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.
June 7, 1999
MindBender
Hotel Bill
Here's a classic puzzle . You can tell it has been around for a long
time because of the dollar amounts involved. A hotel room at a hotel
with a bellboy would be much more than $10 now! Here it is:
Smith gave a hotel clerk $15 for his room for the night. When the
clerk discovered that he had overcharged Smith by $5, he sent a
bellboy to Smith's room with five $1 bills. The dishonest bellboy gave
only $3 to Smith, keeping the other $2 for himself. Smith has now paid
$12 for his room. The bellboy has acquired $2. This accounts for $14.
Where is the missing dollar?
___________________________
Mini-MindBender for Kids
Skateboard
One summer Osha kept a record of how many kilometers she rode on her
skateboard. She won't tell what the number is, but she will give you
these clues:
It is less than 100.
It is more than 44.
If you count by fours, you say the number's name.
The number can be divided evenly by 5 and 8.
How many kilometers did Osha go on her skateboard that summer?
___________________________
...Answer to MindBender
Hotel Bill
Adding the bellboy's $2 to the $12 Smith paid for his room produces a
meaningless sum, since the $2 is included twice. Smith is out $12, of
which the clerk has $10 and the bellboy $2. Smith got back $3, which
when added to the $12 held by the clerk and bellboy, accounts for the
full amount of $15.
This MindBender was modified from a puzzle in Martin Gardner's book,
"Mathematical Magic Show."
___________________________
Answer to Mini-MindBender for Kids
Skateboard
She went 80 kilometers.
The number must be divisible by 40 since it is divisible by 5 and 8 and
5 * 8 = 40. The only multiple of 40 between 44 and 100 is 80. You can
also do a chart of all "multiple of four" numbers between 44 and 100
and then eliminate numbers that are not divisible by 5 or 8. This also
results in the answer of 80.
This MindBender was modified from a puzzle in "The Problem Solver 4"
by Judy Goodnow and Shirley Hoogeboom.
June 14, 1999
MindBender
Palindromic Nora
A college girl has the unusual palindromic name (one that reads the
same forward and backward), NORA LIL ARON. Her boyfriend, a mathematics
major, bored one morning by a dull lecture, amuses himself by trying to
compose a good number cryptogram. He writes his girlfriend's name in
the form of a simple multiplication problem:
..NORA
.....L
.._____
..ARON
He wonders if it's possible to substitute one of the ten digits for
each letter and have a correct product. He is amazed to discover that
it is, and also that there is a unique solution. Neither four digit
number begins with zero. What is the solution?
___________________________
Mini-MindBender for Kids
Flowers
In a certain garden, each flower was either red, yellow, or blue, and
all three colors were represented. A statistician once visited the
garden and made the observation that whatever three flowers you picked,
at least one of them was bound to be red. A second statistician visited
the garden and made the observation that whatever three flowers you
picked, at least one of them was bound to be yellow.
Two logic students heard this and got into an argument. The first
student said: "It therefore follows that whatever three flowers you
pick, at least one is bound to be blue, doesn't it?" The second student
said: "Of course not!"
Which student was right and why?
___________________________
...Answer to MindBender
Palindromic Nora
..2178
.....4
.._____
..8712
Another answer is possible if Nora's middle initial is A:
..1089
.....9
.._____
..9801
2178 and 1089 are the only two four digit numbers with multiples that are
reversible.
There are many ways to arrive at the solution to this MindBender. The
following are some of my thoughts while determining Nora's numbers:
L must be > 1.
N must be < 5 or a 5th digit carry occurs.
If N=4, L must be < 3 or a 5th digit carry occurs. If L=2, A must = 8
or 9 (from the L*N plus a possible carry). But 2*8 or 2*9 (L*A) does
not end in 4.
If N=3, L must be < 4 or a 5th digit carry occurs. If L=2, L*A would
be even, not 3. If L=3, A=9 and 3*9 (L*A) ends in 7 not 3.
