palette of
problems
David Rock and Mary K. Porter
1. The pattern of letters ABBCCCDDDDEEEEE . . .
repeats continuously such that after the final letter Z,
ABBCCCDDDD . . . begins again. Exactly how many times
will the letter V have appeared after the pattern of letters
has been repeated six times?
2. The pattern of letters ABBCCCDDDDEEEEE . . . repeats
continuously such that after the final letter Z, the pattern
increases the occurrence of each letter by one, so that the
second iteration is AABBBCCCCDDDDDEEEEEE. . . . Exactly how many times will the letter V have appeared after
this new pattern of letters has completed five cycles?
3. Coins numbered 1−200 are placed in a hat. What is
the probability that a coin chosen at random from the hat
is a multiple of 6?
4. Coins numbered 1−200 are placed in a hat. What is
the probability that a coin chosen at random out of the hat
is a multiple of 6 or 9?
5. A box contains 20 red balls and 40 green balls. Four
balls are randomly selected from the box without replacement. What is the probability of selecting a red ball followed by a green ball, then another red ball followed by a
green ball?
416
Mathematics Teaching in the Middle School
●
6. Circle 1 contains no constructed diameters; circle 2 has
3 constructed diameters and circle 3 has 8 diameters. If
circle 4 were constructed, it would have 15 constructed
diameters. If this pattern were to continue, how many
diameters would circle 10 have? How many diameters
would circle 50 have? Develop an expression to calculate
the number of diameters in the nth circle.
7. If a + b = c and b + c = d and c + d = e and a = 7
and e = 5, find a + b + c + d + e.
8. Find the sum of the following expression:
1
1
1
1
+
+
+ ⋅⋅⋅ +
(1)(2) (2)(3) (3)(4)
(19)(20)
9. Ernesto drives from his home to the library at a speed of
60 miles per hour. If he drives back home along the same
route at a rate of 40 miles per hour, what is his average
speed for the round trip?
Vol. 14, No. 7, March 2009
Copyright © 2009 The National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved.
This material may not be copied or distributed electronically or in any other format without written permission from NCTM.
Prepared by David Rock, [email protected], Columbus State University, 4225 University Ave., Columbus, GA 31907, and Mary
K. Porter, [email protected], Saint Mary’s College, Notre Dame, IN 46556.
MTMS readers are encouraged to submit single problems or groups of problems by individuals, student groups, or mathematics clubs
to be considered for publication. Send to the “Palette” editor, David Rock, at [email protected]. MTMS is also interested in students’ creative solutions to these problems. Send to “The Thinking of Students” editor, Edward S. Mooney, at [email protected]. Both
problems and solutions will be credited. For additional problems, see the NCTM publication, Menu Collection: Problems Adapted from
“Mathematics Teaching in the Middle School” (stock number 726).
10. Suppose that Ernesto
drives from his home to the
library at a rate of 60 miles per
hour, but this time, his average speed for the round trip
is 50 miles per hour. At what
speed does he travel on his
trip home from the library?
Round your answer to the
nearest tenth.
13. Fred is closing a
bank account and
wants to distribute
the money among his
grandchildren by giving
each of them $20. To do this,
Fred needs an additional $62.
Instead, he gave each grandchild
$17 and had $7 left over. How many grandchildren does Fred have? How much money did he have in
the bank account?
11. Find all possible pairs of numbers a and b that satisfy
all three of these properties:
(a) a is positive.
(b) a = 900b.
(c) T
he product of the reciprocals of a and b is a perfect square between 2 and 10.
12. Loretta’s age now is twice John’s age five years ago.
In three years, the sum of John’s and Loretta’s ages will be
50. How old are Loretta and John today?
14. What is the smallest counting number n so that 375n
is a perfect square?
15. What is the smallest counting number n so that 375n
is a perfect cube?
16. What is the smallest counting number n so that 375n
is both a perfect cube and a multiple of 16?
