101 Word Problems/Solutions
Problem #1: Antifreeze
A quart bottle contains a mixture that is 1/3 antifreeze, 2/3 water. A half gallon bottle contains a
mixture that is 3/4 antifreeze, 1/4 water. The contents of the two bottles are poured into a gallon
jug. What is the ratio (one integer to another, in the form a:b) of antifreeze to water in the jug?
Solution #1: Antifreeze
11:7
Problem #2: Consecutive Birthdays
A woman and her grandson have the same birthday. For six consecutive birthdays, she is an
integral multiple of his age. How old is the grandmother at the sixth of these birthdays?
Solution #2: Consecutive Birthdays
66
Problem #3: Cube Numbers
Find all sets of four positive consecutive integers such that the sum of the cubes of the first three
is the cube of the fourth.
Solution #3: Cube Numbers
3,4,5,6
Problem #4: Don't Spend It All At Once
If I start with $10 and spend all but $3, how much do I have left?
Solution #4: Don't Spend It All At Once
$3. If I spend all but $3, I must have $3 left.
Problem #5: Hands of a Clock
How many times do the two hands of a clock point in the same direction between 6:00 am and
6:00 pm of a single day?
Solution #5: Hands of a Clock
11. It is tempting to give the "obvious" answer 12, but the true answer is 11 as will be found by
counting the individual occasions.
1. Between 6:30 am and 6:35 am.
2. Between 7:35 am and 7:40 am.
3. Between 8:40 am and 8:45 am.
4. Between 9:45 am and 9:50 am.
5. Between 10:50 am and 10:55 am.
6. At Noon.
7. Between 1:05 am and 1:10 am.
8. Between 2:10 am and 2:15 am.
9. Between 3:15 am and 3:20 am.
10. Between 4:20 am and 4:25 am.
11. Between 5:25 am and 5:30 am.
Problem #6: How Old Is My Daughter?
My daughter is twice as old as my son and half as old as I am. In twenty-two years my son will
be half my age. How old is my daughter?
Solution #6: How Old Is My Daughter?
Let us assume my daughter is age x. We are told my daughter is twice as old as my son, so that
my son must be age x/2. We are also told that I am twice as old as my daughter so my age is 2x.
In 22 years time my son will be (x/2 + 22) and I will be (2x + 22). Since he will be half of my
age at that time,
x/2 + 22 = 1/2 (2x+22)
Multiplying both sides by 2
x + 44 = 2x + 22
or
x = 22
My daughter is 22 years old.
Problem #7: The Striking Clock
I woke up one night and heard my clock strike "one." I was too tired to turn on the light to see
what time it was. As I lay there pondering, it occurred to me to speculate how long I would have
to lie awake in order to be sure what was the exact time. My clock strikes the hours and strikes
"one" each half hour. I fell asleep before I solved the problem, but can you work out what is the
longest time I would have to lie awake after hearing the strike "one," to be sure of the time?
Solution #7: The Striking Clock
The strike I heard was either one o'clock or some half hour. If, after another half hour I hear two
more strikes, I know what hour it is, but if I again hear one strike, I may have heard successively
12:30 am and 1 o'clock or 1 o'clock and 1:30 am. After another half hour I either hear 1 strike
(1:30 am) or two strikes (2 o'clock). I may, therefore, have to lie awake a whole hour before I can
be sure of the exact time.
Problem #8: Relations
1. Is it legal for a man to marry his widow's sister?
2. Brothers and sisters I have none, but this man's father is my father's son. Who is "this man?"
Solution #8: Relations
1. Only dead men have widows. So... no, a dead man cannot legally marry anyone.
2. Since I have no brothers or sisters, "my father's son" must be referring to me; I am "this man's
father." So, "this man" must be my son.
Problem #9: School Notes
If 6 boys fill 6 notebooks in 6 weeks and 4 girls fill 4 notebooks in 4 weeks, how many
notebooks will a class of 12 boys and 12 girls fill in 12 weeks?
Solution #9: School Notes
If 6 boys fill 6 notebooks in 6 weeks,
then 12 boys fill 12 notebooks in 6 weeks,
and 12 boys fill 24 notebooks in 12 weeks;
If 4 girls fill 4 notebooks in 4 weeks,
then 12 girls fill 12 notebooks in 4 weeks,
and 12 girls fill 36 notebooks in 12 weeks;
Hence 12 girls and 12 boys fill 24+36=60 notebooks in 12 weeks.
Problem #10: Simple Algebra
Given x = 1 and y = 1, we have
x=y
Multiplying each side by x
x2 = xy
Subtracting y2 from each side
x2 - y2 = xy - y2
Factoring each side
(x + y)(x - y) = y (x - y)
Dividing out the common term ((x - y) we have
x+y=y
Substituting the given values
1+1=1
Or
2=1
What is wrong with this proof?
Solution #10: Simple Algebra
There was nothing wrong up through line
(x + y)(x - y) = y (x - y)
If we substitute the values of x and y we have
(2)(0) = (1)(0)
or
0=0
However, when we divide both sides by (x - y), we break a fundamental rule of mathematics that
we cannot divide out by a fraction which is equal to zero.
Problem #11: A Dinner Party
We were given a formal dinner party for ten (including ourselves) which is a number I always
like because the host can sit at one end of the table and the hostess at the other, and still maintain
the correct alternate male and female around the table. My wife was trying to work out the
seating. "Tom and Jean have not been here to dinner before so they are the guests of honor. Tom
must sit on my right and Jean on your right, but I don't know how I want to seat the others.:
"Well," I said, "I would like Janet on my left. I have a soft spot for her." "You can have her,"
replied my wife, "but I will not have her husband Jack next to me; I think he should be next to
Mary Ann."
Since we do not place husbands and wives next to each other, this determined the seating of
everyone, including Howard's wife Lois, and Mary Ann's husband Bill. Can you work out the
seating arrangement?
Solution #11: A Dinner Party
Numbering the ten places around the table 1 to 10 as shown in the figure below, if I take the head
of the table (1), my wife will sit at (6), Jean will be at (10), Tom at (5), and Janet at (2).
Now my wife will not have Jack next to her so he must be at (9), because (3) would place him
next to his wife. Since Mary Ann is next to Jack, she must be at (8); then Howard must be at (7),
Bill at (3), and Lois at (4).
Problem #12: Dominoes on Board
A board, 9 inches by 7 inches, is marked out into 63 one-inch squares. To make the number of
squares even, the middle square is blocked out by placing a chessman on it. John attempts to
cover the board, other than the middle square, with dominoes which are 2 inches by 1 inch.
In the figure above he has 11 dominoes in place. Since 31 dominoes will be needed in all, more
than one set will be required. Can you completely cover the board, other than the middle square,
without having any domino stick out over the edge of the board, or can you prove it cannot be
done?
Solution #12: Dominoes on Board
The use of a chessman to block out the middle square gives the clue to the solution of this
problem. If the board is marked out as a chess board with alternate black and white squares, with
the middle square black, we find that there are 32 white squares and 30 black squares (not
counting the middle square). Now a domino however placed will cover one white square and one
black square and hence having placed 30 dominoes we shall always be left with 2 white squares
which cannot possibly be covered by a single domino. Hence, there is no possible solution to the
problem.
Problem #13: An Unusual Series
Can you find the fifth term in the following series?
77, 49, 36, 18, ...
Solution #13: An Unusual Series
Each term consists of the first digit of the preceding term multiplied by the second digit of the
same term. Thus,
49 = 7 x 7
36 = 4 x 9
18 = 3 x 6
Hence, the fifth term is 1 x 8 = 8.
Problem #14: Is the Rope Knotted?
A loop of rope is lying on the ground in the position show in the figure below: You are too far
away to see which section of the rope is above or below at each of the crossovers at A, B and C.
If we assume that it is equally likely that either section is on top at each crossover, what is the
probability that the rope is knotted?
Solution #14: Is the Rope Knotted?
We can assume without loss of completeness that at A the rope going from top left to bottom
right is on top. (If it is the other way a mirror solution is produced which does not alter the
probability.)
Then we have four equal possibilities. The segment BOC can be either on top or underneath at
each of the points B and C.
Segment BOC
At B
At C
The rope is
On top
On top
Not knotted
On Top
Under
Under
On top
Knotted
Under
Under
Not knotted
Not knotted
Hence the chance that the rope is knotted is 1/4.
Problem #15: The Two Rugs
A woman has two rugs of the same material, one 10 feet by 10 feet and one a long runner 8 feet
by 1 foot. How can she make a single continuous cut in the 10 foot by 10 foot rug, so as to form
two pieces which, when combined with the 8 foot by 1 foot rug can be stitched into a rug 9 feet
by 12 feet? The two rugs are of plain pattern and have a pile on one side only.
