Puzzles Archive
This is a list of the previous puzzles that have been sent out by
E-mail.
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To see the answer, click and hold your mouse button just to the right
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the answer and make it visible.
November 2, 1999
MindBender
One True
Four men, one of whom was known to have committed a certain crime,
made the following statements when questioned by the police:
1. Archie: Dave did it.
2. Dave: Tony did it.
3. Gus: I didn't do it.
4. Tony: Dave lied when he said I did it.
If only one of these statements is true, who was the guilty man?
___________________________
Mini-MindBender for Kids
405 Fish
In a huge fish tank there are silver fish and blue fish. There are 405 fish
altogether, and there are 125 more silver fish than blue fish. How many
fish of each color are there?
___________________________
...Answer to MindBender
One True
One of statements 2 and 4 is true and the other is false. If statement 2
is true, then Tony is guilty, and statement 3 is also true which is not
possible. Therefore, statement 2 is false and statement 4 is true. Thus,
statements 1 and 3 are also false which means that Gus is guilty.
This MindBender was modified from a puzzle in C. R. Wylie Jr.'s book,
"101 Puzzles in Thought & Logic."
___________________________
Answer to Mini-MindBender for Kids
405 Fish
Subtract the 125 from 405. The result is 280. 280 is the combined equal
number of blue and silver fish (there are also 125 extra silver fish). Divide
280 by 2. The result is 140. 140 is the number of blue fish. Add 125 to
140. The result is 265. 265 is the number of silver fish.
This MindBender was modified from a puzzle in "The Problem Solver 4"
by Judy Goodnow and Shirley Hoogeboom.
November 8, 1999
MindBender
One False
A rerun of last week's MindBender. A small, significant change has
been made.
Four men, one of whom was known to have committed a certain crime,
made the following statements when questioned by the police:
1. Archie: Dave did it.
2. Dave: Tony did it.
3. Gus: I didn't do it.
4. Tony: Dave lied when he said I did it.
If only one of these statements is false, who was the guilty man?
___________________________
Mini-MindBender for Kids
Bank
In a certain bank, the positions of cashier, manager, and teller are
held by Brown, Jones, and Smith, but not necessarily in that order.
The teller, who is an only child, earns the least. Smith, who knows
Brown's sister, earns more than the manager. What position is held
by each person?
___________________________
...Answer to MindBender
One False
One of statements 2 and 4 is true and the other is false. If
statement 2 is true, then Tony is guilty, and statement 1 is also
false which is not possible. Therefore, statement 2 is false and
statement 4 is true. Thus, statements 1 and 3 are also true which
means that Dave is guilty.
This MindBender was modified from a puzzle in C. R. Wylie Jr.'s book,
"101 Puzzles in Thought & Logic."
___________________________
Answer to Mini-MindBender for Kids
Bank
The teller cannot be Brown, since Brown has a sister, while the
teller is an only child. And the teller cannot be Smith, since Smith
does not earn the least money. Therefore, the teller is Jones. Smith
earns more than the manager, so Smith is the cashier. Therefore,
Brown must be the manager.
Or, build and fill a table or matrix. Create a matrix that matches
the names with the positions . An entry in the matrix contains nXm or
nXmp for impossible, or an n!m for definite, where n indicates the
order of filling the matrix and m and p refer to statement numbers.
Anytime an n!m is entered in a matrix entry, all other empty "same
column" and all other empty "same row" entries in the matrix receive
(n+1)X5. Number the statements as follows:
1. The teller is an only child.
2. The teller earns the least.
3. Brown has a sister.
4. Smith earns more than the manager.
5. Anytime an n!m is entered in a matrix entry, all other empty "same
column" and all other empty "same row" entries in that matrix receive
(n+1)X5.
6. The only remaining choice for a row or column.
. . . . Cashier . . Manager . . Teller
Brown . . 7X5 . . . . 8!6 . .. . 1X31
Jones . . 4X5 . . . . 4X5 . . . . 3!6
Smith . . 6!6 . . . . 5X4 . .. . 2X42
This results in the same answer as above.
This MindBender was modified from a puzzle in C. R. Wylie Jr.'s book,
"101 Puzzles in Thought & Logic."
November 15, 1999
MindBender
Secret Ages
When Tom and Betty applied for their marriage license, the first thing
they were asked was their ages. With a natural reluctance to reveal so
important a secret, Betty said that they were both in their twenties,
and wasn't that close enough. The clerk insisted on more specific
information, however, so Tom added that they both had the same
birthday, and 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.
At this the clerk fainted, whereupon the young couple snatched up the
license, hurried off to the preacher's, and lived happily ever after.
