Download Palette of Problems 2 - Narragansett Schools

palette of
problems
David Rock and Mary K. Porter
1. Determine three numbers that fit the following pattern:
2, 8, 4, 10, 5, 11, 5.5, ___, ___, ___
6. The numbers 1 and 9 are two of five counting numbers
that produce a sum of 25. Those same five numbers, when
multiplied, give a product of 945. What are the other three
numbers?
2. Determine three numbers that fit the following pattern:
3, 7, 10, 17, 27, 44, 71, _____, _____, _____
3. If the mean of twenty different positive integers is 20,
what would be the greatest possible value of any one of
these twenty numbers?
4. Suppose you randomly selected an integer from one to
one million, inclusive, and the number turned out to be a
perfect square. What is the probability that the number is
also a perfect cube?
5. Kyle says that there is a 55 percent chance that he
will go to the library tomorrow if it is raining at noon and
a 30 percent chance that he will go to
the library if it is not raining at noon.
The meteorologist on television
forecasts a 40 percent chance
of rain at noon. On the basis
of the information provided,
what is the probability that
Kyle will go to the library?
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Mathematics Teaching in the Middle School
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7. In 1980, a typical telephone
number in the United States contained seven digits. Several areas
of the country now must use
ten-digit telephone numbers.
If the entire country follows,
exactly how many different
ten-digit telephone numbers are
available such that the first digit
cannot be a 0 or 1 and the fourth
digit cannot be a 0?
8. A new operation symbol has been created in mathematics. Your task is to determine how the @ operation works.
Based on the equations below, what would 7 @ 8 equal?
Vol. 13, No. 8, April 2008
Copyright © 2008 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.
1@2=5
3 @ 4 = 25
4 @ 5 = 41
5 @ 6 = 61
7 @ 8 = ___
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).
9. Maki has a special pair of dice: On each die, one face
is labeled tail and the other five faces are labeled head. If
Maki tosses the pair of dice, what is the probability that he
will get exactly one head and one tail?
12. To find my number, take 1 less than the square of the
sixth prime number and subtract this from the least common multiple of 44 and 54. What is my number?
13. Jim paid $5.55 for 1 notepad and 7 pencils. Anne
paid $12.28 for 3 notepads and 2 pencils. How much
does each notepad cost?
10. If Maki tosses his special pair of dice (described
in the previous problem)
twice, what is the probability that he will get a
total of exactly two heads
and two tails?
14. Consider these three facts:
11. Rachel withdrew all of the money from her savings
account at the bank. After giving 10 percent to her sister,
she spent 1/3 of what was left on a book and then gave
the remaining $72.00 to a local animal shelter. How much
money did Rachel’s sister get?
(a) Ben is shorter than Carrie.
(b) Ashley is taller than Duane.
(c) If Ben is the tallest, then Ashley is shorter than
Ben; otherwise, Ben is the second shortest and
Ashley is not the tallest.
Using those three facts, list the four people (Ashley, Ben,
Carrie, and Duane) in order, from shortest to tallest.
15. What is the smallest three-digit number you can obtain
from a product of two or more distinct prime numbers?
16. Erin noticed that both of the numbers 992 and 578
have a special property: The sum of the digits of each
number is equal to 20. For 992, 9 + 9 + 2 = 20, and
for 578, 5 + 7 + 8 = 20. How many three-digit numbers
have this special property?
(Solutions on pages 466–67)
Vol. 13, No. 8, April 2008
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Mathematics Teaching in the Middle School
465
solutions to palette
(Continued from pages 464–65)
(Alternative approaches to those suggested here are encouraged.)
1. 11.5, 5.75, 11.75. The pattern is add
6 to a number in an odd position in
the sequence, and divide a number in
an even position in the sequence by 2.
2. 115, 186, 301. This
�������������������
pattern is an
additive sequence similar to the Fibonacci sequence such that each number
after the 3 and 7 are determined by the
sum of the two previous numbers. 3 +
7 = 10, 7 + 10 = 17, 10 + 17 = 27, 17 +
27 = 44, 27 + 44 = 71, 44 + 71 = 115,
71 + 115 = 186, and 115 + 166 = 301.