Solution #15: The Two Rugs
The 10 foot by 10 foot rug must be cut as shown below, each step being 2 feet long and 1 foot
broad.
This gives two odd-shaped pieces together, a gap will be left in the middle which can be exactly
filled by the 8 foot by 1 foot strip, which is shaded below.
Problem #16: Mexican Table Mats
My son gave me a set of table mats he bought in Mexico. They were rectangular and were made
of straw circles joined together in the form shown in the figure below:
All the circles on the edges of each mat were white and the inner circles red. I noted that there
were 20 white circles and 15 red circles.
I wondered if it were possible to make a mat in this form where the number of white circles
equaled the number of red circles. I found there were two possible solutions. Can you find them
and show that there are only two such solutions?
Solution #16: Mexican Table Mats
Let the number of circles along the top edge of the mat be x and the number of circles along the
side of the mat be y. Then the total number of circles in the mat is
xy
and the number of circles around the edge is
2x + 2y - 4
Since this must be one-half the total number of circles we have
xy = 4x + 4y - 8
xy - 4x - 4y + 16 = 16 - 8
and (x - 4) (y - 4) = 8
Since x and y must be integers, so must (x - 4) and (y - 4) and these must be factors of 8. The
only integer factors of 8 are (8 and 1) and (4 and 2), which give {x = 12, y = 5} and {x = 8, y =
6}.
Hence the mat must be 12 by 5 circles or 8 by 6 circles.
Problem #17: The Curious Sequence
What is the next letter in the following series?
OTTFFSSE
Solution #17: The Curious Sequence
This is on of those nasty catchy puzzles. The real solution is
OTTFFSSEN
being the initial letters of the counting numbers: One, Two, Three, Four, Five, Six, etc.
Problem #18: Colored Labels
Three intelligent women, Alice, Barb and Carol, sit down to try out a test in logical reasoning.
They are so arranged that each can see the color of a label which is either red or blue, attached to
the hats worn by the other two but no one of them can see the color of the label attached to her
own hat. They are told that at least one of the labels is red. If any one of them can logically
deduce the color of the label on her hat, she is to declare it. Carol decides to play this game with
her eyes closed, knowing that the other two women have their eyes open. After a little time
Carol, who has not seen the label on any of the hats declares her label is red. How can she
deduce this?
Solution #18: Colored Labels
Carol argues as follows:
1. If Alice sees two blue labels she will know that her label is red and will declare this and
since a little time has passed without this happening Betty and Carol cannot both have
blue labels.
2. Similarly, if Betty sees two blue labels he will know that his label is red and will declare
this, and hence, Alice and Carol cannot both have blue labels.
3. Carol can now deduce that if her label is blue, both Alice and Betty must have red labels.
Now Carol knows that both Alice and Betty are intelligent. If Betty and seen a red label
on Alice's hat and a blue label on Carol's hat, she would know her hat did not have a blue
label, because in this case Alice would have declared her label to be red.
4. Since a little time has passed and neither Alice nor Betty has declared she could deduce
the color of her label, Carol knows her label cannot be blue, and therefore must be red.
Problem #19: War in the Middle Ages
During a grim medieval battle, 85% of the warriors lost an ear, 80% lost an eye, 75% lost an
arm, and 70% lost a leg. What is the smallest percentages possible of combatants who lost all of
the above?
Solution #19: War in the Middle Ages
Since 85% of the warriors lost an ear and 80% an eye (a total of 165%), at least 65% of them
must have lost both an ear and an eye.
Since at least 65% of the warriors lost both an ear an eye, and 75% lost an arm (a total of 140%),
at least 40% of them must have lost an ear, an eye, and arm and a leg each.
Since at least 40% of the warriors lost an ear, an eye, and an arm and 70% lost a leg (a total of
110%), at least 10% of them must have lost an ear, an eye, and arm and a leg each.
(This problem was set by Lewis Carroll.)
Problem #20: How Many Are We?
How many are we?
You tell us the answer, knowing that the probability that at least two of us have birthdays on the
same day is less than half, but that this would not be the case were we one more in number.
Solution #20: How Many Are We?
Probability that two people have different birthdays: 364/365
Probability that three people have different birthdays: (364/365) x (363/365), etc.
For n people
P = (364/365) x (363/365) x ... x ([653-n+1]/365)
It can be checked that the product of these fractions becomes less than 1/2 when n goes from 22
to 23. We are therefore 22.
Problem #21: Jaws
A seaside resort along the Pacific Ocean is equipped with an electronic shark detection system
which sounds an alarm on the average of 1 day out of 30. There are 10 times more false alarms
than there are undetected sharks. We also know that only three out of four sharks are detected.
Given the above information, what is the percentage of "peaceful" resort days if that term is
defined as days unmarred by alarms or sharks?
Solution #21: Jaws
Let x be the percentage of a shark going undetected. We know that the probability of an alarm is
1/30. Hence the probability of having neither an undetected shark nor an alarm is
1 - (1/30 + x) = 29/30 - x
We have also been told that the probability of false alarm is 10x. Hence the probability of real
alarm is 1/30 = 10x. But this is equal to 3x since 3 times as many sharks are detected as go
undetected
1/30 - 10x = 3x
Hence x = 1/390
Thus the probability of a peaceful day is
29/30 - 1/390 = 0.964
Problem #22: Meteorology
It rains here one day out of three. Our local meteorologists, who are pessimistic by nature, are
wrong in their daily forecasts one time out of two when they should have predicted good weather
but wrong only one time out of five when they should have forecasted rain.
Each morning, Francine leaves home for the day. If she departs without an umbrella and it rains,
she is twice as annoyed as if it had been fair and she had taken her umbrella with her. "Would I
be wiser," she wonders, "to listen to the morning weather forecast and take my umbrella only if
rain is predicted? Or should I consistently carry it? Or should I never take it? What would you
advise Francine to do?
Solution #22: Meteorology
There are four possible situations:
1. Fair weather predicted and occurred: probability 1/2 * 2/3 = 1/3.
2. Fair weather predicted but rain occurred: probability 1/5 * 1/3 = 1/15.
3. Rain predicted and occurred: probability 4/5 * 1/3 = 4/15.
4. Rain predicted but fair weather occurred: probability 1/2 * 2/3 = 1/3.
Let i be the measure of the inconvenience for Francine to carry her umbrella around for one
whole fine day and 2i the measure of inconvenience of not having an umbrella when it rains. If
she systematically carries her umbrella every day, the inconvenience will amount to i on two
days out of three, hence an average inconvenience of 2i/3.
If she never takes her umbrella, the inconvenience will amount to 2i one day out of three, for an
average inconvenience again of 2i/3.
If she decides to follow the advice of the weather forecasters and takes her umbrella only when
rain is predicted why will suffer an inconvenience of i on 1 day out of 3, and ii on 1 day out of
15, which amounts to and average penalty of 7/i/15 (less than the two previous average
penalties). We recommend, therefore, that Francine follow the advice of the weather forecasters.
Problem #23: Amateur Doctor
Aspirin does wonders for my headache and helps my rheumatic knee but it gives me nausea and
upsets my stomach. Herbal medicine cures my nausea and stomach upset but gives me hip pain.
Antibiotics sooth my headache and nausea but irritate my stomach and knee and give me a stiff
neck. Cortisone helps my stiff neck and rheumatic knee but aggravates my hip condition. Hot
compresses work wonders for my upset stomach and stiff neck. I woke up today with a pounding
headache which prevents me from figuring out how best to doctor myself. What do you think I
should do?
Solution #23: Amateur Doctor
I advise you to take aspirin and antibiotics and to apply hot compresses to your neck. Aspirin
relieves the headache but gives you nausea and upsets your stomach; the antibiotics relieve the
nausea but add pains to your knee and give you a stiff neck. The knee is taken care of my the
aspirin. As far as the upset stomach and the stiff neck are concerned, the hot compresses will
take care of them.
Problem #24: Bar Flies
At the same instant that Pete was leaving Harry's Bar to go to the Shamrock Tavern, Johnny was
leaving the Shamrock Tavern to go to Harry's Bar. They walked at a constant speed. When they
met, Pete announced that he had covered 200 meters more than Johnny had. The latter, his mind
befuddled by alcohol, took this as a personal insult and started to beat up Pete, who returned the
punches. When the fighting stopped, the men embraced in tears, then each continued his original
journey, but at half his original speed, as both were slightly hurt. Pete thus arrived at the
Shamrock Tavern in 8 minutes while Johnny took 18 minutes to get to Harry's Bar. How far
apart are the two bars.