When the clerk came to and realized that he would have to complete his
records some way or other, he began to do a little figuring. Before
long, he found out how old Tom and Betty were. Can you find out also?
___________________________
Mini-MindBender for Kids
Box It
Today's Mini-MindBender is not a puzzle at all, but a game. In the
following matrix, two players take turns connecting two dots to draw a
horizontal or vertical line. If you complete a 1X1 box, put your
initial in that box and draw another line. If you can complete a box,
you must complete as many boxes as possible. When the line you draw
does not form a box, it is the other player's turn. The winner is the
one who completes the most boxes.
................
................
................
................
___________________________
...Answer to MindBender
Secret Ages
Tom must be 20, 24, or 28 since he is in his twenties and his age is
a multiple of four. He must be older than Betty since he was twice as
old as she at some time in the past. Therefore, Tom cannot be 20 since
Betty would then not be in her twenties now. The difference in their
ages must be constant. If Tom is 24, Betty was 6 "when Tom was 3 times
..." If Tom is 28, Betty was 7 "when Tom was 3 times ..." When Betty
is 6, Tom must be 9, 12, or 15 to be older than Betty and three times
an earlier age of Betty. When Betty is 7, Tom must be 9, 12, 15, or 18
to be older than Betty and three times an earlier age of Betty. Next
determine Betty's age "when Tom was 2 times ..." Then fill in the
other ages according to the constant difference "when Tom was 3 times
..." Some of the possibilities are then impossible for the reasons
shown. The following shows the possibilities:
.. Now ........ When 3X ...... When 2X
Tom . Betty .. Tom . Betty .. Tom . Betty
24 .... 21 .... 9 ..... 6 .... 6 ..... 3 The only correct answer!
24 .... 18 ... 12 ..... 6 ... 10 ..... 4 Betty is not in her 20s.
10 is not 2X4.
24 .... 15 ... 15 ..... 6 ... 14 ..... 5 Betty is not in her 20s.
14 is not 2X5.
28 .... 26 .... 9 ..... 7 .... 5 ..... 3 5 is not 2X3.
28 .... 23 ... 12 ..... 7 .... 9 ..... 4 9 is not 2X4.
28 .... 20 ... 15 ..... 7 ... 13 ..... 5 13 is not 2X5.
28 .... 17 ... 18 ..... 7 ... 17 ..... 6 Betty is not in her 20s.
17 is not 2X6.
So Tom is 24 and Betty is 21.
This MindBender was modified from a puzzle in C. R. Wylie Jr.'s book,
"101 Puzzles in Thought & Logic."
___________________________
Answer to Mini-MindBender for Kids
Box It
Hints: As you take your turn, look for "free" lines. These are lines
that do not complete three sides of a 1X1 box. Look at each available
move and see if it is a free move. If you must complete three sides of
a box, choose your move to allow your opponent to complete as few
boxes as possible. The board will have smaller and larger areas that
are "connected." A "connected" area is one where any line drawn in
that area allows your opponent to complete the whole area. Make your
move in the smallest "connected" area and your opponent will have to
give you a larger "connected" area at the end of his/her turn. The
matrix can be any size you want it to be. The example was a 3X3 or 9
box matrix. The matrix does not need to be square. Matrices that have
an odd number of lines on each side will never result in a tie because
an odd number times an odd number is always odd, so there will be an
odd number of boxes. You can play this game anytime you want. All you
need is paper and pencil.
The MindBender moderator is the source for this Mini-MindBender.
November 19, 1999
Bonus MindBender
Today
Today is a very special day. It is the very last day of its kind in
your lifetime, no matter how old you are. Try to figure out what
makes 11/11/1999 so special.
___________________________
...Answer to Bonus MindBender
Today
Today, November 19, 1999, is the last day of your life where the
date is specified with all odd digits (11/11/1999). The next
possible such date is January 1, 3111 or November 11, 3111
depending on whether you write January 1, 3111 as 1/1/3111 or as
01/01/3111.
I don't know any source for this problem.
November 22, 1999
MindBender
Ham and Pork
When Adam, Bob, and Carl eat out, each orders either ham or pork.
1.) If Adam orders ham, Bob orders pork. 2.) Either Adam or Carl
orders ham, but not both. 3.) Bob and Carl do not both order pork.
Who could have ordered ham yesterday and pork today?
___________________________
Mini-MindBender for Kids
Coffee/Tea
A market analyst works for an agency with very high standards. The
agency boasts that an employee's first mistake is his last. The
analyst reports that out of 100 consumers interviewed, 78 drink
coffee; 71 drink tea; 48 drink both tea and coffee. Why was the
analyst fired?
