Download ANSWERS FOR MATHEMATICS INVESTIGATIONS

ANSWERS FOR INVESTIGATIONS
ANSWERS FOR MATHEMATICS INVESTIGATIONS
CALCULATOR INVESTIGATION 1.1
Four-Digit Numbers
Number of digits
2
3
4
5
6
7
8
Divisible by 11 (T or F)
T
F
T
F
T
F
T
The property is true for all whole numbers with an even number of digits.
COMPUTER INVESTIGATION 1.2
Triangular Numbers
1.
The following numbers are the units digits in the first 40 triangular numbers. The units digits for the first
through the ninth triangular numbers are repeated in reverse order for the 10th through the 18th triangular
numbers. Also the units digits for the first 20 triangular numbers repeat in the same order for the next 20
triangular numbers.
1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1, 0, 0, 1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1, 0, 0
2.
Yes. Each group of twenty digits ends in two zeros. That is, there are zeros in the 19th and 20th positions;
39th and 40th positions; 59th and 60th positions; etc. The sequence of units digits will never have more
than 2 consecutive zeros.
3.
For the Nth triangular number, divide N by 20 and let R be the remainder. Thus, R = 0, 1, 2, 3, . . . , 19. If
R = 0, the units digit of the Nth triangular number is 0; if R = 1, the units digit of the Nth triangular number
is 1; if R = 2, the units digit of the Nth triangular number is 3; if R = 3, the units digit of the Nth triangular
number is 6; etc. In general, for any remainder R, the units digit of the Nth triangular number equals the
units digit of the Rth triangular number, for R equal 1 to 19. Since the remainder when dividing 5147 by
20 is 7, the units digit of the 5147th triangular number is 8, which is the units digit of the 7th triangular
number.
4.
The 40th triangular number is the sum 1 + 2 + 3 + . . . + 39 + 40. This is illustrated in the figure by the top
40 rows. The 20th triangular number is illustrated by the top 20 rows of the figure and is 1 + 2 + . . . + 20
= 210. The figure shows that the 40th triangular number is 202 + 2 × 210 (one 20 × 20 array and two 1 by
20 staircases). Similarly, the figure shows that the 60th triangular number is 3 × 202 + 3 × 210 (three 20
× 20 arrays and three 1 by 20 staircases). The next 20 rows of the figure will contain three 20 × 20 arrays
and one 1 by 20 staircase and so the 80th triangular number is 6 × 202 + 4 × 210. Similarly, the 100th
triangular number is 10 × 202 + 5 × 210, and the 120th triangular number is 15 × 202 + 6 × 210.
ANSWERS FOR INVESTIGATIONS
COMPUTER INVESTIGATION 1.3
Palindromic Sums
1.
Yes
2.
The following list contains all the two-digit numbers that require two or more steps.
2 steps: 19, 28, 37, 39, 46, 48, 49, 55, 57, 58, 64, 66, 67, 73, 75, 76, 82, 84, 85, 91, 93, and 94
3 steps: 59, 68, 77, 86, and 95
4 steps: 69, 78, 87, and 96
6 steps: 79, 88, 97, and 99
24 steps: 89 and 98
3.
Yes, for two-digit numbers that go to palindromic numbers in 2 steps, the sums of their digits are either 10,
12, or 13. After the first four of these numbers (19, 28, 37, 39), the sums of 10, 12, and 13 repeat for the
remaining two-digit numbers in the above list. All two-digit numbers that require 3 steps have a digit sum
of 14; for 4 steps the sum is 15; for 6 steps the sums are 16 and 18; and for 24 steps the sum is 17.
The sums of the digits of two-digit numbers range from 1 to 18. For a two digit number ab with a + b ≤ 9
or a + b = 11, the number goes to a palindrome in one step, as shown by the following equations.
a(10) + b
+ b(10) + a
(a+b)10 + (a+b) = 10a + 10b + a + b = 11a + 11b = 11(a + b)
For a + b = k with k ≤ 9, then ab goes to the palindromic number 11k in one step. If a + b = 11, then ab
goes to the palindrome 11(11) = 121 in one step.
4,5. There are only eight three-digit numbers that require exactly 23 steps to go to palindromic numbers. They
are listed here. The sum of the digits for each of these numbers is 16.
187 286 385 484 583 682 781 880
All other three-digit numbers except the following 13 require less than 23 steps. (Note: The Mathematics
Investigator software is programmed to stop the process of reversing and adding numbers when sums are
reached that have over 100 digits.)
196, 295, 394, 493, 592, 689, 691, 788, 790, 879, 887, 978, 986
The sums of the digits in each of these 13 numbers are either 16, 23, or 24. However, there are other threedigit numbers whose digit sum is 16, 23, or 24 which do go to palindromic numbers. The 13 numbers that
do not go to palindromic numbers in 23 or fewer steps can also be classified by looking at the middle (tens)
digit of the number and the outer two digits: those with a middle digit of 9 and outer digits whose sum is 7;
those with a middle digit of 8 and outer digits whose sum is 15; and those with a middle digit of 7 and outer
digits whose sum is 17. The only three-digit numbers that satisfy these conditions are the 13 listed above.
6.
There are 236 four-digit numbers that do not go to palindromic numbers in fewer than 220 steps. These
numbers are shown in increasing order in the following list. There are many patterns in these numbers.
Notice that in the first few rows the differences between consecutive numbers alternate between 2 and 88
ANSWERS FOR INVESTIGATIONS
with the exception of a few numbers. The digits of the numbers in the top row of the table also follow
another pattern: the inner two digits of each number have a sum of 13 and the sum of the two outer digits
alternates between 6 and 8. The sum of the inner digits of the first number in the second row (1997) is 18
and the sum of the two outer digits is 8. The next 12 numbers follow the pattern of the numbers in the first
row of the table, and then there is another number (2996) whose inner digit sum is 18 and whose outer digit
sum is 8. These first few rows suggest that we look at the sums of inner digits and the sums of outer digits
for the 236 numbers.