(Solutions on pages 444–46)
Vol. 14, No. 7, March 2009
●
Mathematics Teaching in the Middle School
417
solutions to palette
(Continued from pages 416–17)
(Alternative approaches to those suggested here are encouraged.)
ANSWERS
1. 132
2. 120
3. 33/200, or 0.165
4. 11/50, or 0.22
5. 260/5133, or approximately 0.05
6. 99 diameters; 2499 diameters;
n2 – 1
7. 14
8. 19/20
9. 48 mph
10. Approximately 42.9 mph
11. a = 15, b = 1/60 or a = 10, b = 1/90
12. Loretta is 26, and John is 18.
13. 23 grandchildren; $398
14. 15
15. 9
16. 576
SOLUTIONS
1. The first letter, A, appears 1 × 6 =
6 times. The second letter, B, appears
2 × 6 = 12 times. The third letter, C,
appears 3 × 6 = 18 times. Since V is
the twenty-second letter, V appears
22 × 6 = 132 times.
2. During the first iteration of the pattern, the first letter, A, appears 1 time;
the second letter, B, appears 2 times;
the third letter, C, appears 3 times; and
so on. During the second iteration, the
occurrence of each letter increases by 1.
Therefore, A appears 2 times, B appears
3 times, and C appears 4 times. Since V
is the twenty-second letter, V appears 22
times during the first iteration, 23 times
during the second iteration, 24 times
during the third iteration, 25 times during the fourth iteration, and 26 times
during the fifth iteration for a total of
22 + 23 + 24 + 25 + 26 = 120 times.
3. We want to find the number of
multiples of 6 that are less than 200.
Every sixth counting number is divis444
ible by 6; 20 ÷ 6 = 33 1/3. This means
that there are 33 multiples of 6 less
than 200, with 6 × 33 = 198 being the
last. Therefore, you have 33 chances
out of 200 numbers to select a multiple of 6, or 33/200, or 0.165.
4. The goal is to identify the number of
multiples of 6 and 9 less than 200. Since
some numbers are both a multiple of 6
and 9, we do not want to count them
twice. There are 33 multiples of 6 less
than 200, as found in number 3 above.
Similarly, 200 ÷ 9 = 22 2/9, so there are
22 multiples of 9 less than 200, with 198
being the last. However, we only want
to consider the multiples of 9 that we
have not already counted as a multiple
of 6. Consider that 9 is the first multiple
of 9 and 18 the second, then every other
multiple of 9 is also a multiple of 6,
which means that we want to count 22
÷ 2 = 11 multiples of 9. Therefore, there
are 33 multiples of 6, and 11 additional
multiples of 9, for a total of 44 numbers
out of 200 that can be chosen. The
probability is 44/200, or 11/50, or 0.22.
5. The first ball selected must be a red
ball, which has a probability of 20/60.
This leaves 19 red balls, 40 green balls,
and 59 total balls. The probability of
selecting a green ball next is 40/59. This
leaves 19 red balls, 39 green balls, and 58
total balls. The probability of selecting
a red ball next will be 19/58, leaving 18
red balls, 39 green balls, and 57 total
balls. The probability of selecting a green
ball next is 39/57. Since this probability
is conditional, we multiply each probability. Therefore, the probability will be
592,800
 20   40   19   39 
 60   59   58   57  = 11, 703, 240
260
=
≈ .05.
5133
Mathematics Teaching in the Middle School
1
+
●
Vol.
1 14, No.
1 7, March
1 120091
+
= + +
(1)( 2) ( 2)( 3) ( 3)( 4 ) 2 6 12
6. The pattern of constructed diameters is 0, 3, 8, 15, . . . Using addition,
the pattern is add 3, add 5, add 7,
add 9, and so on. Each number is the
square of the circle number minus 1,
or n 2 – 1.