If N=2, L must be < 5 or a 5th digit carry occurs. If L=2, A=4 or 5,
but 2*4 or 2*5 (L*A) does not end in 2. If L=3, A=6 or 7 or 8, but 3*6
or 3*7 or 3*8 (L*A) does not end in 2. If L=4, A=8 or 9. If A=9, 4*9
ends in 6 not 2. If A=8, OR*4 + 3 = RO. Therefore, RO is odd, so O is
1, 3, 5, 7, or 9. If O>=3, 4*3 has a carry, so O=1. 4<=R<=7, since 4*1
plus a carry = R. The only product of 14, 15, 16, or 17 with 4 plus
the carry of 3 that ends in 1 is 4*17+3 and that works. Therefore, R=7.
Final answer is 2178*4=8712.
If N=1, the only products of two digits ending in 1 are 3*7, 7*3, and
9*9. If L=7, L*N is 7 or 8 or 9, not 3. If L=3, L*N is 3 or 4 or 5,
not 7. If L=9, A would also be 9, so not possible. But let's try
NORA*A=ARON with A=9. OR*9+8=RO. If O is >=2, A*O has a carry, so O is
< 2. If O=1, 9*10+8 is not = 01; 9*11+8 has a carry; and 9*12 or
9*(any number greater than 12) has a carry. Therefore O=0. 9*R+8 must
end in 0 so 9*R must end in 2, so R=8. The answer would then be
NORA*A=ARON is 1089*9=9801.
This MindBender was modified from a puzzle in Martin Gardner's book,
"Mathematical Magic Show."
___________________________
Answer to Mini-MindBender for Kids
Flowers
Smullyan's answer:
The first student was right, and here is why. From the first statistician's
report, it follows that there cannot be more than one yellow flower,
because if there were two yellows, you could pick two yellows and one blue,
thus having a group of three flowers that contained no red. This is
contrary to the report that every group of three is bound to contain at
least one red flower. Therefore, there cannot be more than one yellow
flower. Similarly, there cannot be more than one blue flower, because if
there were two blues, you could pick two blue flowers and one yellow and
again have a group of three that contained no red. And so from the first
statistician's report it follows that there is at most one yellow flower
and one blue. And it follows from the report of the second statistician
that there is at most one red flower, for if there were two reds, you could
pick two reds and one blue, thus obtaining a group of three that contained
no yellow. It also follows from the second report that there cannot be more
than one blue, although we have already deduced this from the first report.
The upshot of all this is that there are only three flowers in the entire
garden -- one red, one yellow, and one blue! And so it is of course true
that whatever three flowers you pick, one of them must be blue.
My answer (a little different way of saying the same thing):
There is at least one flower of each color, R, Y, and B.
If there were 2Y, 2Y and 1B conflict with the first report, so there is
only 1Y.
If there were 2B, 2B and 1Y conflict with the first report, so there is
only 1B.
If there were 2R, 2R and 1B conflict with the second report, so there is
only 1R.
So there are only three flowers, one of each color.
This MindBender was modified from a puzzle in Raymond Smullyan's book,
"To Mock A Mockingbird."
June 21, 1999
MindBender
Cryptogram
This problem is best viewed in a mono-space font like Courier.
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: C equals E
MTHE ECSAIZ-DWMC SAKLAECD'E KGEHSWU AGMLGM XWE TWDZUN IAMCXADMTN.
I thought this was a very difficult Cryptogram so two more clues, if
desired, are below.
For two more clues, highlight the first two arrows. (but first try to solve without any extra clues).
Second Cryptogram Clue: ...D equals R
Third Cryptogram Clue: ...A equals O
___________________________
Mini-MindBender for Kids
What Is It
Y X W V U T S R Q P O N M L K
It's a current problem. What is it?
___________________________
...Answer to MindBender
Cryptogram
THIS SECOND-RATE COMPOSER'S MUSICAL OUTPUT WAS HARDLY NOTEWORTHY.