3. 210. Use the smallest possible
nineteen numbers to determine the
largest possible twentieth number. In
other words, (1 + 2 + 3 + 4 + 5 + 6 +
7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 +
15 + 16 + 17 + 18 + 19 + x)/20 = 20;
190 + x = 400; x = 210.
6. 3, 5, 7. Since the product ends in
5, one of the factors must be 5. The
known addends are 1, 5, and 9, which
add to 15; the remaining two addends
must sum to 10: 1 × 5 × 9 = 45 and
945/45 = 21. The factors of 21 that
give a sum of 10 are 3 and 7.
7. 7,200,000,000 different numbers.
Since the first digit cannot be a 0 or
1, there are only 8 possible choices.
Since the fourth digit cannot be 0,
there are only 9 possible choices.
There are 10 possible choices for each
of the remaining digits. Therefore,
8 × 10 × 10 × 9 × 10 × 10 × 10 × 10 ×
10 × 10 = 7,200,000,000.
8. 113. The @ operation requires you
to find the sum of the squares of each
number.
a tail is 1/6. Maki can get exactly one
head and one tail in two possible ways:
heads on the first die and tails on the
second die (the probability is 5/6 ×
1/6 = 5/36) or tails on the first die and
heads on the second die (the probability
is 1/6 × 5/6 = 5/36). Thus, the probability of getting exactly one head and
one tail is 5/36 + 5/36 = 10/36 = 5/18.
10. 25/216. There arexxx
six ways in
which Maki can get exactly
xxx two
heads and two tails: HHTT, HTHT,
HTTH, TTHH, THTH, and
THHT. On each die, the probability
of a head (H) is 5/6, and the probability of a tail (T) is 1/6, so the probability of any
5 one
5 of
1 those
1 six
25ways is
× × × =
.
25
6 5 6 5 6 1 6 1 1296
× × × =
.
6 6 6 6 1296
Since there are six such ways, then the
probability of getting a total of exactly
two heads and two tails is
xxx
25
25
6×
=
.
25
1296 21625
6×
=
.
1296 216
11. $12.00. Let x represent the amount
$72.00had
0 in
Another possibility is that the @
of money Rachel originally
x=
= $120.00.
$
72
.0
00 is
0.60
operation requires that you find the
her savings account. Then
0.10x
x=
= $120.00.
.60 and
product of the two numbers, double it,
the amount she gave to her0sister,
5 0.90x
5 1 is the
1 amount
25 Rachel had left
and add 1. Therefore:
× × × =
.
5. .40
6 after
6 6giving
6 money
1296 to her sister. There1@2=2×1×2+1=5
fore, Rachel spent (1/3)(0.90x) on a
$4.37
P(Kyle goes to the library and it rains)
3 @ 4 = 2 × 3 × 4 + 1 = 25
book and gave (2/3)(0.90x)
==
$72.00
p=
0.23. to
$4.37
19
= .55 × .40 = .22
4 @ 5 = 2 × 4 × 5 + 1 = 41
the animal shelter. Solve
p = the equation
= 0.23.
19 x, the
P(Kyle goes to the library and it does
5 @ 6 = 2 × 5 × 6 + 1= 61
(2/3)(0.90x) = $72.00 to find
not rain)
7 @ 8 = 2 × 7 × 8 + 1 = 113
original amount that Rachel had:
25
25
= .30 × .60 = .18
6×
=
.
Note the equivalency of the two meth1296 0.60x
216 = $72.00,
The probability that Kyle will go to the
ods. Let a and b be the two consecutive
so
library is the probability that Kyle goes
numbers. Then b = a + 1. So, a2 + b2 =
$72.0
00
x=
= $120.00.
to the library and it rains or the proba2 + (a + 1)2 = a2 + a2 + 2a + 1 = 2a2 +
0.60
ability that Kyle goes to the library and
2a + 1 = 2a(a + 1) + 1 = 2ab + 1.
it does not rain. Since it cannot both
Thus, Rachel’s sister got
rain and not rain at noon, we add the
9. 5/18. On each die, the probability
probabilities: .22 + .18 = .40.
of a head is 5/6, and the probability of
0.10x = 0.10(120) = $12.00.
$4.37
p=
= 0.23.
466 Mathematics Teaching in the Middle School ● Vol. 13, No. 8, April 2008
19
4. .01, or 10/1000, or 1/100. Since
10002 = 1,000,000, there are 1000 perfect squares between 1 and 1,000,000,
because 1 is 12 and 1,000,000 is 10002.