Solution #24: Bar Flies
Let x be the distance between the two bars.
Let d be the distance covered by Pete when he met Johnny.
Let V be Pete's speed and v be Johnny's speed before the fight.
The sum of the distances covered at the time they met is x.
d + (d-200) = x
(1)
Hence x = 2d - 200
When they met, each had walked for the same length of time:
d/V = (d - 200)v
(2)
After the fight, Pete walked for 8 minutes:
(d - 200)(V/2) = 8
Hence V = (d - 200)/4
and Johnny for 18 minutes:
d(v/2) = 18
(3)
Hence v = d/9
(4)
By substituting the values for V and v into equation (2), we have
4d/(d - 200) = 9(d - 200)/d
Hence
5d2 - 3600d + 360,000 = 0
Solving this quadratic in the classic way yields
d = 600 or 120 and x = 2d - 200 = 1000 or 40
However, d must be smaller than x, so only one solution is possible:
d = 600 and x = 1000
The distance between the two bars is exactly 1000 meters.
Problem #25: Subway Escalator
I am in the habit of walking up the subway escalator while it is running. I climb 20 steps at my
normal pace and it takes me 60 seconds to reach the top, whereas my wife climbs only 16 steps
and it takes her 72 seconds to reach the top. If the escalator broke down tomorrow, how many
steps would I have to climb?
Solution #25: Subway Escalator
Let x be the unknown number of steps. When I ride the escalator, I cover x - 20 steps in 60
seconds. When my wife uses the escalator, she covers x - 16 steps in 72 seconds. The escalator
therefore moves at a rate of 4 steps every 12 seconds or 20 steps in 60 seconds. Its total height is
therefore the sum of these 20 steps and of the additional 20 steps I climb: a total of 40 steps.
Problem #26: Lighthouse
How far is the horizon from the top of a 125.7-meter-high lighthouse? (The earth can be
considered spherical with a circumference of 40,000 km.)
Solution #26: Lighthouse
Let S be the top of the lighthouse. Let A be a point on the horizon.
Let O be the center of the earth. Triangle OAS is a right-angle triangle since SA is tangent to the
earth, which we consider to be a perfect sphere.
We thus have OS2 = OA2 + AS2
but OA = 40,000,000/(2pi) meters
and OS = OA + 125.7 = OA + 40pi/OS2 ~= OA*OA + 2 OA = 40pi
Consequently,
AS = sqrt(OS2 - OA2) ~= sqrt(2*40pi.40,000,000/2pi) = 40,000 m The horizon is therefore
approximately 40 km from the top of the lighthouse.
Problem #27: The Demonstration
If we set out by ranks of 10, we will be one short. We will also be one short if we set out by
ranks of 9, 8, 7, 6, 5, 4, 3, and even 2. Yet there are fewer than 5000 participants. How many are
we?
Solution #27: The Demonstration
Let x be the number of demonstrators.
(x + 1) must be a multiple of 2, 3, 4, ..., 9.
(x + 1) is therefore a multiple of the smallest common multiple of these numbers, that is, a
multiple of
2 x 2 x 3 x 3 x 5 x 7 = 2520.
Let x + 1 = k(2520), k being a whole number.
x = k(2520) - 1. Since there are fewer than 5000 demonstrators, k = 1. Hence x = 2519, the
number of marchers.
Problem #28: Year of Birth
Take away from your year of birth the sum of the four numerals that make it up. You will end of
with a number divisible by 9. Why is this?
Solution #28: Year of Birth
Let T, h, t and u be the four digits that make up your year of birth (Thousands, hundreds, tens,
and units).
Your hear of birth can be written:
1000T + 100h + 10t + u.
If you take away
T+h+t+u
you are left with
999T + 99h + 9t
which is divisible by 9.
Problem #29: Spanish Arithmetic
4 + 4 + 4 + 4 + 4 = 20
In Spanish, this can be written:
CUATRO
+ CUATRO
+ CUATRO
+ CUATRO
+ CUATRO
-------VEINTE
How are the 10 numerals represented by the 10 letters A, C, E, I, N, O, R, T, U and V for this
sum to come out right?
Solution #29: Spanish Arithmetic
170469
+ 170469
+ 170469
+ 170469
+ 170469
-------852345
(not yet completed)
Problem #30: A Story of 1's and 2's
Write, side by side, the numeral 1 an even number of times. Subtract from the number thus
formed the number obtained by writing, side by side, a series of 2s half the length of the first
number. You will always get a perfect square. For instance,
1111 - 22 = 1089 = 332
Can you say why this is?
Solution #30: A Story of 1's and 2's
11...1 - 22...2 = 11...1 11...1 - 2(11...1)
------ ------ ------ ------ -----2n times n times n times n times n times
= 11...1 00...0 - 11...1
------ ------ -----n times n times n times
= 11...1 x (100...0 - 1)
------ -----n times n times
= 11...1 x 99...9
------ -----n times n times
= 11...1 x 9 x 11...1
----------n times
n times
= 32 x 11...12
-----n times
=
33...32
-----n times
Example: 11 - 2 = 9 = 32.
Problem #31: How Many Children Were You?
How many children were you?
If you ask me such a question, I will reply that my mother dreamed of having at least 19 children but her
dream did not come true, that my sisters were three times as more numerous than my first cousins, and
that I had half as many brothers as I had sisters.
Solution #31: How Many Children Were You?
The number of my sisters must be a multiple of 3 and a multiple of 2. It must therefore be a multiple of
6.
Total number of children: [(multiple of 6) x (1 + 1/2)] + 1.
Since this number must be less than 19, the only possible solution for the multiple of 6 is 6 exactly. I
therefore have 6 sisters and 3 brothers. We are 10 children in all.
Problem #32: Brother and Sister
"Sister, you have as many brothers as you have sisters." "Brother, you have twice as many sisters as you
have brothers." Can you deduce from this conversation how many children there are in the family?
Solution #32: Brother and Sister
Let x be the number of boys and y the number of girls.
What the boy says to the girl can be written as x = y - 1.
The answer that the sister gives her brother can be written as:
y = 2(x - 1) or y = 2x - 2
Substituting x = y - 1 gives us y = 4 and hence x = 3.
There are therefore seven children in the family.
Problem #33: Ursula and the Cats
If you ask old Ursula how many cats she has at home, she answers sadly: "Four-fifths of my cats plus
four-fifths of a cat." How many cats does this add up to?
Solution #33: Ursula and the Cats
Let n be the number of cats.
We have n = 4/5 n + 4/5. Hence n = 4.
Ursula lives with four cats.
Problem #34: New Math
My son learned how to count in a base different from 10, so that, for instance, instead of writing 136, he
writes 253. In what base does he count?
Solution #34: New Math
Let b be the unknown base. When my son writes 253 in base b, this can be interpreted as
2b2 + 5b + 3 = 136
or
2b2 + 5b - 133 = 0
or
(b - 7)(2b + 19) = 0
Since b is a whole, positive number, the only possible solution is 7; the unknown base is 7.
Problem #35: Chinese Numbers
Think of a number between 1 and 26. Look at the following table of six squares, one square at a time.
+----------+----------+----------+
| 1 4 7| 2 5 8| 3 4 5|
| 10 13 16 | 11 14 17 | 12 13 14 |
| 19 22 25 | 20 23 26 | 21 22 23 |
+----------+----------+----------+
| 6 7 8 | 9 10 11 | 18 19 20 |
| 15 16 17 | 12 13 14 | 21 22 23 |
| 24 25 26 | 15 16 17 | 24 25 26 |
+----------+----------+----------+
Each time the number you have picked belongs to one of the squares, write down the number in the top
left-hand corner. Add together all these numbers.
For example, 16 is in the first, fourth, and fifth squares. If the first numbers of each of these squares are
added together, we get 1 + 6 + 9 = 16, which is the original number we picked. Can you explain this?
Solution #35: Chinese Numbers
Any number n equal to or smaller than 26 must certainly be smaller than 33 = 27. It can therefore be
broken down (in base 3) in the following manner:
n = (0 or 1 or 2) x 30 + (0 or 1 or 2) x 31 + (0 or 1 or 2) x 32
In other words
n = (0 or 1 or 2) + (0 or 3 or 6) + (0 or 9 or 18)
Thus all number that have a "1" in their makeup will be found in the first square, all number that have a
"2" in the second square, a "3" in the third square, a "6" in the fourth square, a "9" in the fifth, and an
"18" in the sixth.