___________________________
...Answer to MindBender
Ham and Pork
From 1.) and 2.), if Adam orders ham, Bob orders pork and Carl orders
pork. This contradicts 3.), therefore, Adam orders only pork. Then
from 2.), Carl orders only ham. So only Bob could order ham yesterday
and pork today.
This MindBender was modified from a puzzle in George J. Summer's book,
"Test Your Logic."
___________________________
Answer to Mini-MindBender for Kids
Coffee/Tea
Of the 78 who drink coffee, 30 (78-48) do not drink tea. Of the 71
who drink tea, 23 (71-48) do not drink coffee. The three
non-overlapping classes of consumers, coffee only (30), tea only
(23), and both (48) total 101. But only 100 consumers were
interviewed. This mistake cost the analyst his job.
This MindBender was modified from a puzzle in C. R. Wylie Jr.'s book,
"101 Puzzles in Thought & Logic."
November 29, 1999
MindBender
$1500
The cashier of a bank pays out $1500, using a certain number of $1
bills, ten times as many $5 bills, a certain number of $10 bills, and
twice as many $50 bills. How many bills of each kind does the cashier
pay out?
___________________________
Mini-MindBender for Kids
6 Numbers
What do the numbers 3, 7, 8, 40, 50, and 60 have in common that NO other
numbers do? Hint: Spell the numbers out.
___________________________
...Answer to MindBender
$1500
Let N be the number of $1 bills and T be the number of $10 bills. Then:
1500 = N + 50N + 10T + 100T
N must be a multiple of 10. If N is 30 or more, the sum is too large
since N + 50N is 1530. So N must be 0 or 10 or 20. If N is 0, no
solution exists since 1500 is not divisible by 110. If N is 20,
N + 50N is 1020. Then T must be 8 or larger to make the total a
multiple of 100, but if T is 8, the total is 1900 (>1500). If T is
larger than 8, the sum would be even bigger. So N is 10. Then the
equation becomes:
1500 = 10 + 500 + 110T
990 = 110T
9 = T
So the cashier paid out 10 $1 bills, 100 $5 bills, 9 $10 bills, and
18 $50 bills.
This MindBender was modified from a puzzle in Pierre Berloquin's book,
"100 Numerical Games."
___________________________
Answer to Mini-MindBender for Kids
6 Numbers
These are the only numbers that have 5 letters - three consonants and two vowels.
I saw this recently, but I don't recall where.
December 6, 1999
MindBender
Cows
Four black cows and three brown cows give as much milk in five days
as three black cows and five brown cows give in four days. Which
kind of cow is the better milk producer, black or brown?
___________________________
Mini-MindBender for Kids
Big Number
I am a big number. I have nine digits, all different. I have no zero
in me. My odd digits increase from left to right and my even digits
increase from right to left. No two odd digits are next to each other.
Neither are any two even digits next to each other. (This last
statement is not required.) What number am I?
___________________________
...Answer to MindBender
Cows
Let B be the amount of milk a black cow produces in one day. Let b
be the amount of milk a brown cow produces in one day. Then the
problem can be stated by the equation:
4*5B + 3*5b = 3*4B + 5*4b or
20B + 15b = 12B + 20b or
8B = 5b or
5b = 8B or
b = 8B/5 = 1.6B so
The brown cows are the better milk producers, by 60%.
This MindBender was modified from a puzzle in C. R. Wylie Jr.'s book,
"101 Puzzles in Thought & Logic."
___________________________
Answer to Mini-MindBender for Kids
Big Number
183,654,729
The MindBender moderator is the source for this Mini-MindBender.
December 13, 1999
MindBender
Bad Balance
Timothy's balance is wrong. One of the arms is longer than the other.
1 kilogram on the left pan balances 8 apples on the right pan.
1 kilogram on the right pan balances 2 apples on the left pan. If all
the apples have the same weight, what is that weight?
___________________________
Mini-MindBender for Kids
A Number
I am a number. My units digit is the number of digits in me. My tens
digit is half of my units digit. My hundreds digit is double my units
digit. My thousands digit is half of my tens digit. What number am I?
___________________________
...Answer to MindBender
Bad Balance
The weights that balance on pans with unequal arms are not equal, but
they are proportional: if 1 kilogram balances 8 apples, 2 kilograms will
balance 16 apples, and so on. Let A be the weight of an apple. Then:
1/(8A) = (2A)/1
16A*A = 1
4A = 1
A = 1/4 or 0.25 kilograms or 250 grams.