1495, 1497, 1585, 1587, 1675, 1677, 1765, 1767, 1855, 1857, 1945, 1947,
1997, 2494, 2496, 2584, 2586, 2674, 2676, 2764, 2766, 2854, 2856, 2944,
2946, 2996, 3493, 3495, 3583, 3585, 3673, 3675, 3763, 3765, 3853, 3855,
3943, 3945, 3995, 4079, 4169, 4259, 4349, 4439, 4492, 4494, 4529, 4582,
4584, 4619, 4672, 4674, 4709, 4762, 4764, 4799, 4852, 4854, 4889, 4942,
4944, 4979, 4994, 5078, 5168, 5258, 5348, 5438, 5491, 5493, 5528, 5581,
5583, 5618, 5671, 5673, 5708, 5761, 5763, 5798, 5851, 5853, 5888, 5941,
5943, 5978, 5993, 6077, 6167, 6257, 6347, 6437, 6490, 6492, 6527, 6580,
6582, 6617, 6670, 6672, 6707, 6760, 6762, 6797, 6850, 6852, 6887, 6940,
6942, 6977, 6992, 7059, 7076, 7149, 7166, 7239, 7256, 7329, 7346, 7419,
7436, 7491, 7509, 7526, 7581, 7599, 7616, 7671, 7689, 7706, 7761, 7779,
7796, 7851, 7869, 7886, 7941, 7959, 7976, 7991, 8058, 8075, 8079, 8089,
8148, 8165, 8169, 8179, 8238, 8255, 8259, 8269, 8328, 8345, 8349, 8359,
8418, 8435, 8439, 8449, 8490, 8508, 8525, 8529, 8539, 8580, 8598, 8615,
8619, 8629, 8670, 8688, 8705, 8709, 8719, 8760, 8778, 8795, 8799, 8809,
8850, 8868, 8885, 8889, 8899, 8940, 8958, 8975, 8979, 8989, 8990, 9057,
9074, 9078, 9088, 9147, 9164, 9168, 9178, 9237, 9254, 9258, 9268, 9327,
9344, 9348, 9358, 9417, 9434, 9438, 9448, 9507, 9524, 9528, 9538, 9597,
9614, 9618, 9628, 9687, 9704, 9708, 9718, 9777, 9794, 9798, 9808, 9867,
9884, 9888, 9898, 9957, 9974, 9978, 9988, 9999
The following table contains the classification of the numbers from the above list according to the sums of
the inner and outer digits. These numbers fall into 12 classes, and every four-digit number that has the
inner and outer digit sums shown in this table is also in the above list of 236 numbers.
Sums of Inner Digits
18
18
16
16
17
14
13
13
8
7
7
5
Sums of Outer Digits
18
8
13
17
17
16
6
8
17
17
13
16
Number of Four-digit Numbers
1
8
18
6
4
15
36
48
18
16
48
18
Total 236
It is natural to wonder if inner digit sums and outer digit sums can be used to classify five-digit numbers.
This may be true. Consider the following patterns. All numbers such as 99883, 38989, 58897, 67996,
99973, etc., whose inner three digit sum is 25, whose outer two digit sum is 12, and which do not have 7 as
the middle (hundreds) digit, do not go to palindromic numbers in less than 220 steps. There are 35
four-digit numbers that satisfy the preceding conditions. All four-digit numbers whose inner three digit
ANSWERS FOR INVESTIGATIONS
sum is 25, whose outer two-digit sum is 12, and whose middle digit is 7, such as 99793, 89794, 59797, etc.,
go to a palindromic number in five steps.
You may wish to pursue this investigation further. It appears to be unexplored territory. So far it has been
determined that: all two-digit numbers go to Palindromic numbers; all but 1.444...% of the three-digit
numbers go to palindromic numbers; all but 2.6222...% of the four-digit numbers go to palindromic
numbers; and all but 6.49111...% of the five-digit numbers go to palindromic numbers.
COMPUTER INVESTIGATION 2.1
Consecutive Numbers
1.
The only numbers less than or equal to 50 that cannot be written as the sum of nonzero consecutive whole
numbers are 1, 2, 4, 8, 16, and 32. Conjectures: 2n for n ≥ 0, cannot be written as a sum of nonzero
consecutive whole numbers. Any whole number that is not a power of 2 can be written as a sum of
nonzero consecutive whole numbers.
2.
Odd numbers can always be written as the sum of two nonzero consecutive whole numbers. In general, any
odd number 2n - 1, for n ≥ 2, can be written as the sum of the two nonzero consecutive numbers n - 1 and n.
Every whole number greater than 3 that can be divided evenly by 3 is the sum of three nonzero consecutive
whole numbers. In general, for any whole number n ≥ 2, 3n is the sum of the three consecutive whole
numbers n-1, n, and n+1.
The numbers 10, 14, 18, 22, . . . can be written as the sum of four consecutive whole numbers.
10
14
18
22
=
=
=
=
1+2+3+4
2+3+4+5
3+4+5+6
4+5+6+7
In general, for any whole number n ≥ 2, 4n + 2 can be written as the sum of the four consecutive whole
numbers n-1, n, n+1, and n+2.
The numbers 15, 20, 25, 30 . . . can be written as the sum of five consecutive whole numbers.
15
20
25
30
=
=
=
=
1+2+3+4+5
2+3+4+5+6
3+4+5+6+7
4+5+6+7+8
In general, for any n ≥ 3, 5n can be written as the sum of the five consecutive whole numbers n-2, n-1, n,
n+1, and n+2.
3.
Whenever a whole number can be written as the sum of nonzero consecutive whole numbers, it can be
represented by columns of tiles of increasing heights, as shown on the next page. The staircase at the top
of the columns will always represent a triangular number. If the staircase represents the nth triangular
number, then once the staircase is removed the remaining rectangle will have a width of n. In the
ANSWERS FOR INVESTIGATIONS
illustration at the right, the 1-to-5 staircase represents the 5th
triangular number and the rectangle is 5 × 8. This example
suggests the following method for determining a sum of
nonzero consecutive numbers for a given whole number: For
any whole number K, subtract the nth triangular number and
determine if the difference is divisible by n. If it is, say
(K - nth triangular number) = nH, where H is the height of
the rectangle, then
K = H+1 + H+2 + H+3 + . . . + H+n
For the figure above, n = 5 and H = 8, and the five columns have heights of 9, 10, 11, 12, and 13. This
method of subtracting triangular numbers can be used to determine all the different sums of nonzero
consecutive numbers that equal a given number. Let's use this method to determine all such sums that
equal 117. The following table shows the results of systematically subtracting triangular numbers
beginning with the second triangular number. Notice that the table does not begin with the first triangular
number, because this case corresponds to just one column of tile, and thus does not involve two or more
nonzero consecutive whole numbers.