The number n can therefore be found by adding the numbers at the top left-hand corner of the squares
where n appears.
Problem #36: The Smallest Possible
Which is the smallest number which, when divided by 2, 3, 4, 5 and 6 will give 1, 2, 3, 4 and 5 as
remainders, respectively?
Solution #36: The Smallest Possible
Let n be this unknown number. Since n divided by 2 leaves a remainder of 1, n + 1 must be divisible by 2.
Since n divided by 3 leaves a remainder of 2, n + 1 must be divisible by 3. Similarly, n + 1 must be
divisible by 4, 5, and 6.
The smallest common multiple of 1, 2, 3, 4, 5, and 6 is 60.
Therefore, n + 1 = 60. Hence n = 59.
Problem #37: For Those Under 16
Tell me in which columns your age appears and I will guess your age by adding the first number of the
corresponding columns. Is this magic? squares, one square at a time.
+----+----+----+----+
| 2 | 8 | 4 | 1 |
| 3 | 9 | 5 | 3 |
| 6 | 10 | 6 | 5 |
| 7 | 11 | 7 | 7 |
| 10 | 12 | 12 | 9 |
| 11 | 13 | 13 | 11 |
| 14 | 14 | 14 | 13 |
| 15 | 15 | 15 | 15 |
+----+----+----+----+
Solution #37: For Those Under 16
In is not a matter of magic but simply the characteristics of a number written in base 2. Any whole
number less than 16 can be written as
x020 + x121 + x222 + x323
where each xn is either a 0 or a 1.
When xn = 1, the corresponding age is in the column headed 2n.
When xn = 0, the corresponding age is not in the column headed 2n.
Example: If you are 13 years old, your age will appear in the last three columns because
13 = 1.20 + 1.21 + 0.22 + 1.23 = 8 + 4 + 0 + 1
Problem #38: Product
The product of four consecutive whole numbers is 3024. What are these numbers?
Solution #38: Product
3024 ends in neither a 5 nor a 0; therefore, none of the four numbers given can be divisible by 5 or by
10. If all four numbers were greater than 10, their product would exceed 10,000, which is not the case.
The four numbers must either be {1,2,3,4} or {6,7,8,9}. In the first case, the product is 24. The answer
can only be {6,7,8,9}, as can be easily verified.
Problem #39: Riverside Drive
Would you like to have dinner with me tonight? Don't go to the wrong house. I live in one of the 11
houses along Riverside Drive. When I am at home, facing the river, and I multiply the number of houses
on my left by the number of houses on my right, I get a number which is greater by 5 units than the
number that my neighbor to my left would get if he did the same thing. Where along the Drive do I live?
Solution #39: Riverside Drive
Let r be the number of houses to the right of my house and l the number of houses to the left of my
house when I look toward the river. We have the following simultaneous equations:
r + l = 10
rl - (r+1)(l-1) = 5
Hence, r + l = 10 and r - l = 4.
thus r = 7 and l = 3.
My house is therefore the fourth from the left when facing the river.
Problem #40: Summit Meeting
Two delegations are to meet at the top floor of a skyscraper whose elevators can hold up to nine people
at a time. The first delegation to arrive makes up a certain number of elevator loads, filling each one
except the last, which has space for five more people. The second delegation does the same, not using
one-third of the last elevator load.
At the start of the meeting, each member of each delegation shakes hands with each member of the
other delegation, and each time, a photo is taken. Knowing that the photographer was using films with
nine exposures on each, how many unexposed frames will he have left on his last film?
Solution #40: Summit Meeting
Let d1 be the number of the members in the first delegation and d2 the number of the members in the
second.
d1 = (multiple of 9) + 4
d2 = (multiple of 9) + 6
Number of photographs:
d1 x d2 = (multiple of 9) + (4 x 6)
= (multiple of 9) + (18 + 6)
= (multiple of 9) + 6
The photographer therefore exposed six frames on his last film.
There are three frames left.
Problem #41: A Bag Of Marbles
A group of children share marbles from a bag. The first child takes one marble and a tenth of the
remainder. The second child takes two marbles and a tenth of the remainder. The third child takes three
marbles and a tenth of the remainder. And so on until the last child takes whatever is left. Knowing that
all the children end up with the same number of marble, how many children were there and how many
marbles did each one get?
Solution #41: A Bag Of Marbles
Let n be the number of children, x each child's share of marbles, and N the total number of marbles. We
have N = nx.
First child's share: 1 + (N-1)/10 = x.
Last child's share: x.
Penultimate child's share: n - 1 + x/9 = (N/x) - 1 + x/9 = x.
Consider the expression for the first child's share: N = 10x - 9.
Substituting into the equation for the penultimate child's share gives us
(10x - 9)/x - 1 + x/9 = x
which can be written
8x2 - 81x + 81 = 0 or (x - 9)(8x - 9) = 0
The number of marbles taken by each child being a whole number, the only possible solution is x = 9;
hence N = 81 and n = 9.
There were nine children. Each got nine marbles.
Problem #42: The Five Numbers
Can you find five consecutive whole number that are all positive and such that the sum of the squares of
the two largest numbers is equal to the sum of the squares of the three smallest?
Solution #42: The Five Numbers
Let n be then number in the middle of the series. We have the relationship
(n + 1)2 + (n + 2)2 = n2 + (n - 1)2 + (n - 2)2
Expanding and simplifying gives us
n(n-12) = 0
The only acceptable solution is n = 12. The five consecutive whole numbers are
10, 11, 12, 13 and 14
Problem #43: 100!
How many zeros are there at the end of 100! ?
Solution #43: 100!
The number of zeros at the end of a number is equal to the number of times that number is a multiple
of 10. However, 5 and 2 are factors of 10. The number of zeros will therefore be equal to the smaller of
the following two numbers: number of times 2 appears as a factor and number of times 5 appears as a
factor in the breakdown of the original number into prime factors.
Here, of course, 2 appears as a factor more often than 5. Let us calculate the number of times 5 appears
as a factor: 100! has 20 numbers that are multiples of 5. Some of them (4 of them) are even multiples of
25: 25, 50, 75, and 100. In the breakdown of 100! into prime factors, one will find 5 raised to the power
24 (20 + 4 = 24). There are therefore 24 zeros at the end of 100!.
Problem #44: House of Cards
Do you know how to build card houses?
The first floor is easy:
(2 cards)
The second floor is easy, too:
(7 cards)
Here's the third:
(15 cards)
And so forth.
How many cards does it take to build a 47-story house?
Solution #44: House of Cards
Note first that in all houses of cards, every level has three cards more than the level immediately above
it.
The number of cards for a 47-story house is therefore
2
+ 2 + 3
+ 2 + 3.2
+ 2 + 3.3
+ ...
+ 2 + 3.(n-1)
+ ...
+ 2 + 3.(47-1)
---------------------47.2 + 46.47/2 x 3 = 3337
Problem #45: Lunch With Friends
Ten couples meet for lunch. After a cocktail they go into the dining room in a random order, one by one.
How many people must enter the dining room to assure that:
1. There is at least one married couple in the group?
2. There are at least two diners of the same sex?
Solution #45: Lunch With Friends
As soon as 11 people enter the dining room, there will necessarily be one married couple among them.
As soon as three people enter the room, there will necessarily be two diners of the same sex.
Problem #46: Thirty-two Cards
Alfred, Brian, Christopher and Damon play with a deck of 32 cards. Damon deals them out unequally,
then says: "If you want us to have the same number of cards, do exactly as I say. You, Alfred, divide half
of your cards between Brian and Christopher. Then, Brian, you do the same with Christopher and Alfred.
Finally, Christopher, you follow suit with Alfred and Brian." How did Damon distribute the cards?
Solution #46: Thirty-two Cards
At the end of the game each of the four players has 8 cards. Christopher, having just shared half his
cards between Brian and Alfred, must have had 16 cards and Alfred and Brian 4 each. But Brian had just
shared half his cards between Christopher and Alfred: before that share-out, Brian must have had 8,
Alfred 2, and Christopher 14. But this was just after Alfred shared half his cards between Brian and
Christopher.
Hence at the start Alfred had 4 cards, Brian 7, Christopher 13, and Damon (who was not involved in the
three various share-outs) 8.
Problem #47: Checker Pattern
How many turquoise squares will be required to build the twentieth figure in this pattern?
Solution #47: Checker Pattern
761.
One visual representation for the pattern is n2 + (n-1)2 (outer squares + inner squares)
Problem #48: Odd or Even
Does 124FIVE represent an odd number? How can you determine whether a number is odd by looking at
its base-five representation?