An alternate solution method:
The weights on the two sides of the balance are equal if a constant
multiplier factor, F, is applied to one side. Let A be the weight of
an apple. We have:
1*F = 8*A and
2*A*F = 1 so
using the first equation to substitute into the second, we have:
2*A*8*A = 1
16*A*A = 1
A*A = 1/16
A = 1/4 or 0.25 kilograms or 250 grams.
Solving for F instead produces equivalent results:
1*F = 8*A or F/4 = 2*A and
2*A*F = 1 or F*2*A = 1 so
using the first equation to substitute into the second, we have:
F*F/4 = 1
F*F = 4
F = 2 so
8 apples weigh 2 kilograms (F*1 kilogram) so
1 apple weighs 2/8 or 1/4 or 0.25 kilograms or 250 grams.
This MindBender was modified from a puzzle in Pierre Berloquin's book,
"100 Numerical Games."
___________________________
Answer to Mini-MindBender for Kids
A Number
1824.
The MindBender moderator is the source for this Mini-MindBender.
December 20, 1999
MindBender
Holiday Dinner
Timothy is having a holiday dinner with James. Timothy brings five
dishes and James brings three dishes. At the last minute another
friend, Robert, comes and eats with them. Robert pays $4 as his share.
If all the dishes have the same value, how should the $4 be divided
between Timothy and James?
___________________________
Mini-MindBender for Kids
Holiday Cryptogram
Today's MindBender is a Cryptogram, or as some call them, a Cryptoquip.
Each letter stands for another. If you think D=K, for example, it
would equal K throughout the puzzle.
Decode the following:
Cryptogram Clue: U equals M
RS: XPP RMZ UJAEOZAEZY NSPWZYN
J MSBZ RMXR KSH XPP ZAFSK RMZ OZNR SD MSPJEXKN CJRM XPP RMZ DHA, DXUJPK, XAE DYJZAEN!
DYSU: RMZ UJAEOZAEZY USEZYXRSY
___________________________
...Answer to MindBender
Holiday Dinner
Since Robert paid $4, the total cost of the meal must be 4*3 = $12.
Eight dishes have been eaten. Each one costs $12/8 = $1.50. Timothy
brought 5*1.50 = $7.50 worth of food. His share was $4, so he receives
$3.50. James brought 3*1.50 = $4.50 worth of food. His share is $4, so
he receives $0.50 or 50 cents.
This MindBender was modified from a puzzle in Pierre Berloquin's book,
"100 Numerical Games."
___________________________
Answer to Mini-MindBender for Kids
Holiday Cryptogram
This should have been a fairly easy cryptogram. The pattern RMXR
is very often the word THAT, especially when RMZ also occurs (THE).
The XPP pattern is often ALL. Single letter words (J) must be I or
A. And the pattern XAE where the first letter is A is often AND.
TO: ALL THE MINDBENDER SOLVERS
I HOPE THAT YOU ALL ENJOY THE BEST OF HOLIDAYS WITH ALL THE FUN, FAMILY, AND FRIENDS!
FROM: THE MINDBENDER MODERATOR
The MindBender moderator is the source for this Mini-MindBender.
December 27, 1999
MindBender
Bargain Shopping
Mrs. Davis, Mrs. Jones, and Mrs. Smith, whose first names are Dorothy,
Helen, and Mary, though not necessarily respectively, went shopping for
bargains after Christmas. While shopping, Mary spent twice as much as
Helen and Helen spent three times as much as Dorothy. If Mrs. Davis
spent $3.85 more than Mrs. Smith, what is each woman's full name and
how much did they each spend?
___________________________
Mini-MindBender for Kids
Santa's Socks
Santa is tired after a long night's work and his feet are cold. He
goes to look for a warm pair of socks. His sock drawer is a mess.
Even though Santa just wants warm feet, he does like to have his
socks match. In his drawer, he has twenty black socks, fifteen red
socks, thirty-two green socks, five purple socks, two brown socks and
five blue socks. Santa is in a hurry. The light is burned out in his
closet and he can't see the color of the socks. The socks are of the
same texture and material. Only the colors vary. How many socks must
he remove and take to the other room to look at in the light, to be
sure that he has at least one matching pair?
___________________________
...Answer to MindBender
Bargain Shopping
The first names, in order of amount spent, are Mary, Helen, Dorothy.
They spent, in order, 6X, 3X, and X. Mrs. Davis compared to Mrs.