Number
117
117
117
117
117
117
117
117
117
Triangular Number
2nd, 3
3rd, 6
4th, 10
5th, 15
6th, 21
7th, 27
8th, 36
9th, 45
13th, 91
Difference Sums of Consecutive Numbers
117-3 = 114 2 × 57 = 114, so 117 = 58 + 59
117-6 = 111 3 × 37 = 111, so 117 = 38 + 39 + 40
117-10 = 107 4 does not divide 107
117-15 = 102 5 does not divide 102
117-21 = 96 6 × 16 = 96, so 117 = 17 + 18 + . . . + 22
117-27 = 90 7 does not divide 90
117-36 = 81 8 does not divide 81
117-45 = 72 9 × 8 = 72, so 117 = 9 + 10 + . . . + 17
117-91 = 26 13 × 2 = 26, so 117 = 3 + 4 + . . . + 15
GRAPHING CALCULATOR INVESTIGATION 2.2
Graphs of Functions
1.
2.
The graphs are straight lines. In general, the graph of the equation y = kx for any whole number k, has an
increasing slope as k increases.
1
The graphs are straight lines. In general, the graphs of the equation y = n x for whole numbers n > 0, have
decreasing positive slopes which approach zero as n increases.
3.
The graphs of y = x + b are parallel lines which pass through the points (0,b) for whole numbers b.
4.
The graphs of y = kx2 for k = 1, 2, 3, 4, . . . are parabolas that open upward and are symmetric to the
1
y-axis. As k increases, the two branches of the curve become steeper. The graphs of y = kx2 for k = 2 ,
ANSWERS FOR INVESTIGATIONS
1 1
1
3 , 4 , . . . are also parabolas that open upward and are symmetric to the y-axis. As the fractions n
decrease, the two branches of the curve flatten out.
The graphs of y = x2 + k are the same as the graph of y = x2, except that their positions are changed so that
the y-intercept is (0, k).
1
The graphs of y = kx3 are shown here for k = 10 , 1,
and 5. For whole numbers k, as k increases the two
1 1
branches of the curve become steeper. For k = 2 , 3 ,
1
4 , . . . , the two branches of the curve flatten out as the
fractions decrease.
The graphs of y = x3 + k for all values of k that are whole numbers or fractions are the same as the graph of
y = x3, except that their positions are changed so that the y-intercept is (0, k).
COMPUTER INVESTIGATION 2.3
Differences of Squares
1.
The following numbers which are less than 50 cannot be written as the difference of two square numbers.
1, 2, 4, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, 46
Conjecture (1): All odd numbers greater than 1 can be written as the difference of two square numbers.
Conjecture (2): Every even number greater than 4 that is not divisible by 4 cannot be written as the
difference of two square numbers.
Conjecture (3): Every even number greater than 4 that is divisible by 4 can be written as the difference of
two square numbers.
2.
Conjecture (4): Every odd number can be expressed as n2 - m2, where n + m equals the odd number.
ANSWERS FOR INVESTIGATIONS
3.
Since 1024 and 2076 are divisible by 4, Conjecture (3) predicts that they can be written as the difference of
two square numbers. The computer program shows that
1024
1024
1024
1024
=
=
=
=
402 - 242
682 - 602
1302 - 1262
2572 - 2552
2076 = 1762 - 1702
2076 = 5202 - 5182
Since 2077 is an odd number that equals 1038 + 1039, Conjecture (4) predicts that 2077 = 10392 - 10382.
4.
If 72 is represented as a 3 × 24 array of tiles, then
the method of forming an L-shaped figure of
tiles is to mark off 3 tiles along the upper edge of
the rectangle and then divide the remaining
length into two equal parts. Since 21 cannot be
divided evenly by 2, this method cannot be used
on a 3 × 24 array to obtain 72 as the difference of
two square numbers.
21
3
3
However, since 72 can be represented as a 4 × 18 array, which can be reformed into the L-shaped figure
shown below, 72 = 112 - 72. Similarly, 72 can be represented by a 2 × 36 array, and so it is equal to
192 - 172.
7
4
14
4
4
4
7
7
The forming and reshaping of a rectangular array can be used to illustrate that any whole number
n > 4 which is divisible by 4 can be written as the difference of two square numbers. The rectangle in
Figure (a) represents a number that is divisible by 4. By cutting and repositioning the rectangle, the same
number is represented as a 2 × 2k rectangle in Figure (b). Then by marking off the 2k dimension of the
rectangle into a length of 2 and two equal lengths of k-1, the rectangle can be reformed as in figure (c) to
show that the original number is the difference of (k + 1)2 - (k - 1)2. For example, 44 = 4 × 11, where k =
11. So, 44 = 122 - 102.
k
2
2k
4
k-1
2
2
(a)
k–1
(b)
k-1
(c)
ANSWERS FOR INVESTIGATIONS
CALCULATOR INVESTIGATION 3.1
The Number 6174
1.
Yes, in seven or fewer steps.
2.
The "special number" for three-digit numbers is 495. If this process is applied to three-digit numbers,
whose digits are not all equal, each number will go to 495 in five or fewer steps.
3.
There is no such "special number" for all five-digit numbers. There are several numbers which turn up and
for which cycles are created that return to the given number. Two such cycles are:
61974 → 82962 → 75933 → 63954 → 61974
83952 → 74943 → 62964 → 71973 → 83952
It is interesting that some five-digit numbers lead to the four-digit number 6174 and others lead to the
three-digit number 495.
COMPUTER INVESTIGATION 3.2
Palindromic Differences
1.
Yes, in five or fewer steps. All two-digit numbers with different digits go to 9, and two-digit palindromic
numbers go to 0.
2.
Yes, in five or fewer steps. All three-digit numbers with different first and last digits go to 99, and all
three-digit palindromic numbers go to 0.
3.