Solution #48: Odd or Even
Yes. The number 124FIVE is represented by the base-five pieces that follow. Each piece has an odd
number of units. The case of 124FIVE has an odd number of pieces, 7, so 124FIVE is odd. In general, if the
sum of the digits in base five is odd, the number is odd.
Problem #49: Five Times Five
The digits in this addition problem have been replaced by letters. Replace each letter with a different
digit to obtain a correct sum that is as large as possible. What is the value of "FIVE"?
FIVE
FIVE
FIVE
FIVE
+ FIVE
-----ISLE
Solution #49: Five Times Five
1970.
1. F must equal 1 because anything higher would make the sum greater than four digits.
2.
(not yet completed)
Problem #50: Largest Product
Use each digit 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 exactly once to form two five-digit numbers that when
multiplied produce the largest quantity.
Solution #50: Largest Product
96,420 and 87,531 with a product of 8,439,739,020.
A reasonable first guess is two numbers that begin 97... and 86..., however these do not actually give the
largest product.
Problem #51: Multiple Choice
Given that one and only one answer is correct, which of the following is true?
A. All of the below
B. None of the below
C. One of the above
D. All of the above
E. None of the above
F. None of the above
Solution #51: Multiple Choice
E. Consider the choices. Contradictions between answers, such as between C and D, eliminate A as a
possibility. B is false because if it were correct, then C would also be true. It follows that C is false
because both A and B are false. Similarly, it follows that D is false. Aha! E is true! Of course, that result
makes F false.
Problem #52: A Square Number
For what bases does the number 121BASE represent a square number?
Solution #52: A Square Number
For any base!
In base b, 121 = 1.b2 + 2.b + 1 = (b + 1)2.
Problem #53: Reverse Multiplication
Find all five-digit numbers that are reversed when multiplied by 4.
Solution #53: Reverse Multiplication
21,978 is the only such number. Suppose that ABCDE x 4 = EDCBA. Since 4A < 10 and A is even, A = 2.
Therefore, E = 8. Note that E x 4 must end in 2. It follows that B <= 2 because 23 x 4 > 89. In fact, B = 1
because the two-digit number B2 must be a multiple of 4.
Thus far we have found that 21CD8 x 4 = 8DC12. Observe that D >= 4 and that D8 x 4 ends in 12.
Therefore, D = 7. The only possible value of C is 9.
Problem #54: Odd Product
The product of three consecutive odd numbers is 357,627. What is the smallest of the three?
Solution #54: Odd Product
69. The cube root of 357,627 is slightly less than 71. It seems that 69 x 71 x 73 is a possible product. The
result can be verified. Note that 9 x 1 x 3 = 27, thus making the last digit a 7, which is a good visual
check.
Alternatively, solve a cubic equation by setting up a product of successive odd numbers, say, a-2, a, and
a+2. This setup produces a more workable equation than does 2n-1, 2n+1, and 2n+3.
Problem #55: Carnival Dice
A carnival game offers you the opportunity to bet $1 on a number from 1 through 6 on a single roll of
the two dice. If your number comes up on one die, you win $2 and keep th $1 you be. If it comes on
both dice, you win $5 and keep the $1 you bet. Only if the number does not appear on either die do you
lose your $1 bet. Does this game favor you or the carnival?
Solution #55: Carnival Dice
The game is fair and does not favor either you or the carnival. The chart shows the eleven winning and
twenty-five losing combinations for betting on the number 1. One outcome will win $5 for you, ten will
win $2, and twenty-five will lose $1. Since these outcome are all equally likely, your expected winnings
are:
($2 x 10) + ($5 x 1) - ($1 x 25) = 0
1
1
2
3
4
5
6
2
1,1
2,1
3,1
4,1
5,1
6,1
WIN
3
4
5
6
1,2 1,3 1,4 1,5 1,6--2,2 2,3 2,4 2,5 2,6 L
3,2 3,3 3,4 3,5 3,6 O
4,2 4,3 4,4 4,5 4,6 S
5,2 5,3 5,4 5,5 5,6 E
6,2 6,3 6,4 6,5 3,6 |
|---------LOSE------+
Problem #56: Sum Cubes
23 = 23 + 23 + 13 + 13 + 13 + 13 + 13 + 13 + 13
Only one other integer requires nine cubes of positive integers to represent its value as a sum of cubes.
This other integer is between 200 and 300. Can you find it?
Solution #56: Sum Cubes
239. The sum can be expressed as:
23 = 43 + 43 + 33 + 33 + 33 + 33. + 13 + 13 + 13
Problem #57: Tower of Squares
For this tower of nine squares, determine a line passing through point P that will split the area of the
nine squares into two equal parts.
Solution #57: Tower of Squares
PA has the area of 5 squares below it, and PB has the area of 3 squares below it. Thus the area of /\PAB
is equal to the area of 2 squares. We have to choose Q so that AQ:QB as 1:3 to get a line with the area of
4.5 squares below it.
Problem #58: Sum to 30
Find different ways to express 30 as the sum of two or more consecutive integers.
Solution #58: Sum to 30
The expressions 4 + 5 + 6 + 7 + 8, 6 + 7 + 8 + 9, 9 + 10 + 11, (-3) + (-2) + (-1) + ... + 6 + 7 + 8, (-5) + (-4) + (3) + ... + 7 + 8 + 9, and (-8) + (-7) + (-6) + ... + 9 + 10 + 11 are possibilities.
Three of the foregoing six sums consist of only positive integers. Any sum of consecutive positive
integers (a+1) + (a+2) + ... + (a+n) is unchanged by placing the sum (-a) + (-a+1) + ... + (a-1) + a in front of
it. Therefore, the three "positive only" sums ensure that three more possibilities are feasible.
Consider the factors of 30: 1 x 30, 2 x 15, 3 x 10, and 5 x 6. All odd factors greater than 1 produce
consecutive integral sums, as follows. Take 3 x 10, for example. Three consecutive integers with 10 in
the middle will add to 30, which produces 9 + 10 + 11.
Observe that 30 = 0.5 x 60. The consecutive integers 0 and 1 average 0.5. Another sum can be formed by
taking thirty pairs of integers each totaling 1. The sum is given by (-29) + (-28) + (-27) + ... + 28 + 29 + 30.
The result is actually the sum based on putting integers in front of 30, which comes from 30 x 1. Note
that 1.5 x 40, 2.5 x 24, and 7.5 x 8 correspond to some of the other foregoing sums.
Problem #59: The King's Fortune Telling Scale
The king has a fortune telling scale. For one penny the scale spits out a slip of paper with a fortune on it
along with the weight of the king or whatever is on the scale. The king also has five bags of gold that he
has given to five trusted friends for safe keeping. Rumor has it that one of his five trusted caretakers is
not to be trusted and has been asking a counterfeiter to make phony gold. All the king could find out
about the counterfeit gold was that one of the king's five bags is now filled with counterfeit gold and
that the weight of that bag is different from the others. He does not know whether it weighs more or
less than each of the others.
The king calls back all of his bags of gold. Using his fortune telling scale, the king wants to find out:
 which bag contains counterfeit gold, and
 how much it weighs, exactly.
Being a very economical king, he wants to find this out using the least number of pennies possible. His
wizard says it can be done with just 3 pennies. No one else can see how. Anyone can do it with 5
pennies. Some say they can do it with 4 pennies. Can you figure out how to do it with 3 pennies?
(Assume that the fortunes the scale spits out offer no clues to which bag contains the counterfeit gold.)
Solution #59: The King's Fortune Telling Scale
Five pennies:
Weigh each bag once.
Four pennies:
(not yet completed)
Three pennies:
Solution provided by Alan Fung.
Let's name the five bags of gold as A, B, C, D and E. One of them is odd and we don't know if the odd bag
is lighter or heavier.
The first two weights are like these:
A + B + C = 3X ----------------(weight #1)
A + D = 2Y ----------------(weight #2)
X and Y are average unit weight of the combination. That is, they are roughly the weight of each bag of
gold.
If X = Y, it means the weight ratio among A, B, C and D are the same. (You may check it by adjusting one
of them as lighter or heavier, X will not equal to Y). Because A, B, C and D are the same, E must be odd.
So the final weight will be to weigh E and that will be the answer.
If X < Y, it is because "B or C is lighter" or "A or D is heavier".
If X > Y, it is because "B or C is heavier" or "A or D is lighter". The final weight will be:
C + D + E = 3Z ----------------(weight #3)
The reasons behind picking C, D and E is because now we know E has to be normal, we need to pick E as
reference and two bags from the lighter and heavier possibilities. I tried to pick A but it wouldn't work
because I shouldn't pick the same thing with the same coefficient that appears in all three equations. By
the same token, C and B are interchangeable.
