council members (Anthony, Bernard,
Claude, David, and Edwin) are about to elect one of them mayor. They
sit in alphabetical order clockwise around a round table. In the first
voting round, each council member votes for the one council member who
votes for his neighbor on the left. (The terms "each council member"
and "his" refer to the same council member.) Of course, no one is
elected. But who voted for whom?
___________________________
Mini-MindBender for Kids
Miles
Amy and Marcus had walked a total of 10.5 miles in their school
walk-a-thon. Amy had walked twice as far as Marcus. How many miles
did Amy and Marcus each walk?
___________________________
...Answer to MindBender
Votes
Let us consider Anthony's case. He can't vote for himself because
he would then need to vote for his left neighbor also.
Anthony can't vote for Bernard, because Bernard is Anthony's left
neighbor and that would mean that Anthony has to vote for himself
also, which is contradictory.
Anthony can't vote for Claude, because Claude would have to vote for
Anthony's left neighbor Bernard, Bernard for Anthony (who voted for
Claude, Bernard's left neighbor), leaving David and Edwin to vote
for each other, which is impossible.
Anthony can't vote for Edwin, because Edwin would have to vote for
Anthony's left neighbor Bernard, Bernard for Anthony (Edwin votes
for whoever votes for Edwin's left neighbor, Anthony), leaving Claude
and David to vote for each other, which is impossible.
Anthony can only vote for David, who votes for Bernard, who votes for
Edwin, who votes for Claude, who votes for Anthony.
This MindBender was modified from a puzzle in Pierre Berloquin's book,
"100 Games of Logic."
___________________________
Answer to Mini-MindBender for Kids
Miles
Amy - 7 miles. Marcus - 3.5 miles.
This MindBender was modified from a puzzle in "The Problem Solver 4"
by Judy Goodnow and Shirley Hoogeboom.
October 25, 1999
MindBender
Five Stores
Beth spent all her money in five stores. In each store, she spent $1
more than half of what she had when she came into that store. How
much did Beth have when she entered the first store?
___________________________
Mini-MindBender for Kids
ACE
How many ways can you read ACE off the following diagram? You can
move horizontally, vertically, or diagonally.
..... A
... A C A
. A C E C A
... A C A
..... A
___________________________
...Answer to MindBender
Five Stores
Beth entered each store with twice as much as the amount which is $1
more than she had when she left. Since she was broke when she left
the fifth store, she entered it with:
2 * (0 + 1) = $2
Likewise she entered:
the fourth store with 2 * (2 + 1) = $6
the third store with 2 * (6 + 1) = $14
the second store with 2 * (14 + 1) = $30
the first store with 2 * (30 + 1) = $62
Beth started with $62.
This MindBender was modified from a puzzle in Pierre Berloquin's book,
"100 Numerical Games."
___________________________
Answer to Mini-MindBender for Kids
ACE
Work backwards. Starting from the E, there are four Cs that can be
reached. From each of those Cs, there are 3 As that can be reached.
4 * 3 = 12. Thus ACE can be read off in 12 ways.
This MindBender was modified from a puzzle in Pierre Berloquin's book,
"100 Numerical Games."
Next Page
Other Pages
If you have any riddles or MindBenders to add please
e-mail me or
e-mail Mike Avery
PaintSaint helped make this page. check out his
Riddles page.