Conjectures will vary. One conjecture suggested by the data from #1 and #2 is that all non-palindromic
four-digit numbers will go to 999 in five or fewer steps. This conjecture is not true. (see #4,5)
4,5. All four-digit numbers go to the palindromic numbers 0, 99, 909, or 999, except those that lead to the cycle
2178 → 6534 → 2178 → 6534 → . . .
It appears that all five-digit numbers go to a palindromic number except those that lead to the cycles
2178 → 6534
21978 → 65934
It appears that all six-digit numbers go to a palindromic number except those that lead to the cycles
2178 → 6534
21978 → 65934
219978 → 659934
ANSWERS FOR INVESTIGATIONS
It appears that all seven-digit numbers go to a palindromic number except those that lead to either the
above three cycles or the following cycle:
2199978 → 6599934
It is interesting that all the numbers in the preceding cycles are divisible by 11.
COMPUTER INVESTIGATION 3.3
Number Chains
1.
19 goes to 19 in one step. Any number greater than 19 eventually leads to 19 or to the number chain with
all whole numbers from 1 to 18.
2,3. Multiplying the units digit by 3 and adding the remaining digits produces one number chain with all whole
numbers from 1 to 28 or it produces the number 29 which goes to itself.
4.
Multiplying the units digit by 4 and adding the remaining digits produces nine different number chains with
whole numbers less than or equal to 39. The numbers 13, 26, and 39 go to themselves in number chains
which have only one number while the other numbers less than 39 are contained in the following number
chains:
1 → 4 → 16 → 25 → 22 → 10 → 1
2 → 8 → 32 → 11 → 5 → 20 → 2
3 → 12 → 9 → 36 → 27 → 30 → 3
6 → 24 → 18 → 33 → 15 → 21 → 6
7 → 28 → 34 → 19 → 37 → 31 → 7
14 → 17 → 29 → 38 → 35 → 23 → 14
The following results are obtained for multipliers 5 through 9. Multiplier 5: All numbers go to 49 or to one
of two other number chains. Multiplier 6: All numbers either go to 59 or to the chain of numbers from 1 to
58. Multiplier 7: All numbers either go to 23, 46, or 69, which go to themselves, or to one of three other
number chains. Multiplier 8: All numbers either go to 79 or to one of six other chains. Multiplier 9: All
numbers either go to 89 or to one of two other chains.
5.
The following numbers go to themselves in one step: 11, 22, 33, 44, . . . 88, 99. This is easy to see, since
for any two-digit number kk, 10k + k is the two-digit number kk. All other numbers are contained in one of
45 chains which each have just two numbers. Here are two examples of chains with two numbers:
17 → 71 → 17 and 26 → 62 → 26. In general, consider the two-digit number ab, with a ≠ b. For the
multiplier 10 we have 10b + a which equals the two-digit number ba. Then using the multiplier of 10 once
again, 10a + b equals the original two-digit number.
6.
The following table shows the multipliers 2 through 15 and the numbers that go to themselves in one step.
Several conjectures are possible. Notice the multipliers that produce just one large chain, other than the
single number chain that begins with 10n - 1 for each multiplier n. These multipliers, 2, 3, 6, 11, 12, and
15, suggest that maybe the multipliers 20, 21, and 24 will each have just one large chain, other than the one
number that goes to itself in one step. The multiplier 10, that produces many chains with just two numbers
ANSWERS FOR INVESTIGATIONS
each, suggests another conjecture: Will multipliers of 20, 30, 40, or 50 produce similar chains? Another
conjecture is suggested by the multipliers 4, 7, 10, and 13. These multipliers produce chains that have
more than one number that goes to itself in one step, and in each case these numbers are multiples of the
smallest number that goes to itself in one step. When there is more than one number that goes to itself in
one step, will these numbers be multiples of the smallest number?
Multipliers Numbers that go to themselves in one step Additional number chains
2
3
4
5
6
7
8
9
10
11
12
13
14
15
19
29
13, 26, 39
49
59
23, 46, 69
79
89
11, 22, 33, 44, 55, 66, 77, 88, 99
109
119
43, 86, 129
139
149
1 chain, 1-18
1 chain, 1-28
6 chains of 6 numbers
2 chains
1 chain, 1-58
3 chains
6 chains
2 chains
45 chains
1 chain, 1-108
1 chain, 1-118
6 chains
3 chains
1 chain, 1-148
CALCULATOR INVESTIGATION 3.4
Sums and Differences of Squares
1.
Yes, this pattern holds for all two-digit numbers.
2.
No, 12 does not divide 102 + 22. However, the pattern holds for the following two-digit numbers: 25, 36,
and the multiples of 10 (10, 20, 30, ..., 90). For example, 25 divides 202 + 52; 36 divides 302 + 62; and 10
divides 102 + 02.
3.
4.
Yes, the pattern holds for all three-digit numbers.
(10 + 3)(10 - 3) = 102 - 32, which is divisible by 13, since 13 is equal to (10 + 3). The difference between
any two squares, a2 - b2, can be factored:
a2 - b2 = (a + b)(a - b)
Thus, 1002 - 342 = (100 + 34)(100 - 34), which shows why 1002 - 342 is divisible by 134.
Similarly, this pattern holds for four-digit numbers, five-digit numbers, etc. For example:
1345 divides 10002 - 3452
because 10002 - 3452
= (1000 + 345)(1000 - 345)
= (1345)(655).
ANSWERS FOR INVESTIGATIONS
COMPUTER INVESTIGATION 4.1
Frequency of Primes
1.
There are 25 primes less than 100. The following tables contain the numbers of primes for intervals up to
2000. While there are some increases from interval to interval, over all, the numbers of primes appear to be
decreasing.
1-100
100-200
200-300
300-400
400-500
500-600
600-700
700-800
800-900
900-1000
25
21
16
16
17
14
16
14
15
14
1000-1100 1100-1200 1200-1300 1300-1400 1400-1500 1500-1600 1600-1700 1700-1800 1800-1900 1900-2000
16
12
2.
15
11
17
12
15
12
12
13
The following tables contain the numbers of twin primes in intervals of 100. While there are some
increases from interval to interval, over all, the numbers of twin primes appear to be decreasing. Looking
at the first two intervals of 1000 numbers there are 34 twin primes less than 1000 and 26 twin primes
between 1000 and 2000. Possible conjectures: (1) No interval of 1000 beyond 1000 contains more than 34
twin primes; (2) Each interval of 100 beyond 1100 contains less than 5 pairs of twin primes.