This Cryptogram came from the Minneapolis Star Tribune newspaper.
___________________________
Answer to Mini-MindBender for Kids
What Is It
Y2K
The MindBender moderator is the source for this "What Is It" problem.
June 28, 1999
MindBender
Biscuits
I once set out a bowl of biscuits for my four dogs. First, the oldest
one came by and ate half of them and one more. Then, the second dog
came by and ate half of what he found and one more. Then, the third
one came by and ate half of what he found and one more. Then, the
fourth and littlest one came by and ate half of what he found and one
more, and the biscuits were then all gone. How many biscuits were in
the bowl at the start?
___________________________
Mini-MindBender for Kids
Coins
Suppose you and I have the same number of coins. How many must I give
you so that you have ten more than I?
___________________________
...Answer to MindBender
Biscuits
This problem is most easily solved by working it backwards. How many
biscuits did the last dog find in order that by eating half of them
and one more, all were gone? The only possibility is two. The dog
before that must have found six. The dog before that must have found
fourteen. The first dog must have found thirty.
This problem is also easily solved by simple algebra. Let W be the
number of biscuits the last dog found; X be the number of biscuits the
third dog found; Y be the number of biscuits the second dog found; Z
be the number of biscuits the first dog found. W-W/2-1=0 results in
W=2. X-X/2-1=2 results in X=6. Y-Y/2-1=6 results in Y=14. Z-Z/2-1=14
results in Z=30.
This MindBender was modified from a puzzle in Raymond Smullyan's book,
"Satan, Cantor, and Infinity."
___________________________
Answer to Mini-MindBender for Kids
Coins
The answer is five, not ten.
This problem is also easily solved by simple algebra. Let X be the
number of coins that we each have at the start; Y be the number of
coins that I give to you. (X+Y)-(X-Y)=10 results in Y=5. X can be any
number as long as X>=Y or X>=5.
This MindBender was modified from a puzzle in Raymond Smullyan's book,
"Satan, Cantor, and Infinity."
July 5, 1999
MindBender
Cryptarithmetic
This problem is best viewed in a mono-space font like Courier.
This is a Cryptarithmetic puzzle. It represents a long division problem,
where each missing digit has been replaced by an *. You are to
determine all the digits in the long division.
........**8**
....+--------
.** | *******
......***
......---
.........**
.........**
.........--
..........***
..........***
..........---
............1
___________________________
Mini-MindBender for Kids
Age
Tom has lived one-fourth of his life as a boy, one-fifth of his life as
a youth, one-third of his life as a man in his prime, and 13 years in
his advancing age. How old is Tom?
___________________________
...Answer to MindBender
Cryptarithmetic
This answer is best viewed in a mono-space font like Courier.
After the first product is subtracted and after the second product is
subtracted, it is necessary to bring down two digits from the dividend.
Hence, the quotient has two 0's: *080*.
The divisor times 8 is a two-digit number. The divisor can't be more
than 12, since 8 X 13 = 104. But if the divisor were less than 12, it
couldn't give a three-digit number (first product) when multiplied by a
digit. So the divisor is 12 exactly.
The divisor (12) only gives a three-digit number when multiplied by 9.
Therefore, the quotient begins with 9 and the dividend begins with
108 = 9 X 12.
Similarly, the last digit of the quotient is 9; the last product, like
the first, is 108 and is subtracted from 109 to give 1. The second
product is, of course, 96.
The total answer then becomes:
........90809
....+--------
.12 | 1089709
......108
......---
.........97
.........96
.........--
..........109
..........108
..........---
............1
This MindBender was modified from a puzzle in Pierre Berloquin's book,
"100 Numerical Games."
___________________________
Answer to Mini-MindBender for Kids
Age
Tom is 60. This can be found by solving the following algebraic
equation:
X/4 + X/5 + X/3 + 13 = X
This MindBender was modified from a puzzle in Raymond Smullyan's book,
"Satan, Cantor, and Infinity."