Now to make the solution easier to understand, I will start with the solution and work out the weight
comparisons.
If X < Y
If B is lighter, X < Z and Y = Z
If C is lighter, X = Z and Y > Z
If A is heavier, X > Z and Y > Z
If D is heavier, X < Z and Y > Z
If X > Y
If B is heavier, X > Z and Y = Z
If C is heavier, X = Z and Y < Z
If A is lighter, X < Z and Y < Z
If D is lighter, X > Z and Y < Z
Because all results from weights comparisons are unique, we covered all possible solutions.
For the second part of the question, to know how exactly does the odd bag weigh, we can use simple
substitution to work it out. For example, if we found B is odd, B = 3X - 2Y.
Problem #60: Knights of the Round Table
King Arthur liked to invite his knights over for parties around the round table. When the king had a gift
that he could only give to one knight, he had them play a game that went like this:
First, King Arthur numbered the chairs around the table. At the start, every chair was occupied by a
knight. (King Arthur himself did not sit at the table.) Then he stood behind the knight in chair 1 and said,
"You're In." Next, he moved to the knight in chair 2 and said, "You're Out," and that knight left his seat
and went off to stand at the side of the room to watch the rest of the game. Next he moved to the
knight in chair 3 and said, "You're In." Then he said, "You're Out" to the knight in the chair 4, and that
knight left his seat and went to the side of the room.
He continued around the table in this matter. When he came back around to the knight in chair 1, he
said either "You're In" or "You're Out," depending on what he had said to the last knight. (If the last
knight was "In," then the first knight was now "Out, " and vice versa.)
The king kept moving around and around the table, alternately saying, "You're In" or "You're Out" to the
knights that remained at the table. (If a chair was now empty, he just skipped it.) He continued until only
one knight was left sitting at the table. That knight was the winner.
If you were a knight, which chair number would you try to sit in at King Arthur's table?
The number of knights at the table was always changing, so you should come up with a general rule.
Solution #60: Knights of the Round Table
(not yet completed)
Problem #61: Photographers and Cannibals
Three National Geographic photographers and three cannibals are traveling together through a jungle
when they come to a river. The largest boat available can carry only two people at a time. The
photographers are safe only if on each side of the river there are equal numbers of photographers and
cannibals or there are more photographers than cannibals; otherwise, the photographers become
dinner. How can they all get across?
(Tokens such as pennies and nickels may be helpful in solving this problem.)
Solution #61: Photographers and Cannibals
A cannibal and a photographer cross.
The photographer returns.
Two cannibals cross.
One cannibal returns.
Two photographers cross.
One cannibal and one photographer return.
Two photographers cross.
One cannibal returns.
Two cannibals cross.
One cannibal returns.
Two cannibals cross.
Problem #62: Handshakes
Is the number of people in the world who have shaken hands with an odd number of people odd or
even?
(Dead people count.)
Solution #62: Handshakes
Even.
If you were to ask everyone in the world how many hands he or she has shaken, the total would be an
even number because each handshake would have been counted twice -- once each by the two people
who shook hands. A group of numbers whose sum is even cannot contain an odd number of odd
numbers.
Problem #63: Bookworm
A set of encyclopedias consists of volumes that have 1/8-inch covers and one inch of pages. The set is
arranged in order on a shelf from left to right. If a bookworm starts at the first page of Volume I and eats
its way through to the last page of Volume II, how far does it travel?
Solution #63: Bookworm
Only 1/4 inch -- the thickness of the two covers. Put a paperclip on the first page of one book, and
another on the last page of a second book. Put both books on a shelf together, the first on the left and
the second on the right, and you will see why this surprising answer is correct.
Problem #64: Match Game
Using six identical matches, make exactly four equilateral triangles.
Solution #64: Match Game
The solution is a three-dimensional tetrahedron.
Those who have trouble with this one assume they must solve it in only two dimensions. Once you
realize that the use of a third dimension is permissible, it's easy.
Problem #65: Where There's a Will
A father wishes to divide a square piece of land among his five sons. One son is his favorite, and he
wants to give him one quarter of the land, as shown:
How can he divide the remaining land into four plots of equal size and shape?
Solution #65: Where There's a Will
Problem #66: Rope Trick
Is it possible to pick up a piece of rope, one hand holding each end, and tie a knot in the rope without
letting go of either end?
Solution #66: Rope Trick
not completed
Problem #67: Alphabet Soup
Put the rest of the alphabet in its proper place:
A
B D
C
EF HI KLMN
G J
O
Solution #67: Alphabet Soup
A
B D
C
EF HI KLMN
T VWXYZ
G J
PQR U
O
S
The top row contains letters that are made up of straight lines; the middle row, letters formed by
combinations of straight and curved lines; the third row, letters formed only by curves.
Problem #68: All Wet
A recipe calls for 4 cups of water. You have only a 3-cup and a 5-cup container. How can you measure
out four cups of water?
Solution #68: All Wet
Here are two solutions:
I.
a. Fill the 3-cup container and pour it into the 5-cup container.
b. Fill the 3-cup container again and from it fill the 5-cup container, leaving 1 cup of water in the 3cup container.
c. Empty the 5-cup container and pour the remaining cup from the 3-cup container into it.
d. Now fill the 3-cup container and add it to the 1 cup that is already in the 5-cup container.
The result is the desired 4 cups.
II.
a. Fill the 5-cup container.
b. Fill up the 3-cup container using the water in the 5-cup container, leaving 2 cups in the 5-cup
container.
c. Empty the 3-cup container.
d. Pour the remaining 2 cups from the 5-cup container into the 3-cup container.
e. Fill the 5-cup container again.
f. Pour water from the 5-cup container into the 3-cup container (that already has 2 cups of water
in it) until the 3-cup container is full.
Now, 4 cups of water remain in the 5-cup container.
Problem #69: The Telltale Number
Write a ten-digit number so that the first digit tells how many zeros there are in the number, the second
how many ones, the third how many twos, and so forth.
Solution #69: The Telltale Number
6,210,001,000
Problem #70: Something Fishy
A fish weighs ten pounds plus half of its weight. How much does it weigh?
Solution #70: Something Fishy
Twenty pounds.
Here is an algebraic solution:
W = 10 + 1/2 W
W - 1/2 W = 10
1/2 W = 10
W = 20
Problem #71: The Baffling Bicycle
A bicycle climbs a certain hill at 10 miles per hour and returns at 20 miles per hour. What is its average
speed for the entire trip?
Solution #71: The Baffling Bicycle
The apparent answer -- 15 miles per hour -- is wrong. The correct answer is 13 1/3 miles per hour
because speed is determined by dividing the distance by time. Notice, incidentally, that the answer is
the same no matter how long the hill is.
Problem #72: Figure Eight
An eight-digit number contains two 1's, two 2's, two 3's, and two 4's. The 1's are separated by one digit,
the 2's by two digits, the 3's by three digits, and the 4's by four digits. What is the number?
Solution #72: Figure Eight
41,312,432
Problem #73: The Balancing Brick
A brick balances evenly with three quarters of a pound and three quarters of a brick. What does the
whole brick weigh?
Solution #73: The Balancing Brick
Three pounds. If the brick balances with three quarters of a brick plus three quarters of a pound, then
one quarter of a brick must weigh three quarters of a pound. Thus a whole brick weighs four times as
much, or three pounds.
Problem #74: The Scrambled Salesman
An egg salesman was asked how many eggs he had sold that day. He replied, "My first customer said, 'I'll
buy half your eggs and half an egg more.' My second and third said the same thing. When I had filled all
three orders I was sold out and I had not had to break a single egg all day."
Solution #74: The Scrambled Salesman
Seven. He sold four eggs to the first customer, two to the second, and one to the third. The problem is
solved most easily if you start with the last customer and work backwards.
Problem #75: Wrong Number
Thirteen per cent of the people in a certain town have unlisted phone numbers. You select three
hundred names at random from the phone book. What is the expected number of people who will have
unlisted numbers?
Solution #75: Wrong Number
None. An unlisted phone number doesn't appear in the phone book.
Problem #76: Magic Numbers
This is a good way to mystify a friend. Ask someone to pick a number, any number at all, but not to tell
you what it is. Have him silently multiply it by five, add five to the product, multiply the resulting number
by two, add two to it, and tell you the result. Instantly you are able to tell him the number he started
with.
What is the secret and why does it work?