Smith can either be Mary compared to Helen, Helen compared to Dorothy,
or Mary compared to Dorothy. Helen compared to Dorothy is not possible
since $3.85 is not divisible by 2 ($3.85 would be 3X - X = 2X). Mary
compared to Helen is not possible since $3.85 is not divisible by 3
($3.85 would be 6X - 3X = 3X). So Mary is compared to Dorothy. Mary is
Mary Davis. Helen is Helen Jones. Dorothy is Dorothy Smith. Dorothy
spent $0.77 (6X - X = 5X = $3.85 so X = $0.77). Helen spent 3 times
that or $2.31. Mary spent twice that or $4.62. Apparently, not many
bargains were available.
This MindBender was modified from a puzzle in C. R. Wylie Jr.'s book,
"101 Puzzles in Thought & Logic."
___________________________
Answer to Mini-MindBender for Kids
Santa's Socks
Santa needs to take at least seven socks out to the light so he can
see the colors and patterns. There are six different colors or
patterns and if he pulls out six or less, he could end up with all
different socks. However, if he pulls out seven socks, he will get at
least one matching pair.
This MindBender was modified from a puzzle on riddleaday.com. To
subscribe to riddleaday.com, simply send an empty email to:
subscribe@riddleaday.com
January 3, 2000
MindBender
Hard Cider
How can you measure one liter of hard apple cider out of a barrel, if
all you have is a 3-liter pitcher and a 5-liter pitcher?
___________________________
Mini-MindBender for Kids
Gloves
Gloria has six pairs of pink gloves and five pairs of yellow gloves
all mixed up in her drawer. In complete darkness, how many gloves
does she need to take from the drawer to be sure she gets one pair?
___________________________
...Answer to MindBender
Hard Cider
Fill the 3-liter pitcher. Pour its contents into the 5-liter pitcher.
Refill the 3-liter pitcher. Pour from it into the 5-liter pitcher to
fill the 5-liter pitcher. The 3-liter pitcher now contains 1-liter of
hard apple cider. Just don't drink it all at once.
This MindBender was modified from a puzzle in Robert M¸ller's book,
"The Great Book of Math Teasers."
___________________________
Answer to Mini-MindBender for Kids
Gloves
Twelve. She could draw all eleven left hand (or right hand) gloves
the first eleven picks, but then the next pick must make a pair. So
she needs to draw twelve gloves to be sure of getting a pair.
This MindBender was modified from a puzzle in Adam Hart-Davis' book,
"Amazing Math Puzzles."
January 10, 2000
MindBender
Difficult Cryptogram
Today's MindBender is a Cryptogram, or as some call them, a Cryptoquip.
Each letter stands for another. If you think Q=U, for example, it
would equal U throughout the puzzle. Decode the following:
Cryptogram Clue: Q equals U
LYRS KZCIQZ IKH BRP ZKCIPZ VCBQC LRYSIC AKUUPH VYZ UQRAIKUBC.
This is a very difficult Cryptogram. More clues may
be desirable so a second, third, and fourth clues, if desired, are
below. Don't use them if you don't have to.
Concentrate on the word VCBQC with Q=U and two
equal characters (C). Very few English words follow that pattern.
For three more clues, highlight the first three arrows. (but first try to
solve without any extra clues).
Second Cryptogram Clue: ...I equals H
Third Cryptogram Clue: ...Z equals R
Fourth Cryptogram Clue: ...V equals S
___________________________
Mini-MindBender for Kids
Shakes
Ricky has ten friends over to celebrate his birthday. Each person
shakes hands with every other person just once. How many hand
shakes took place?
___________________________
...Answer to MindBender
Difficult Cryptogram
KING ARTHUR HAD ONE RATHER STOUT KNIGHT CALLED SIR LUNCHALOT.
This Cryptogram came from the Minneapolis Star Tribune newspaper,
but many months ago.
___________________________
Answer to Mini-MindBender for Kids
Shakes
The first person shook hands with ten people. The second person shook
hands with nine additional people since we already counted the second
person shaking hands with the first. Similarly, the third person shook
hands with eight additional people. Continue like this until the tenth
person who has only one person left to shake hands with. Therefore:
10+9+8+7+6+5+4+3+2+1 = 55 hand shakes took place.
This total is also available as 10*11/2 = 55
This is an example of the general rule for the sum of all integers
from 1 to N:
Sum of all integers from 1 to N = N*(N+1)/2
We also could say that all 11 people at the party shook hands with 10
others (11*10 = 110) and since each hand shake takes two people, we
counted each shake twice. Therefore, 110/2 = 55 hand shakes took place.
This MindBender was modified from a puzzle in Robert M¸ller's book,
"The Great Book of Math Teasers."
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