1-100
100-200
200-300
300-400
400-500
500-600
600-700
700-800
800-900
900-1000
8
7
4
2
3
2
3
0
5
0
1000-1100 1100-1200 1200-1300 1300-1400 1400-1500 1500-1600 1600-1700 1700-1800 1800-1900 1900-2000
5
1
3.
3
2
4
0
4
2
2
3
The following tables contain intervals of 100 with the numbers of pairs of consecutive primes whose
difference is 4. The numbers of pairs of such primes appear to be decreasing as such intervals continue.
For example, in the intervals 1 - 500, 500 - 1000, 1000 - 1500, and 1500 - 2000, the numbers of pairs of
primes whose difference is 4 is 23, 14, 11, and 10, respectively.
1-100
100-200
200-300
300-400
400-500
500-600
600-700
700-800
800-900
900-1000
7
5
3
4
4
0
3
3
5
3
1000-1100 1100-1200 1200-1300 1300-1400 1400-1500 1500-1600 1600-1700 1700-1800 1800-1900 1900-2000
3
0
2
1
5
3
3
1
2
1
ANSWERS FOR INVESTIGATIONS
4.
5.
The following table shows pairs of consecutive primes whose differences are the first few even numbers.
These pairs contain the smallest possible primes for these differences. Notice how large the primes must be
for a difference of 16. A famous conjecture by A. dePolignac states that every even number is the
difference of two consecutive primes in an infinite number of ways. In order for the difference between
two primes to be an odd number, one prime would have to be even. Since 2 is the only even prime, the
only two consecutive primes whose difference is an odd number are 2 and 3.
3,5
7,11
23,29
89,97
139,149
211,223
317,331
1831,1847
523,541
2
4
6
8
10
12
14
16
18
The next table shows the primes between the first few whole numbers and their doubles. From this data we
might conjecture that between every whole number greater than 5 and its double, there are at least two
primes, or between every whole number greater than 8 and its double there are at least three primes.
2 and 4
3 and 6
4 and 8
5 and 10
6 and 12
7 and 14
8 and 16
9 and 18
10 and 20
11 and 22
3
5
5,7
7
7,11
11,13
11,13
11,13,17
11,13,17,19
13,17,19
COMPUTER INVESTIGATION 4.2
Factorizations
1.
Three factors:
Four factors:
4 = 2 × 2; 9 = 3 × 3; 25 = 5 × 5; 49 = 7 × 7
6 = 2 × 3; 8 = 2 × 2 × 2; 10 = 2 × 5; 14 = 2 × 7; 15 = 3 × 5; 21 = 3 × 7; 22 = 2 × 11;
26 = 2 × 13; 27 = 3 × 3 × 3; 33 = 3 × 11; 34 = 2 × 17; 35 = 5 × 7; 38 = 2 × 19;
39 = 3 × 13; 46 = 2 × 23; 51 = 3 × 17; 55 = 5 × 11; 57 = 3 × 19; 58 = 2 × 29;
62 = 2 × 31; 65 = 5 × 13; 69 = 3 × 23; 74 = 2 × 37; 77 = 7 × 11; 82 = 2 × 41;
85 = 5 × 17; 86 = 2 × 43; 87 = 3 × 29; 91 = 7 × 13; 93 = 3 × 31; 94 = 2 × 47;
95 = 5 × 19
Five factors:
16 = 2 × 2 × 2 × 2; 81 = 3 × 3 × 3 × 3
Possible conjectures:
(1) Square numbers have an odd number of factors. The numbers less than 100 with either 3 or 5 factors
are square numbers. The remaining square numbers less than 100 are 1, 36, and 64, and their numbers of
factors are 1, 9, and 7, respectively.
(2) If a number is the product of two different primes or three equal primes, it has four factors. This is true
for all numbers less than 100 that have four factors.
(3) If a number is the product of four equal primes, it has five factors. This is true for 16 and 81. It is also
true for 625 = 5 × 5 × 5 × 5. These three numbers are the only numbers less than 1000 with five factors.
ANSWERS FOR INVESTIGATIONS
2.
Here is the completed table.
Number of Factors
2
3
4
5
6
7
8
9
10
Number
Prime Factorization
2
4
6
16
12
64
24
36
48
2
2×2
2×3
2×2×2×2
2×2×3
2×2×2×2×2×2
2×2×2×3
2×2×3×3
2×2×2×2×3
The smallest numbers with 11, 12, and 13 factors are 210 = 1024, 22 × 3 × 5 = 60, and 212 = 4096,
respectively.
Conjecture: For any prime p, the smallest number having p factors is 2p-1 .
3,4. Here is the extended table for each of the smallest numbers having from 11 to 20 factors.
Number of Factors
11
12
13
14
15
16
17
18
19
20
Number
1024
60
4096
192
144
120
65,536
180
262,144
240
Prime Factorization
210
22 × 3 × 5
212
26 × 3
24 × 32
2×2×2×3×5
216
22 × 32 × 5
218
24 × 3 × 5
The following steps produce all the numbers in the above tables except the number 24 which has 8 factors and
the number 120 which has 16 factors. For numbers with 8 and 16 factors, this method produces the numbers 30
and 210, respectively, which are not the smallest numbers having these numbers of factors.
Step 1 Factor n: n = p1 × p2 × p3 × . . . × pn , where the pi are not necessarily all distinct primes.
Step 2 Build the smallest number: 2a-1 × 3b-1 × 5c-1 × 7d-1 × 11e-1 × . . . , where a = largest pi ; b = next
largest pi ; etc.
ANSWERS FOR INVESTIGATIONS
Example: Obtain the smallest number having 18 factors.
Step 1: 18 = 2 × 3 × 3
Step 2: 22 × 32 × 5 = 180
COMPUTER INVESTIGATION 5.1
Integer Differences
1.