Solution #76: Magic Numbers
The secret: To find the number selected, simply delete the last digit from the number given you. Then
subtract one from the remaining number. Reason:
(not yet completed)
Problem #77: Case of the Counterfeit Coins
You have ten stacks of ten silver dollars each. They are identical, except that one stack consists entirely
of counterfeit dollars. You know the weight of an authentic dollar, and you also know that a counterfeit
dollar weights one gram less. How many weighings are needed to reveal which stack is counterfeit?
Solution #77: Case of the Counterfeit Coins
Only one. Weigh one coin from the first stack, two coins from the second, and so forth. The number of
grams by which the total is light will correspond to the number of the counterfeit stack.
Problem #78: Strange Series
The following number is the only one of its kind. Can you figure out what is so special about it?
8,549,176,320
Solution #78: Strange Series
It is the only one that contains all the numerals in alphabetical order.
Problem #79: Boxed In
Three boxes contain two coins each. One contains two nickels, one contains two dimes, and one
contains a dime and a nickel. All three boxes are mislabeled. (Each contains the label of one of the other
two boxes.) If you are permitted to take out only one coin at a time, how many must you take out in
order to be able to label all three boxes correctly?
Solution #79: Boxed In
Only one. Take it from the box labeled "Dime and Nickel." Since you know all three boxes are
mislabeled, the box contains either two nickels or two dimes. Put the correct label on that box after
inspecting one of the coins from the box. Then simply switch the two remaining labels.
Problem #80: Weighty Question
Which weighs more, a pound of gold or a pound of lead? (You may think you have heard this one before,
but you probably haven't.)
Solution #80: Weighty Question
A pound of lead. Lead is weighed in the standard measure, in which 7,000 grains equal a pound. Gold is
measured in Troy weight, in which 5,7600 grains equal a pound.
Problem #81: Early Bird
A chauffeur always arrives at the train station at exactly five o'clock to pick up his boss and drive her
home. One day his boss arrives an hour early, starts walking home, and is picked up by the chauffeur on
the way out to the train station. They arrive at home twenty minutes earlier than usual. How long did
she walk before she met her chauffeur?
Solution #81: Early Bird
For fifty minutes. She saved the chauffeur ten minutes of traveling time each way and thus was picked
up at 4:50 pm rather than the usual time.
Problem #82: Take a Letter
What three different digits are represented by X, Y, and Z in this addition problem?
XZY
+XYZ
---YZX
Solution #82: Take a Letter
X=4, Y=9 and Z=5.
In the middle column, adding Y to Z gives a digit of Z. This means Y must be either a 0 or a 9. (It can be a
9 if the addition in the far right column causes carrying a 1 into the middle column.) Since Y is the first
digit in the number on the bottom, it cannot be a zero. So, Y=9.
The two X's in the left column plus a possible 1 carried from the middle column must add up to Y, which
is 9. Since 9 is odd, a 1 must have been carried over from the middle column, leaving X+X=8. So, X=4.
The value of Z can be determined from the right column: Since Y=9 and X=4, Z must be 5 because it is
the only digit that can be added to 9 that puts a 4 in the one's place (9+5=14).
Problem #83: Water and Milk
There are two glasses on the table, one containing water and the other one milk. They both contain
exactly the same amount by volume. If you take a teaspoon of water and mix it into the milk and then
take a teaspoonful from the milk glass and mix it with the water, both glasses become contaminated.
But which is the more contaminated? Does the water now contain more milk than the milk does water
or the other way round?
Solution #83: Water and Milk
They are both equally contaminated. The water contains exactly as much milk as the milk contains
water. The most elegant proof for this celebrated little puzzle is as follows: It does not matter how many
transfers are made between the glasses or whether the contents are stirred. Provided that the volumes
in the two glasses are equal, then any water in the water glass must be in the milk, there is nowhere else
it can be. The milk that it has replaced must be in the water glass. The water glass therefore contains as
much milk as the milk contains water.
Problem #84: Four Sheep
Farmer Giles has four sheep. One day, he notices that they are standing in such a way that they are all
the same distance away from each other. That is to say, the distance between any two of the four sheep
is the same. How can this be so?
Solution #84: Four Sheep
The sheep are standing on the four corner points of an equal-sided pyramid (a tetrahedron). Or to put it
another way, three are on the points of an equilateral triangle and the other is on a mound of earth in
the center.
Problem #85: Do Helmets Increase Head Injuries?
At the beginning of the first World War, the uniform of the British soldiers included a brown cloth cap.
They were not provided with metal helmets. As the war went on, the army authorities and the War
Office became alarmed at the high proportion of men suffering head injuries. They therefore decided to
replace the cloth headgear with metal helmets. From then on, all soldiers wore the metal helmets.
However, the War Office was amazed to discover that the incidence of head injuries then increased. It
can be assumed that the intensity of fighting was the same before and after this change. So why should
the recorded number of head injuries per battalion increase when men wore metal helmets rather than
cloth caps?
Solution #85: Do Helmets Increase Head Injuries?
The number of recorded head injuries increased, but the number of deaths decreased. Previously, if a
soldier had been hit on the head by a piece of shrapnel, it would have pierced his cap and probably
killed him. This would have been recorded as a death, not a head injury. After helmets were issued it
was more likely that a fragment of shrapnel would cause an injury rather than death. Thus, the
incidence of head injuries increased, while the incidence of deaths decreased.
Problem #86: Sons
I. Mrs. Jones has two children. At least one is a boy. What are the chances that both are boys?
II. Mrs. Brown has two children. The younger one is a boy. What are the chances that both are boys?
Solution #86: Sons
For two children, there are only four possible combinations:
Older Younger
-----------------A Girl Girl
B Girl Boy
C Boy Girl
D Boy Boy
Each of these combinations is equally likely (i.e. there is a one in four chance that any two-child family
will have one of the above combinations).
For Mrs. Jones, the possibilities are narrowed down to A, B or C and of these, only A means that both are
both boys. Therefore, the chance that both her children are boys is one in three.
For Mrs. Brown, A and C are the only possibilities. There is thus a one in two chance that both her
children are boys.
Problem #87: Dinner for Three
An ancient Arabic puzzle goes like this: A hunter met two shepherds, one of whom had three loaves of
bread and the other, five loaves. All the loaves were the same size. The three men agreed to share the
eight loaves equally between them. After they had eaten, the hunter gave the shepherds eight bronze
coins as payment for his meal. How should the two shepherds fairly divide this money?
Solution #87: Dinner for Three
The shepherd who had three loaves should get one coin and the shepherd who had five loaves should
get seven coins.
If there were eight loaves and three men, each man ate two and two-thirds loaves. So the first shepherd
gave the hunter one third of a loaf and the second shepherd gave the hunter two and one-third loaves.
The shepherd who gave one-third of a loaf should get one coin and the one who gave seven-thirds of a
loaf should get seven coins.
Problem #88: One Clock
In the days before watches were invented, clocks were valuable items. There was a man who had one
clock in his house. It kept good time, but one day he found that it had stopped. He had no idea what the
correct time was. He walked to the next valley to visit his friend who had a clock showing the right time.
He spent a little while chatting with his friend then he walked home. He did not know the exact length of
the journey before he started. How did he manage to set his one clock correctly on his return?
Solution #88: One Clock
He wound his clock and set it at some particular time before he left. He noted the exact time of his
arrival at and departure from his friend's house. He noted the time showing on his clock when he
returned. He walked at the same pace on the two journeys. The elapsed time on his clock is the duration
of the two journeys plus the length of his visit to his friend. Knowing the time he spent with his friend,
he subtracts this from the elapsed time on his clock and divides the result by two in order to calculate
the duration of the journey. He adds this to the exact time he left his friend's house in order to set his
clock at the correct time. For example, assume he set his clock at 12, arrived at his friend's at 6:30, and
left at 7:30. When he returned, his clock showed 4:00. Then his journey took one and one half hours and
the correct time is 9:00.
Problem #89: The Mongolian Postal Service
The Mongolian Postal Service has a strict rule stating that items sent through the post must not be more
than 1 meter long. Longer items must be sent by private carriers, and they are notorious for their
expense, inefficiency, and high rate of loss of goods.
Boris was desperate to send his valuable and ancient flute safely through the post. Unfortunately, it was
1.4 meters long and could not be disassembled as it was one long hollow piece of ebony. Eventually he
hit on a way to send it through the Mongolian Postal Service. What did Boris do?
Solution #89: The Mongolian Postal Service
Boris placed the flute diagonally in a suitcase that measured 1 meter by 1 meter. This suitcase was quite
acceptable to the postal officials its sides measured 1 meter. From corner to corner, it measures 1.414
meters -- the square root of two.