Circle No. 1
2 3 4 10
n −1
3 8 15 99 2499
2
0
50
7. If c + d = e and e = 5, then c + d = (5)
and c = 5 – d. Using b + c = d, we get
b + (5 – d) = d or b = 2d + 5. If a + b =
c and a = 7, we get (7) + b = c. Substituting b = 2d – 5, we get
7 + (2d −
­­ 5) = 5 −
­­­ d
2 + 2d = 5 − d
2d = 3 − d
3d = 3
d=1
If d = 1 and e = 5, then from c + d = e
we find that c + (1) = 5 or c = 4. From
a + b = c and a = 7 and c = 4, we find
that (7) + b = (4) and b = −3. Therefore, a = 7, b = −3, c = 4, d = 1, e = 5,
592,800
 20their
 39 
  40sum
  19
and
is 14.
 60   59   58   57  = 11, 703, 240
8. One strategy for solving260
this prob=
≈ .05.
lem is to try a simpler problem
5133 and
look for a pattern.
1
1
1
1 1 1
+
+
= + +
(1)( 2) ( 2)( 3) ( 3)( 4 ) 2 6 12
6 2 1
= + +
12 12 12
9 3
= =
12 4
Notice that1the factors
1 3 and
1 4 of the
1
+
+
+
denominator
in
the
final
term
make
(1)( 2) ( 2)( 3) ( 3)( 4 ) ( 4 )( 5)
up the numerator and denominator in
1 1 1 1
the answer. Trying another
= problem
+ + +
2 6 12 20
will verify this rule:
30 10 5 3
= + + +
60 60 60 60
48 4
= =
60 5
5133
3000
1 1 1
1
1
1
x=
= 42.85714...
= + +
+
+
50
x
60
(
+
)
=
120
70x
(11)( 2) 1( 2)(13) 1( 3)( 4 ) 2 6 12
+
= + +
50 x + 3000 = 120 x
6 2 1
( 3)( 4 ) 2 6 12
= + +
 13000
  1 = 701x
12 12 12
6 2 1
= .
= + +

 a   b  3000
ab
12 12 12 = 9 = 3
x=
= 42.85714...
12 4
9 3
70
= =
12 4
1
900 b ) b =
d +d
2d
(
1
1
1
1
1
1
1




=
4
+
+
+
d d
t1 + t 2
 a   b  = ab . 1
(11)( 2) ( 21)( 3) ( 3)( 4 ) ( 4 )( 5)
1
+
+
+
+
b2 =
1 1 1 1
d + d 60 2dx
( 2)( 3) ( 3)( 4 ) ( 4 )( 5)
4(900 )
= + + +
d + d = 22dd
t1 + t 2= = d d
1 1 1 1 2 6 12 20
1 1
1
d
t1 + t 2 xd d + 60
= + + +
900 b ) b b==
300 10 5 3
=
(
+x d
60
0   40   19
592=,80
+
 20
2  63912
+ + +
4 3600 60
x
=
6060
x 260
d x
30  57
10  511, 703
3 60
0   59   58
, 24060 60 60
1
2
d
=
= + + + 48 4
2d 60d
b 2 a= = 900(1/60) = 15, so
Since a = 900b,
60 60 602606=0 =
= = xd
41(900 )
+ )60d
+ 60
d ( xxd
.05. 5
2
a = 15 and900
b =b1/60.
60 x +60 x
48 4 = 5133 ≈60
=
60
x
60
x
= =
602xd
1
9 1
60 5
=
b=
2
d
=
1
60) x 
Therefore, the answer
to
Case 2: If 1/ab =2 9, then
1/9.
3600ab =60
=2dd ()x + 60
1
b =
= (2
1
1
1
1 1 11 + 1 + 1 +⋅⋅⋅+
d (xd+( x60+)60 ) 
Since
a
=
900b,
then
(900b)b
= 1/9.