July 12, 1999
MindBender
Binary Cryptarithmetic
This problem is best viewed in a mono-space font like Courier.
This is a Cryptarithmetic puzzle. It represents a multiplication
problem, where each missing digit has been replaced by an *. You are
to determine all the digits in the multiplication. The problem is in
the binary system, where one is "1," two is "10," three is "11," and
so on. Note: In the binary system, 0 X 0 = 0, 0 X 1 = 0, and 1 X 1 = 1,
just as in our decimal system. Also 0 + 0 = 0, 0 + 1 = 1, and
1 + 1 = 10, that is, 0 and 1 to "carry."
.....****
......***
.________
.....****
...****
.________
..***1***
___________________________
Mini-MindBender for Kids
Siblings
A boy had as many brothers as sisters. His sister Grace had twice as
many brothers as sisters. How many brothers and sisters were there in
the family?
___________________________
...Answer to MindBender
Binary Cryptarithmetic
This answer is best viewed in a mono-space font like Courier.
On each line the leftmost digit must be 1 or it would not have been
given. The second intermediate product is two spaces to the left of the
first instead of one. Therefore, the multiplier has a 0 in the middle
and is 101. Since only 1 can be carried from the sum of two digits from
intermediate products, the second digit of the product must be 0, with
a 1 above it. To produce the 1 to be carried, the third digit of the
product must also be 0, with a 1 above it. In turn, the fourth digit
of the product (given as 1) must have two 1s above it so that 1 will
be carried. The fifth digit of the product must be 0 with two 1s above
it so that 1 will be carried. The answer is:
.....1111
......101
.________
.....1111
...1111
.________
..1001011
In the decimal system, 15 X 5 = 75.
This MindBender was modified from a puzzle in Pierre Berloquin's book,
"100 Numerical Games."
___________________________
Answer to Mini-MindBender for Kids
Siblings
There were four brothers and three sisters.
Simple algebra helps us again. Let X be the number of brothers (and
also sisters) that the boy had. Then his sister Grace has one more
brother (because of adding the boy) and one less sister (because of
subtracting herself). From this and the fact that Grace has twice as
many brothers as sisters, we have:
X + 1 = 2 * (X - 1) or
X + 1 = 2X - 2 or
3 = X
meaning that the boy had 3 brothers and three sisters. Counting the
boy, there were 4 boys and 3 girls in the family.
This MindBender was modified from a puzzle in Raymond Smullyan's book,
"Satan, Cantor, and Infinity."
July 19, 1999
MindBender
Mile Markers
Tim leaves town, driving at a constant speed. After a while, he passes
a mile marker displaying a two digit number. An hour later he passes a
mile marker displaying the same two digits, but in reversed order. In
another hour he passes a third mile marker, with the same two digits
(backward or forward, he won't say) separated by a zero.
What is the speed of Tim's car in miles per hour?
___________________________
Mini-MindBender for Kids
Breakfast
Grandfather is a very hard-boiled customer. In fact, his eggs must be
boiled for exactly 15 minutes, no more, no less. One day he asks you to
prepare breakfast for him, and the only timepieces in the house are two
hourglasses. The larger hourglass takes 11 minutes for all the sand to
descend; the smaller, 7 minutes. What do you do? Grandfather is growing
impatient.
___________________________
...Answer to MindBender
Mile Markers
The number of miles on the first mile marker can be written as 10A+B.
The number of miles on the second mile marker is 10B+A. On the third
mile marker, the number can be either 100A+B or 100B+A. Since Tim's
speed is constant, the distances between the second and third mile
markers are equal. The maximum case would be 19 for the first marker
and 91 for the second. In this case, the final marker would be 91 plus
the distance from marker two to marker three which is the same as the
distance from marker one to marker two. That would make the final
marker 91+(91-19)=163. (This, however, doesn't work as a solution.)
