Puzzle Solutions
Puzzle Solutions
1. Train.
A train one mile long travels at the rate of one mile a minute through a tunnel which is one mile
long. How long will it take the train to pass completely through the tunnel?
Solution. It will take 2 minutes. When the engine of the train reaches the end of the tunnel,
the rest of the train is still in the tunnel. Therefore it takes one minute for the engine to clear
the tunnel and another minute for the rest of the train to clear the tunnel.
2. Count the Bricks.
One side of a chimney on a flat roof looks
like the figure on the right. No bricks are
cut. The small squares represent the ends
of bricks. How many bricks are used in
making the entire chimney?
Solution. There are 70 bricks in the chimney.
One can certainly count the bricks by
brute force, but the easier solution is to
observe what the top of the chimney looks
like when looked at from above. See the
figure at right. The top layer has 10
bricks and there are 7 rows of bricks, so
there are 70 bricks total.
3. Next Number.
The following sequence of numbers has two possible replacements for the question mark. Can
you find them both?
17, 19, 23, 29, ?
Solution. The two values for ? would be 37 and 31. We can make two observations. The
first is that the 19 − 17 = 2, 23 − 19 = 4, and 29 − 23 = 6 and we see that if we add 8 to 29 we
would continue the pattern and ? = 37.
The second observation is that 17, 19, 23, 29 are not only prime numbers, but they are consecutive
prime numbers. Therefore, the next prime in line is 31 and ? = 31
4. The Chemist
A chemist had been working on discovering a liquid in which all substances will dissolve. One
day he announced he was able to make a full bottle of this liquid. What’s wrong with his claim?
Solution. He’s lying! If he truly discovered a liquid in which all substances will dissolve, nothing could hold it. His bottle would dissolve.
5. How Many Quarters?
Place a quarter on a flat surface. Place another quarter to the right and just touching the first
one. Keep going around the circle (see figure below), with each quarter touching the center one
and the one just placed. By the time you place the last possible quarter, how many quarters will
you have placed around the center one?
Solution. Six quarters are needed. Clearly, if you have enough quarters you can solve the
problem by placing the quarters around the center quarter.
But using mathematics is more fun. In the following figure, we see that the lines joining the
centers of the quarters form an equilateral triangle, since each side of the triangle is equal to the
diameter of a quarter.
Since the angles of an equilateral triangle are all equal to 60◦ and there are 360◦ in a circle, we
end up needing six triangles,
6. A Number Pyramid
Suppose the numerical pyramid below kept going. Without writing down the following rows,
what would the sum of the numbers in the tenth row be?
1
3
7
13
5
9
15
11
17
19
Solution. The tenth row would sum to 1000. If we sum the numbers in each row, we get:
1, 8, 27, 64. These numbers represent: 13 , 23 , 33 , 43 . It appears then that each row represents
the cube of its row number. Hence the tenth row should sum to 103 = 1000.
7. Secret Code.
Joey sent his friend Kevin a secret coded message using the standard code A = 1, B = 2, . . .,
and so on. However in one coded word he forgot to leave space between the numbers. What he
sent was 312125. Can you find what this word is? (Actually there are two possibilities.)
Solution. The words are CABLE and CLAY. If we separate the number as 3 1 2 1 2 5 and
let A = 1, B = 2, C = 3 and E = 5, we get CABABE. This is not a word. We need to look at
other possible separations. We can’t have 31 since there are only 26 letters in the alphabet, so
the 3 and 1 have to be separated. So the possible valid numbers are 3, 1, 2, 12, 21, 25 and 5, with
A = 1, B = 2, C = 3, E = 5, L = 12, U = 21 and Y = 25. We try the various combinations to
see that 3 1 2 12 5 gives CABLE and 3 12 1 25 gives CLAY.
8. The Bleachers.
When the bleachers at a high school basketball game were full they held 559 people. Assuming
each row held the same number of people, how many rows were there?
Solution. There are 13 rows of 43 seats. Clearly, we would not have one row of 559 seats, so
we see if we can factor 559. With a little effort we see that 559 = 13 × 43. Now there is no other
factoring for 559, so we either have 13 rows of 43 or 43 rows of 13. The latter seems unlikely for
a gym, so we choose 13 rows of 43 seats.
9. Divisibility.
What is the smallest base ten number containing only 0’s and 1’s that is divisible by 6?