9
900
(
)
+
+
=
+
+
(11)( 2) ( 2)( 3) 1 ( 3)( 4 )
(19)( 20 ) d + d60 x 2d
1
( 2) ( 2)( 3) 1 ( 3)(
60
x
+ 4 ) 2 + 6 12 +⋅⋅⋅+
1 1
120 x = 60 x 
1
(1)( 2) ( 2)( 36) ( 32)( 4 ) 1
(19)( 20 )
900 b 2b==
==
t1 (2
+2td2)  d 60 xd 
=
+
=d + d+ +2d
9 8100 90
x= +(2
260
d )d60
( x + 60 ) 
2d
12 =12 12 =
 d ( x +x60 ) 
1
2d 3d
120 xrate to2dbe 50
t1 +9t 22d3 d d
2d
d + d is 19/20.
b2 =
He wants his average
+
+
120
x
=
=
=
== = 60 40 120 120
9(900 )
+ 60must
2d 43d
miles per hour, =
sox we
xd find60the
d
t1 + t 2 d d
12
+
+
x
+
60
+
120 x 60 x 60 x
2d distance 1that
20 
1
1
9.60
Let 40
d represent
the
value of x by solving
120 =120
50 =
= ( 2d )
b=
=

from
x + 60
5d 
8100 90
2d
2d 1 travels
 120
 5dhis
1 =Ernesto
1 home to
= ( 2d+Because
) 1 +120
=
the+
library.
distance
=
rate
5d4 ) ( 4 )( 5)
120 x d ( x + 60 )
(1)( 2) 5d( 2)( 3) (3)(
x
50
=120120
× 120
time, d = (60 mph)(t
Since a = 900b, a = 900(1/90) = 10, so
2401), where t1 is
50
(
x
+
60
)
=
50 =x +x 60 60 x
1 1= 1 = 48
1 mph
the
time it =takes
a = 10 and b = 1/90.
+ for
+ 5his+ trip to the
240
 60 x 
50 x + 3000 = 120xx+ 60
2t =6 d/60.
12 Note
20 that d is
=library.=Thus,
48 mph
=
(2
2
d
)
or


5
 d ( x + 60 ) 
301 10 5 travels
3
x x
50( x3000
+ 60=) =70120
also the distance
from
12. Make a table where x represents
= + Ernesto
+ +
50( x + 60 ) = 120 x
60 If60
60
50 x + 3000 =3000
120 x= 120 x
the library to60
home.
t2 represents
John’s age five years ago.
x = = 120=x 42.85714...
50 x + 3000
48
4
x
+
60
the time it =takes=for his trip home
3000 = 70
70 x
3000 = 70 x
60 then
5 d = (40 mph)
from the library,
Five
Three
3000
= 42.85714...
(t2), so t2 = d/40. Therefore, the averYears
Now
Years
from
 1   1  xx==1 3000
=
42
.
85714
..
.
70
age rate for the
1 complete
1 round
1 trip
1  a   b  = ab . 70 120 x
Ago
Now
+
+
+⋅⋅⋅+
50 =
is calculated(1from
)( 2) average
( 2)( 3) rate
( 3)(=4 )(to- (19)( 20 )
Loretta 2x – 5 2x
2x + 3
x + 60
1
1
1




tal distance)/(total time) as follows:
On the 1trip
home
from
the library,
1
1



.
=
 a should
  ab . approximately
x
John
x+5
x+8
Ernesto
 ab )bbb= 1=travel
900
(
ab ) = 120 x
2d
2d
d +d
50
x
60
(
+
42.9 miles per hour.
4
=
=
2d 3d
t1 + t 2 d d
50
x
+
3000
= 120 x
Let x represent John’s age five years
+
+
2
11
=
b
60 40 120 120
1
=43000
11. Note
a )=is 70
positive
and a
ago. Now his age is x + 5, and in three
(9(90000bthat
)b )bb since
x
(
900
=4
2d
 120 
=
900b,
then
b
must
also
be
positive.
years it will be x + 8. Loretta’s age now
4
=
= ( 2d )
11x = 3000
1 = 42.85714...