Incidentally, if his flute had measured 1.7 meters, he could have fitted it across the diagonal of a box
whose sides were 1 meter long. The diagonal of a cube is the square root of the three times its side -1.73 meters.
From a theoretical mathematical viewpoint, there is no reason why this process cannot be extended
indefinitely. If Boris could construct a four-dimensional box with 1-meter sides, then he could get a 2meter flute in it (square root of 4) and a 25-dimensional construction could contain a 5-meter flute while
still meeting the rules of having no side longer than 1 meter.
Problem #90: The Amorous Commuter
John Jones lives in Maidenhead. He has one girlfriend in Reading and another in Slough. He has no car
and therefore takes a train whenever he goes to see them.
Trains stopping in Maidenhead can go either east or west. If they are westbound, they will go to
Reading. If they are eastbound, they will go to Slough. There are an equal number of trains going in each
direction.
John likes his two girlfriends equally. Because he finds it hard to choose between them, he decides that
when he goes to the station, he will take the first arriving train, regardless of whether it is going east or
west. After he has done this for a month, he finds that he has visited the girlfriend in Slough 11 times as
often as he visited the girl in Reading. Assuming that he arrived an the Maidenhead station at random
times, why should the poor girl in Reading have received so little attention?
Solution #90: The Amorous Commuter
The Slough trains depart from Maiden head at five past every hour. The Reading Trains depart at ten
past the hour.
If John arrives at any time between ten past the hour and five past the next hour, then the first train to
arrive will be bound for Slough. He will catch the Reading train only if he happens to arrive between five
past and ten past the hour. It is therefore 11 times more likely that he will catch the Slough train than
the Reading train.
Problem #91: Short Roads
There are four main towns in Lateralia. We will call them A, B, C and D. They lie at the corners of a tenmile square. In order to improve communications between the towns, the Lateralian Department of
Transport decided to build a new road linking all four towns together. Because they had very little
money, it was decided that the new road system should be as short as possible and still allow access
from any one town to any other. The engineers came up with three designs shown below.
Number one uses 40 miles of road, number two uses 30 miles of road, and number three uses 28.3 miles
of road. The designers naturally recommend plan number three because it employed the smallest road
area and, therefore, cost the least. However, when they submitted their plan to the Minister of Finance,
he accused them of extravagance and quickly pointed out a better design that required even less total
road surface. What was his superior solution?
Solution #91: Short Roads
The shortest solution is shown. It represents some 27.3 miles of road in total and therefore saves a
precious mile in road-building expense compared to the two diagonals. Someone starting from A would
have a shorted journey to D but a longer one to B or C.
Problem #92: A Weighty Problem I
A shopkeeper wants to be able to dispense sugar in whole pounds ranging from one pound up to 40
pounds. He has a standard, equal-arm balance weigh scale. Being of an extremely economical outlook,
he wants to use the least possible number of weights to enable him to weigh any number of pounds
between 1 and 40. How many weights does he need and what are they?
Solution #92: A Weighty Problem I
If the weights can be placed in either of the scale pans, then you can solve the problem with weights of
1, 3, 9, and 27 pounds only. With that combination any weight from 1 to 40 pounds can be measured. If
the weights can be placed in one scale pan only, then you need the weights 1, 2, 4, 8, 16, and 32 pounds,
which enables you to measure any weight up to 63 pounds. This latter solution is really an example of
counting with a binary number system where any number can be expressed as the sum of 2 raised to
various powers. For example, 63 expressed in binary form is 111111.
Problem #93: A Weighty Problem II
You have an equal-arm balance scale and twelve solid balls. You are told that one of the balls has a
different weight from all the others, but you do not know whether it is lighter or heavier. You can weigh
the balls against each other in the scale balance. Can you find the odd ball and tell if it is lighter or
heavier in only three weighings?
Solution #93: A Weighty Problem II
Let us call the balls A, B, C, D, ..., L. Start by weighing four against four. If they balance, then weigh any of
the remaining three against any three of the good balls. If they balance then we know the odd one is the
remaining ball and we can identify whether it is heavier or lighter in the final weighing. If the three
against three do not balance then we take the three containing the odd ball and weigh an one against
another.
If the first weighing of four against four does not produce a balance, then the second weighing involves
three against three with balls switched between the two pans and a good ball introduced. So:
If
A+B+C+D>E+F+G+H
We try
A + B + E against C + F + J
If
A + B + E = C + F + J,
then we know that either D is heaver or G or H is lighter, so we weight G against H.
If
A + B + E > C + F + J,
then we know that either F is lighter or A or B is heavier, so we weight A against B.
If
A + B + E < C + F + J,
then we know that either E is lighter or C is heavier, so we weight either against a good ball (e.g. K
against E).
An alternative second weighing is A + B + E against C + D + F, which follows similar lines to the above.
Problem #94: Eight Years Old
A girl was eight years old on her first birthday. How could that be?
Solution #94: Eight Years Old
She was born on February 29, 1896. The year 1900 was not a leap year (only centuries divisible by 400
are leap years), so the next February 29 fell in 1904 when she was eight. She was twelve on her second
birthday.
Problem #95: Cover That Hole
A manhole is a hole which allows someone to gain access to the sewers or other pipes which are below
ground. Our local town council recently decided that all the town's manhole covers should be changed
from square to round ones. We are used to the town council making silly decisions, but this time they
were absolutely right. Why?
Solution #95: Cover That Hole
A square or rectangular manhole cover can fall down the hole, while a round manhole cover cannot. The
square cover will fit down the diagonal of the hole (unless the rim it sits on is very large) but no matter
how you turn a circle it never measures less than its diameter. So for safety and practicality all manhole
covers should be round.
Alex Freuman also notes: "In addition to the advantage that a circular manhole cover cannot fall into a
manhole, one can roll a circular manhole cover rather than lifting it."
Problem #96: Pond Problem
A man wishes to reach the island in the middle of an ornamental lake without getting wet. The island is
20 feet from each edge of the pond (see diagram) and he has two planks each 19 feet long. How does he
get across?
Solution #96: Pond Problem
Hey lays the planks as shown in this diagram:
Problem #97: River Problem I
A man came to a river carrying a fox, a duck, and a bag of corn. There was a boat in which he could ferry
of the three items across the river at any one time. He could not leave the fox alone with the duck, nor
the duck alone with the corn, so how did he get all three across?
Solution #97: River Problem I
First the man took the duck across, then he came back and took the fox over. He left the fox on the far
side of the river and returned with the duck. He then left the duck on the near side and took the corn
over. Then he returned and took the duck across. Pretty straightforward, eh?
Problem #98: River Problem II
This time the man reached the river with a fox, a duck, and a bag of corn, but this fox ate corn as well as
ducks! There was the same boat as before in which he could take only one of the three with him. He
could not leave the fox with either the corn or the duck, and of course, the duck would gladly eat the
corn if they were left together. How did he get all three across?
Solution #98: River Problem II
The man tied the duck to the back of the boat wit a rope. The duck swam along behind the boat as the
man ferried the fox and corn over in turn.
Problem #99: The Bicycles and the Fly
Two boys on bicycles 20 miles apart, began racing directly toward each other. The instance they started,
a fly on the handle of one bicycle started flying straight toward the other cyclist. As soon as it reached
the other handle bar it turned and started back. The fly went back and forth this way, from handle bar to
handle bar, until the two bicycles met.
If each bicycle had a constant speed of 10 miles per hour, and the fly flew at a constant speed of 15
miles per hour, how far did the fly fly?
Solution #99: The Bicycles and the Fly
Each bicycle travels at 10 miles per hour, so they will meet at the center of the 20-mile distance in
exactly one hour. The fly travels at 15 miles per hour, so at the end of the hour it will have gone 15
miles.
Problem #100: Where does the square go?
Paste a sheet of graph paper on a piece of cardboard. Draw the square shown in figure 1, then cut along
the lines to make five pieces. When you rearrange these same five pieces, in the manner shown in
Figure 2, a hole will appear in the center of the square!
The square in Figure 1 is made up of 49 smaller square. The square in Figure 2 has only 48 small squares.
Which small square has vanished and where did it go?
Solution #100: Where does the square go?
When the two largest pieces are stitched, each small square that is cut by the diagonal line becomes a
trifle higher than it is wide. This means that the large square is no longer a perfect square. In has
increased in height by an area that is exactly equal to the area of the hole.
Problem #101: Siblings
Todd and Kristina
(not yet completed)
Solution #101: Siblings
(not yet completed)