Six. In the 1930's Professor E. Ducci of Italy discovered that this process will always produce four zeros.*
Some sets of four numbers require more steps than others. For example, 1, 2, 5 and 10 requires nine
squares before obtaining all zeros. This suggests other questions to investigate: Is there a set of four
integers such that 10 squares are required to obtain all zeros? Is there a set of four integers such that k
squares are required to obtain all zeros for every whole number k? Is there a largest whole number k such
that any set of four integers will require less than k squares before producing all zeros?
2
No. For example, 3, 2, and 2 will lead to the numbers 1, 0, and 1, but never to all zeros.
Conjecture: If the three integers at the vertices of the original triangle are not all equal, eventually an inner
triangle will be obtained with one zero and two equal nonzero whole numbers.
3.
Conjecture: For any set of k integers at the vertices of a k-sided polygon, k zeros will eventually be
produced if k = 2n, for n ≥ 2. That is, if k = 4, 8, 16, . . . . For all other values of k > 2, the set of k
numbers will not always lead to zero.
LABORATORY INVESTIGATION 5.2
Paper Folding
1.
Fold triangular corner A (see figure at left below) onto B and mark off 2 1/2 inches on the 8 1/2 inch strip.
This leaves a 2 1/2 inch strip with a length of 6 inches.
8"
1
2 "
2
A
B
C
6"
1
8 "
2
D
3"
*Honsberger, Ross Ingenuity In Mathematics (Washington, D.C., Mathematical Association of America, 1970),
pp. 80-83.
ANSWERS FOR INVESTIGATIONS
2.
Fold the 6 inch length in half.
3.
Fold the 3 inch length from #2 onto the 11 inch edge on the right side of the original sheet so that the lower
edge of the original sheet is placed on the right edge of the paper and region C is matched to region D (see
figure above). The remaining edge on the right side of the sheet has a length of 8 inches.
4.
Repeatedly folding the 8 inch length in half produces lengths of 4 inches, 2 inches, and 1 inch.
5.
Fold the 8 1/2 inch length in half to obtain a 4 1/4 inch length, and then fold triangular corner E to F and
mark off 4 1/4" on the right edge of the original sheet (see figure below). Fold the 11 inch length on the
right edge of the original sheet in half twice to obtain a 2 3/4" length. Then fold the paper by bringing up
the 2 3/4" length so that it is placed at the end of the 4 1/4" length to obtain a total length of 7".
1
4 "
4
E
F
1
4 "
4
3
2 "
4
6.
The only whole number lengths less than 11 inches that have not been obtained are 5, 9, and 10 inches.
The lengths of 1 inch and 2 inches from #4 were obtained along an 11 inch edge, so the remaining lengths
along this edge are 10 inches and 9 inches. A 5 inch length can be found by folding the 6 inch length from
#1 onto the 11 inch length on the right side of the original sheet.
LABORATORY INVESTIGATION 5.3
Fraction Patterns
1.
As the number of equal parts to a bar increase, the size of the parts decrease. For example,
1/3 > 1/7 > 1/10 > 1/12.
Looking down the left side of the tower shows a decreasing sequence of fractions:
1/2 > 1/3 > 1/4 > 1/5 > . . . .
Looking down the right side of the tower shows an increasing sequence of fractions:
1/2 < 2/3 < 3/4 < 4/5 < 5/6 < . . . . This can be explained by the fact that the additional amounts needed to
make a whole bar are decreasing, so the fractions are getting closer and closer to 1.
Beginning with the top bar, every other bar has a line down the center:
1/2 = 2/4 = 3/6 = 4/8 = 5/10 = 6/12 . . . .
ANSWERS FOR INVESTIGATIONS
Beginning with the thirds bar, every third bar has vertical lines which line up:
1/3 = 2/6 = 3/9 = 4/12; and 2/3 = 4/6 = 6/9 = 8/12.
Beginning with the fourths bar, every fourth bar has vertical lines which line up:
1/4 = 2/8 = 3/12; and 3/4 = 6/8 = 9/12.
2.
Yes, when subtracting unit fractions whose denominators are consecutive whole numbers, the difference is
equal to the product of the two fractions. In general, for any whole number n ≠ 0.
1
1
1
1
−
=
×
n
n +1
n
n +1
3.
Yes. This relationship continues to hold. For example, the bars show that
4
1
1
4
8 = 9 + 2 × 9
These equations are a special case of a more general relationship that can be obtained by using the equation
in #2.
1
1
1
1
1
1
1
1
−
= ×
or
=
+
×
n
n +1 n n +1
n n +1 n n +1
and multiplying both sides of this equation by the nonzero whole number r, we obtain
1
r
r
r
=
+
×
n+1
n
n
n+1
For n = 8 and r = 1, 2, 3, and 4, we have
1
1
1
1
8 = 9 + 8 × 9
2
2
2 1
8 = 9 + 8× 9
3
3
3 1
8 = 9 + 8× 9
4
4
4
1
=
+
×
8
9
8
9
ANSWERS FOR INVESTIGATIONS
COMPUTER INVESTIGATION 6.1
Repeating Decimals
1.
a.
Fractions of the form 1/N, where the only prime factors of N are 2 and 5, will have a terminating decimal.
If N = 2n5m, then the number of digits in the terminating decimal is the maximum of n and m. For
example, 1/500 has a terminating decimal with 3 digits because 500 = 22 × 53 and the maximum value of
the exponents is 3: 1/500 = .002.
b.
If the prime factorization of N has one or more factors of 2 or 5 and at least one other prime factor not
equal to 2 or 5, then the decimal for 1/N will have a nonrepeating part and a repeating part. As in part a,
the length of the nonrepeating part is the maximum of the exponents of 2 and 5 in the prime factorization of
N.
c.
If the prime factorization of N has no factors of 2 or 5, then the decimal for 1/N will be purely periodic.
For example, if N = 273, then since 273 = 3 × 7 × 13, the decimal for 1/273 is purely periodic.
d.
For some primes p ≠ 2 or 5, 1/p will have p-1 digits in the repetend. The following table shows the primes
less than 50 and the length of the repetend for 1/p. There are six primes less than 50 such that the repetend
for 1/p has p - 1 digits.
Prime p
3
7
11
13
17
19
23
29
31
37
41
43
47
Length of repetend of 1/p
1
6
2
6
16
18
22
28
15
3
5
21
46
In general, the length of the repetend for a prime p ≠ 2 or 5, is the smallest value of k such that 10k - 1 is
divisible by p. For example, 1/11 has a repetend of two digits because for k = 2, 11 divides 102 - 1 = 99.