2
5d
 5d 
The product
of
the
reciprocals
of
a
is twice John’s age five years ago, so her
=
bb =
1
2=
70
=
b
3600
60
4
(
900
)
120
and b is written as4(900 )
age now is 2x, and in three years it will
240
be 2x + 3. In three years, the sum of
1
1
=
= 48 mph
b1== 13600
 1 1==601
their ages will be (2x + 3) + (x + 8) =
5
2 b1
= .
900 b =   3600
 a 9  b  ab 60
50, or 3x + 11 = 50. Therefore, 3x = 39,
10. Let d, t1, and t2 be defined as in
and x = 13. Since x represents John’s
11
2
b 2b =
1
the solution to the previous problem,
The only
perfect
squares
that
are
age five years ago, Loretta is now 26
900
2=
) 1
9009b009=(b9900
b =22 = 4 and
( 109)are
and let x represent Ernesto’s rate on
between 2 and
years old, and John is 18 years old.
11 4 1
2
the trip home from the library. His
32 = 9, so webb must
have
1/ab
=
4,
=2=
1 = 1
2 ) 90
(b900
average rate for the complete round
or 1/ab = 9. b =98100
13. Let g represent the number of
=)
9(900
4(900 )
trip can be calculated average rate =
Fred’s grandchildren and x, the
1
1
1 ab
b == 4, then
= 1=11/4 . 1
(total distance)/(total time), as
Case 1: If 1/ab
amount of money he has to distribute.
b = 8100
=
b = =90
90
follows:
Since a = 900b, then8100
(900b)b
=
1/4.
If he gives $20 to each child, then he
3600 60
Vol. 14, No. 7, March 2009
900 b 2 =
1
9
●
Mathematics Teaching in the Middle School
445
would need $20g. However, he needs
$62 more to do this, which leads to
x + 62 = 20g. If Fred gives only $17 to
each grandchild, he would distribute
$17g and have $7 remaining. This
leads to a second equation: x = 17g + 7.
Substituting 17g + 7 for x in the first
equation gives (17g + 7) + 62 = 20g, or
3g = 69, or g = 23. Therefore, Fred has
23 grandchildren. The amount of
money he had originally was 23(17) +
7 = $398, or 23(20) – 62 = $398.
(Ed. note. For an in-depth discussion
of problems such as this one, see
“Math Roots: The Chinese Remainder Theorem” in this issue.)
14. A perfect square can be written
as a 2, where a can be either prime or
composite. We must find a counting
number for a that is a multiple of 375.
Written in prime factored form, 375 is
3 • 53. We can reach 32 • 54, which has
all even exponents and can be written
as (3 • 52)2, a perfect square. We reach
this by multiplying by 3 • 5, or 15.
To check, calculate 375n = 375(15) =
5625, which is 752.
15. A perfect cube, in prime factored
form, is a product of prime numbers
in which each prime is raised to a
power that is a multiple of 3. Written
in prime factored form, 375 is 3 • 53.
We must multiply this by the smallest counting number that will yield a
product in which each prime factor is
raised to a power that is a multiple of
3, so use n = 32 = 9. This gives the following product: 375n = 375(9) =
(3 • 53)(32) = 3353 = (3 • 5)3 = 153,
which is a perfect cube.
16. Similar to the solution to the
previous problem, a perfect cube is a
NCTM’s
product of prime numbers in which
each prime is raised to a power that
is a multiple of 3. Written in prime
factored form, 375 is 3 • 53 and 16 =
24. Because 375 and 16 are relatively
prime, to obtain a number that is
a multiple of both 375 and 16, we
multiply (3 • 53)(24) to obtain 24 • 3 •
53. We next multiply this product by
the smallest counting number that
will yield a product in which each
exponent is a multiple of 3, so use
2232 = 36. This gives the following
product:
375n = (375)(16)(36)
= (3 • 53)(24)(2232)
= 26 • 33 • 53
= (2 • 3 • 5)3
This is a perfect cube. Thus, the
smallest such counting number is n =
(16)(36) = 576 = 26 • 32. l
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