This maximum case does show that the hundreds digit of the last marker
must be 1. Is it 100A or 100B that equals 100? It has to be 100A since
the number on the first marker (where A stands for tens) must be
smaller than the number on the second marker (where B stands for tens).
Thus, A = 1. Then, since the distance between the first and second
marker equals to the distance between the second and third markers, we
have:
(10B+A) - (10A+B) = (100A+B) - (10B+A) and since A=1 from above,
(10B+1) - (10+B) = (100+B) - (10B+1) or
10B+1-10-B = 100+B-10B-1 or
18B = 108 or B =6. The mile markers read 16 61 106 and Tim's speed is
45 miles per hour.
This MindBender was modified from a puzzle in Pierre Berloquin's book,
"100 Numerical Games."
___________________________
Answer to Mini-MindBender for Kids
Breakfast
You must begin by turning over both hourglasses. Turning over only one
can only return you to the initial situation. Now there are two simple
solutions:
1) After turning over both hourglasses, put the egg into the boiling
water right after the smaller one (7 minutes) empties. Let the larger
one run out (4 minutes) and turn it over (11 more minutes). After it
runs out again, take the egg out of the water. The egg will have been
boiled exactly 15 minutes (4 + 11), but this solution has one flaw: the
whole procedure takes 22 minutes, and your grandfather is both
impatient and a retired mathematician. You need an optimal solution.
2) Put the egg into the water at the instant you turn both hourglasses
over. Now, after 7 minutes the small hourglass is inverted, and there
are 4 minutes left in the large one. When the large one runs out four
minutes later, there are 3 minutes left in the top of the small one,
but more important, there are 4 minutes worth of sand in the bottom!
With a sigh of relief, you invert the small hourglass at this instant,
wait for it to run out, and present your grandfather with his breakfast.
This solution takes exactly 15 minutes, and you can do no better.
This MindBender was modified from a puzzle in "The Chicken from Minsk"
by Yuri B. Chernyak and Robert M. Rose.
July 26, 1999
MindBender
Fountain
You go to a fountain which delivers an unlimited amount of water. You
bring two empty containers, one of 7 liters, the other of 11 liters.
How many operations are required to fill one of the containers with
exactly 6 liters of water? In these container games, in each operation
one container must be completely filled or completely emptied.
___________________________
Mini-MindBender for Kids
Hotel Keys
You arrive at a hotel at the same time as nine other guests. You all
want rooms and luckily there are just ten rooms available. However,
the hotel clerk has made a mistake and mixed up the room keys. There
are no identifying marks on the keys, and the ten rooms are locked.
All ten of you are very tired. What is the maximum number of trials
(the worst case) required to sort out all of the keys? Notice that the
object is to sort out all the keys, not to open all the doors.
___________________________
...Answer to MindBender
Fountain
Ten operations are needed. The table below shows how many liters are
in each container after each operation. In the first operation, the
first container is filled. In the second and third, the second
container receives the water from the first, which is then refilled.
In the fourth and fifth operations, the second container receives as
much as possible from the first and is then emptied. Continue in this
way through all ten steps.
The first number is the amount of water in the 7 Liter container. The
second number is the amount of water in the 11 Liter container.
7 0
0 7
7 7
3 11
3 0
0 3
7 3
0 10
7 10
6 11
This MindBender was modified from a puzzle in Pierre Berloquin's book,
"100 Numerical Games."
___________________________
Answer to Mini-MindBender for Kids
Hotel Keys
45 trials. There is only one trick here: if you have n keys and you
know that one of them will open the door, you need try at worst n-1 of
these keys, since if the first n-1 keys fail, the remaining key is the
correct key. No more than nine trials are required to determine the
correct key for the first room, since if the first 9 fail, you know
the remaining key is the right one. Thus, the worst case is
9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 + 0 = 45.
This MindBender was modified from a puzzle in "The Chicken from Minsk"
by Yuri B. Chernyak and Robert M. Rose.
August 2, 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."
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