Similarly, the repetend for 1/41 has 5 digits because 41 divides 105 - 1 = 99,999.
2.
The prime factorization of N can be used to determine the length of the repetend of the decimal for 1/N.
Consider the repetend for 1/119: 119 = 7 × 17; the repetends for 7 and 17 are 6 and 16, respectively; the least
common multiple of 6 and 16 is 48; and the repetend for 1/119 has a length of 48. Let's consider the length of
the repetend for 1/357. Since 357 = 3 × 7 × 17 and the lengths of the repetends for 1/3, 1/7, and 1/17 are 1, 6,
and 16, respectively, the length of the repetend for 1/357 is the LCM(1, 6, 16) which is 48. So, 1/357 has a
repeating decimal with 48 digits in the repetend.
Conjecture (1): For N = p1 × p2 × . . . × pn , where all pi are different and not equal to 2 or 5, the length of the
repetend of 1/N is LCM (k1, k2, . . . , kn), where ki is the length of the repetend of 1/pi .
Conjecture (2): If N = pn , where p is a prime greater than 5, then the length of the repetend of 1/N is the length of
the repetend of 1/p times pn-1. As examples, the length of the repetend of 1/72 is 6 × 7 = 42 and the length of the
repetend of 1/73 is 6 × 7 × 7 = 294. If p = 3, then the length of the repetend of 1/pn is pn-2 . For example, the
length of the repetend of 1/34 is 32 = 9.
Conjecture (3): If N is the product of powers of m distinct primes, px × qy × . . ., none of which equals 2 or 5, then
the length of the repetend of 1/N is the least common multiple of the repetends of the m numbers
1/px, 1/qy, . . ., as given by Conjecture (2).
ANSWERS FOR INVESTIGATIONS
Examples of Conjecture (3):
1
1
has a repetend of length 336, because the lengths of the repetends of 17 and 2 are 16 and 42,
17 × 7
7
respectively, and LCM (16, 42) = 336.
1
2
1
2
has a repetend of length 66, because the length of the repetend of
7 ×11
1
of 7 is 6; and LCM (22, 6) = 66.
1
has a repetend of length 18, because the lengths of the repetends of
4
3 x7
LCM (9, 6) = 18.
1
112
is 2 × 11; the length of the repetend
1
and 7 are 9 and 6, and
3
1
4
For prime numbers greater than 486, Conjecture (2) needs to be modified. For further details on the topic of repeating
decimals, see Neal Jacobs, "More on Repeating Decimals", Mathematics Teacher 68 (March 1975): 249-252.
LABORATORY INVESTIGATION 6.2
Digit Draw
1.
83.52 (9.6 × 8.7)
2.
.98
3.
Yes. 9.6 × 8.7 ≠ 9.7 × 8.6
4.
.03
(0.1 × 9.8)
(3.01 - 2.98, 4.01 - 3.98, 5.01 - 4.98, 6.01 - 5.98, and 7.01 - 6.98).
COMPUTER INVESTIGATION 6.3
Palindromic Decimals
1.
Yes, every two-digit decimal leads to a palindromic number in four or fewer steps except one, which
requires 10 steps.
2.
The decimals 2.9, 3.8, 4.7. 5.6, 6.5, 7.4, 8.3, and 9.2 each require two steps to become a palindromic
number, whereas, 29, 38, 47, 56, 65, 74, 83, and 92 each require only one step to become a palindromic
number. The sum of each pair of digits in these numbers is 11. The other two-digit decimal requiring more
steps than the corresponding whole number is 9.9.
ANSWERS FOR INVESTIGATIONS
3.
For each of the following three-digit decimals and their corresponding whole numbers, the number of steps
to become a palindromic number is written in parentheses after each number. One possible conjecture
based on this sample: Most three-digit decimals go to palindromic numbers in 23 or fewer steps. For these
12 decimals, six require more steps to become a palindromic number that the corresponding whole number.
3.49 (1) 96.3 (1) 8.17 (1) 67.3 (3) 47.9 (3) 3.89 (2) 54.8 (4) 89.7 (7) 7.76 (3) 9.99 (23) 99.9 (23) 98.9 (4)
349 (3) 963 (2) 817 (2) 673 (3) 479 (2) 389 (3) 548 (3) 897 (4) 776 (2) 999 (16) 999 (16) 989 (19)
4.
The following four decimals will not lead to palindromic numbers in 100 or fewer steps: 88.9, 89.8, 8.98,
and 9.88. The corresponding whole numbers, 889, 898, and 988 each require six steps to become a
palindromic number.
5.
The only three-digit decimals that will not lead to a palindromic number in 100 or fewer steps, 88.9, 89.8,
8.98, and 9.88, suggest that we consider decimals such as 88.99, 89.89, 8.989, 9.889, etc., to find four-digit
decimals that will not go to a palindromic number. Investigating these types of decimals leads to the
following decimals that do not go to a palindromic number in 25 steps or fewer: 89.99, 989.9, 9.999, 99.99,
999.9. All other four-digit decimals whose digits have a sum of 34 or 35 go to palindromic numbers in less
than 20 steps.
6.
One type of five-digit decimal to try are those who digits are 9s or combinations of 8s and 9s. For example,
the following decimals do not go to palindromic numbers in 25 or fewer steps: 9.9999, 99.999, 999.99,
9999.9, 99.899, 98.999, 999.89. There are other such five-digit decimals.
LABORATORY INVESTIGATION 6.4
Pythagorean Theorem
2.
1.
No. The following figure indicates that no matter
how square A is placed on square C the four pieces
from square B will not fit to cover the remaining
portion of square C.
B
A
c
b
B
a
c
b
a
B
A
ANSWERS FOR INVESTIGATIONS
GRAPHING CALCULATOR INVESTIGATION 7.1
Trend Lines
1.
a.
If the correlation coefficient for a trend line is close to +1, the points of the scatter plot will cluster
along a line with a positive slope. The correlation coefficient for the trend line on the given scatter plot
is .82, to the nearest hundredth.
b.