Solution. The number is 1110. To be divisible by 6 the number has to be divisible by 3 and
by 2. To be divisible by 3 the sum of the digits must be divisible by 3, so we need three 1’s. To
be divisible by 2, the number must end in 0, 2, 4, 6, 8, so our number can only end in 0 and the
smallest number has to be 1110.
10. Buying a Hat and Coat.
If a hat and coat costs $110 and the coat costs $100 more than the hat, how much does the coat
cost?
Solution. The coat costs $105. h + c = $110 and c = h + $100. Hence 2h = $10 and h = $5,
so c = $5 + $100 = $105.
11. Back Where You Started.
Take a two-digit number and square it getting a three-digit number with distinct digits. Reverse
the three distinct digits of this square. Take its square root. Reverse the two digits of that
number, and you have the original two-digit number. What is that number?
Solution. The number is 13 or 31. 132 = 169. Reversing the digits gives 961 and the square
root of 961 is 31. Reversing these digits gives 13. Since we are told the square of the number is
three digits, we know its reverse must be a number less than 1000 and its square root less than
32. Therefore we need only check 11, 12 ,13, 21, 22, 31. 11, 12, 21, 22 are eliminated in that their
squares have repeated digits which are not allowed.
12. Three Player Game.
A game is played by 3 players in which the one who loses must double the amount of money that
each of the other 2 players has at the time. Each of the players lose one game at at the conclusion
of three games each man has $16. How much money did each man start with?
Solution. The players started with $26, $14 and $8. This problem can best be worked
backwards. We will determine how much money each player had before each game. Let the
players be designated as A, B and C. Further, let A lose the first game, B the second and C the
third.
A
B
C
After the third game
$16 $16 $16
Before the third game which C lost
$8 $8 $32
Before the second game which B lost $4 $28 $16
Before the first game which A lost
$26 $14 $8
13. Tree Stump.
An evergreen tree is in the shape of an equilateral triangle. As it happens, the shape of the tree
makes it possible to draw a circle around the tree as shown in the figure below. How tall is the
stump of the tree in relation to the entire tree?
Solution. The stump is 41 the height of the tree. Equilateral triangles have some interesting
properties. One that we will draw upon is that the altitudes, medians and angle bisectors are all
the same line. Since the tree is in the form of an equilateral triangle, if we draw the altitudes to
each side, we will divide the triangle into six equal right triangles. And since the altitudes are
also angle bisectors, the triangles are of the form 30◦ − 60◦ − 90◦ .
Now 30◦ − 60◦ − 90◦ right triangles have the property that the side opposite the 30◦ angle is onehalf the hypotenuse. Thus in triangle OBD, OD = 12 OB. Since OB is the radius of the circle,
OD = 21 r and then DG = 12 r. The stump is therefore 14 the length of the total height of the tree.
14. Calendar Dates.
The date July 14, 1998 has the interesting property that if you write the date in the numerical
form 7/14/98, it has the property that the month multiplied by the day is equal to the year.
During the 1950s, which years had no dates of this form?
Solution. The years ending in 53, 58 and 59 had no dates of this form. Since 53 and
59 are the only primes in the list from 50 to 59, those are eliminated. 58 = 2 × 29 so the only
possibility would be 2/29/58; however 1958 is not a leap year, so February cannot have 29 days.
15. Calendar Dates Again.
A set of three whole numbers {a, b, c} form what we call a Pythagorean triple if a2 + b2 = c2 . This
is a property of right triangles given in the Pythagorean Theorem; that is, the sum of the squares
of the two legs of a right triangle is equal to the square of the hypotenuse. The date March 4,
2005, written in numerical form 3/04/05 is a Pythagorean triple, since 32 + 42 = 52 . Can you
find some other Pythagorean triples in the 21st Century?
Solution. 6/8/10, 5/12/15 and 10/24/26 are some examples. It turns out that 10/24/26
is the last one possible.
16. The Striking Clock.
If it takes 7 seconds for a clock to strike 7, how long does it take to strike 10?
Solution. 10 21 seconds. When the clock strikes 7, it makes its first strike right at 7 and
then the remaining 6 strikes are evenly spaced over 7 seconds. Hence since these 6 strikes take 7
seconds, each strike takes 76 seconds, or 1 16 seconds. When striking 10, the clock will make 9 evenly
spaced strikes and we know each strike take
seconds.