For a number to be a perfect square and
a perfect cube, it must be a perfect 6th
power. Only 10 are possible—16, 26,
36, 46, . . . , 106—since 106 = 1,000,000.
Therefore, the probability is 10/1000.
1 @ 2 = 12 + 22 = 5
3 @ 4 = 32 + 42 = 25
4 @ 5 = 42 + 52 = 41
5 @ 6 = 52 + 62 = 61
7 @ 8 = 72 + 82 = 113
12. 1020. Using the prime factorizations
of 44 (22 × 11) and 54 (2 × 33), we can
find the least common multiple of 44
and 54: 22 × 33 × 11 = 1188. One less
than the square of the sixth prime number is 132 – 1 = 169 – 1 = 168. Thus, my
number is 1188 – 168, or 1020.
13. $3.94. Let n represent the cost of
each notepad, and let p represent the
cost of each pencil. From Jim’s purchases, we know that
so
1n + 7p = $5.55,
xxx
n = $5.55 – 7p.
From Anne’s purchases, we know that
3n + 2p = $12.28.
5 Since
5 1 1
25
× × × =
.
6 6 6 6 1296
n = $5.55 – 7p,
by substitution we have
so
3($5.55 – 7p) + 2p = $12.28,
25
25
6×
=
.
3($5.55
–
7p)
1296 216+ 2p = $12.28
= $16.65 – 19p.
Therefore,x =
$72.000
= $120.00.
0.60
19p = $16.65 – $12.28
= $4.37,
so
p=
$4.37
= 0.23.
19
Thus, pencils cost $0.23 each. Since
n = $5.55 – 7p
= $5.55 – 7($0.23)
= $5.55 – $1.61
= $3.94,
each notepad costs $3.94.
14. Duane, Ben, Ashley, Carrie. By
(a), Ben is shorter than Carrie, so
we know that Ben is not the tallest. By (c), Ben is the second shortest and Ashley is not the tallest. By
(b), Ashley is taller than Duane, so
Ashley is neither the shortest nor the
tallest. Ben is the second shortest,
so Ashley must be the third shortest
person (taller than Ben, but not the
tallest person). That leaves Carrie and
Duane—one is tallest and the other
is shortest. By (b), Duane is not the
tallest, so Duane must be the shortest
and Carrie must be the tallest.
nations will yield exactly six different
numbers, giving a total of 24 more
three-digit numbers: 398, 389, 938,
983, 839, 893, 497, 479, 947, 974, 794,
749, 596, 569, 956, 965, 695, 659, 587,
578, 875, 857, 785, and 758. Thus, we
have a grand total of 12 + 24, or 36,
three-digit numbers in which the sum
of the digits is exactly 20. l
15. 102. Start with 100 and work up:
100 = (2)(2)(5)(5). Although each
factor is prime, they are not distinct
primes because 2 and 5 are used twice;
101 is prime, so it is the product of
only one prime and 1; 102 = (2)(3)(17)
so 102 is our solution, since each factor
is prime and used only once.
16. 36 different three-digit numbers.
We will first systematically list all
combinations of three digits in which
the sum is 20: 2, 9, 9; 3, 9, 8; 4, 9, 7; 4,
8, 8; 5, 9, 6; 5, 8, 7; 6, 8, 6; and 6, 7, 7.
Note that if 0 or 1 were one of the
digits, then the sum of the three
digits would be less than 20, so no
combination can include 0 or 1. All
possible combinations in which the
sum is 20 have been identified, so our
list is complete. We will now count
the number of different three-digit
numbers that can be formed using the
combinations listed. The combination
2, 9, 9 yields exactly three different
numbers: 299, 929, and 992. Similarly,
each of the other combinations in
which a digit is used twice (namely, 4,
8, 8; 6, 8, 6; and 6, 7, 7) yields exactly
three different numbers (namely, 488,
848, 884, 866, 686, 668, 677, 767,
and 776), so these four combinations
together yield a total of 12 different
three-digit numbers. Each of the other
four combinations (namely, 3, 9, 8; 4,
9, 7; 5, 9, 6; and 5, 8, 7) contains three
distinct digits. Each of these combiVol. 13, No. 8, April 2008
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Mathematics Teaching in the Middle School
467