7
6
seconds. Therefore, we have 9 ×
7
6
=
21
2
= 10 12
17. Family Matters.
How many brothers and sisters are there in a family in which each boy has as many sisters as
brothers but each of the girls has twice as many brothers as sisters?
Solution. 4 boys and 3 girls. The first statement tells us that there is one more boy than
there are girls. The second statement tells us there are an even number of boys. One could also
solve the problem by reason and test. Since b = g + 1, we could test for g = 1, g = 2, g = 3, . . .
until we found that g = 3 and b = 4 satisfied the conditions.
An algebraic solution would be as follows. The number of boys (b) must equal one more than the
number of girls (g). That is, b = g + 1. On the other hand, each girl sees twice as many brothers
as sisters. Since she has g − 1 sisters, we have b = 2(g − 1). Hence g + 1 = 2(g − 1) or g = 3 and
therefore b = 4.
18. Same Sum.
Arrange the numbers 1 through 9 in the square below so that each of the three columns has the
same sum.
Solution. There are many solutions. Here is an example.
Now if you can arrange the numbers such that all the rows and all the columns have the same
sum, you will have a magic square. Can you do that?
Solution. Again, there are more than one solution. Here is one example.
19. Follow the Dots.
Nine dots are arranged in a square formation in 3 rows of 3. Draw 4 straight lines, the second
beginning where the first ends, the third beginning where the second ends, and the fourth beginning where the third ends so that each dot is on at least one line.
Solution. You need to think outside the box!
20. The Gold Chain.
A man had no money but he had a gold chain which contained 23 links. His landlord agreed to
accept 1 link per day in payment for rent. The man, however, wanted to keep the chain as intact
as possible because he expected to receive a sum of money with which he would buy back what
he had given the landlord. Of course open links can be used as payment too and “change” can
be made with links already given to the landlord.
What is the smallest number of links which must be opened in order for the man to be able to
pay his rent each day for 23 days?
Solution. Two links, the 4th and the 11th, must be cut to pay the rent for 23 days.
If the 4th and 11th links are cut, he will then have two single links, one chain of 3 links, one
chain of 6 links and a chain of 12 links. Every whole number from 1 to 23 can be represented by
a combination of 1, 1, 3, 6, and 12.
Day
1
2
3
4
5
6
7
8
9
10
11
12
Payment
1
1+1
3
1+3
1+1+3
6
1+6
1+1+6
3+6
1+3+6
1+1+3+6
12
Day
13
14
15
16
17
18
19
20
21
22
23
Payment
1 + 12
1 + 1 + 12
3 + 12
1 + 3 + 12
1 + 1 + 3 + 12
6 + 12
1 + 6 + 12
1 + 1 + 6 + 12
3 + 6 + 12
1 + 3 + 6 + 12
1 + 1 + 3 + 6 + 12
21. Measuring.
A container holds exactly 8 quarts of liquid and is filled to the top. Two empty containers are
also available, one of which holds 5 quarts and the other 3 quarts. Without the use of any other
measuring devices, the liquid can be divided into equal parts of 4 quarts each simply by pouring
from one container to another. How is this done?
Solution. The following table shows how this can be done.
3qt
0
0
3
0
2
2
3
0
5qt
0
5
2
2
0
5
4
4
8qt
8
3
3
6
6
1
1
4
22. Rabbits and Chickens.
A man was asked how many rabbits and chickens he had in his yard. He replied, “Between
the two there are 60 eyes and 86 feet.” Although this reply was not exactly responsive, cn you
determine how many rabbits and chickens the man had?
Solution. 13 rabbits and 17 chickens. Since there are 60 eyes, we know there are 30 animals.
So there are r = 30 − c rabbits. Each rabbit has 4 legs and each chicken has 2 legs. Using 30 − c
for r, we have 86 = 4(30 − c) + 2c which gives c = 17 and therefore r = 13.
23. A Box of Marbles.
There are 4 black , 8 red and 5 white marbles in a box. How many marbles must one take out
(without looking at them) to be sure that there are at least 2 of one color among the marbles
selected?
Solution. Need to select 4 to be certain of getting a match. It is possible to draw 3
marbles without getting a match; however, the fourth marble drawn has to match one of the
others.
24. Making Money?
A man bought something for $60 and sold it for $70. Then he bought it back for $80 and resold
it for $90. How much profit, if any, did he make?
Solution. He made $20. He made a profit of $10 on the first sale and $10 on the second sale.