If the point (13, 14) is moved to (13, 10.4), the new value of r to the nearest hundredth is .84. The four
points of the plot that are shown in the following table were moved onto the line y = 2.6 + .6x and the
resulting value of r to the nearest hundredth is .92. Note: the points are being moved onto y =
2.6 + .6x, however as each point is moved there will be a slightly different trend line with a new
equation. You may wish to obtain the new equation as each point is moved to the vicinity of
y = 2.6 + .6x.
Original point
(17, 15)
(12, 7)
(10, 6)
(7, 9)
New point
(17, 12.8)
(12, 9.8)
(10, 8.6)
(7, 6.8)
Resulting value of r
.844
.872
.895
.918
The revised plot, after moving five points onto the trend line y = 2.6 + .6x is shown below. Moving
these five points produces a slightly different trend line whose equation is y = 2.9 + .6x. This new
plot provides an example of a set of points whose linear correlation is approximately .92.
2.
The following scatter plots are from section 7.1. The value of the linear correlation coefficient r to the
nearest hundredth is shown under each plot.
ANSWERS FOR INVESTIGATIONS
3.
The trend line for a scatter plot with just two points will
be the line containing these points and its correlation
coefficient will be +1 or -1, depending on whether the
line has a positive slope or a negative slope. The line
containing the points (1, 5) and (4, 14) is shown at the
right. If we include the point (3, 9) so that the plot has
three points, the linear correlation coefficient is .97 to
the nearest hundredth. Adding points to the plot that
are close to the trend line results in small changes to the
value of r and adding points that are not close to the
trend line results in larger changes to the value of r.
4.
The curve of best fit for this plot is an exponential function. Its correlation coefficient to the nearest
hundredth is .96, whereas, the linear correlation coefficient is only .91.
COMPUTER INVESTIGATION 7.2
Standard Deviations
1.
The mean increases by the same amount that each measurement is increased and the range and standard
deviation remain the same. For example, the mean of the old test scores is 70.5 and the mean of the new
test scores is 75.5. However, the range for both sets of scores is 37 and the standard deviation to the
nearest hundredth for both sets of scores is 10.77.
2.
The mean and standard deviation to two decimal places are 6.25 and 3.90. 50% of the data are within 1
standard deviation of the mean. If the data is changed to 1, 1, 6, 6, 6, 6, 12, 12, 200 the new mean and
standard deviation to two decimal places are 27.78 and 61.00, and 88.8% of the data are within 1 standard
deviation of the mean.
3.
The following measurements are given to two decimal places.
Set A: Mean = 71.83 and standard deviation = 18.37.
Set B: Mean = 72.00 and standard deviation = 24.31
ANSWERS FOR INVESTIGATIONS
The means for sets A and B are very close to each other, but the much smaller standard deviation for Set A
indicates that the data for this set is more clustered about the mean than the data for Set B.
4.
Answers will vary. The purpose of this question is to encourage students to experiment with creating data
which clusters about the mean and for which small amounts of data are at the high and low extremes.
5.
The following ten numbers have a mean of 20 and a standard deviation of approximately 8.50. Since plus
or minus two standard deviations only includes eight of these numbers, only 80% of the data is within plus
or minus two standard deviations.
1, 20, 20, 20, 20, 20, 20, 20, 20, 39
The 19 numbers below have a mean of 18 and a standard deviation of 8.49. In this example, 78.9% of the
measurements are within plus or minus two standard deviations.
1, 1, 1, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 39
COMPUTER INVESTIGATION 7.3
Dice Roll Simulation
1.
The following tables show the results obtained from the Dice Roll Simulation program.
Experiment
1
2
3
4
5
6
7
8
9
10
11
12
13
Number of rolls to
obtain double sixes
27
30
29
21
18
31
5
20
6
12
50
9
59
Experiment
14
15
16
17
18
19
20
21
22
23
24
25
Number of rolls to
obtain double sixes
14
8
94
45
60
20
38
137
23
29
12
37
The mean of the number of rolls for the 25 experiments is 33.4 to the nearest tenth. Theoretically, the mean
number of rolls to obtain double sixes is 36.
2.
The next two tables show the number of rolls for each of 25 experiments with three dice before obtaining a
sum of 15 or more.
Experiment
1
2
3
4
5
6
7
8
9
10
11
12
13
Number of rolls to
obtain a sum > 15
6
17
21
18
4
22
6
8
3
5
24
10
23
ANSWERS FOR INVESTIGATIONS
Experiment
14
15
16
17
18
19
20
21
22
23
24
25
Number of rolls to
obtain a sum > 15
8
5
14
7
16
2
20
12
27
6
5
6
The mean of the number of rolls for the 25 experiments is 11.8 to the nearest tenth. Theoretically, the
mean number of rolls to obtain a sum greater than or equal to 15 is 10.8 to the nearest tenth.
3.
a.
These tables show the results obtained from 25 experiments with two dice. Each sequence of rolls ended
when either a 7 or 11 was obtained.
Experiment
1
2
3
4
5
6
7
8
9
10
11
12
13
Number of rolls to
obtain 7 or 11
1
3
2
4
16
2
16
5
10
1
1
3
1
Experiment
14
15
16
17
18
19
20
21
22
23
24
25
Number of rolls to
obtain 7 or 11
3
4
8
3
8
1
4
7
3
3
6
4
The mean number of rolls for the 25 experiments is 4.8 to the nearest tenth. Theoretically, the mean
number of rolls with two dice to obtain a 7 or an 11 is 4.5 .
b.
These tables show the results obtained from 25 experiments with three dice. Each sequence of rolls
ended when either a 7 or an 11 was obtained.
Experiment
1
2
3
4
5
6
7
8
9
10
11
12
13
Number of rolls to
obtain 7 or 11
2
9
10
4
5
6
10
9
3
5
2
4
3
Experiment
14
15
16
17
18
19
20
21
22
23
24
25
Number of rolls to
obtain 7 or 11
4
4
5
8
2
10
4
1
4
3
1
1
The mean number of rolls for the 25 experiments is 4.8 to the nearest tenth. Theoretically, the mean
number of rolls with three dice to obtain a 7 or an 11 is 4.5 to the nearest tenth.
ANSWERS FOR INVESTIGATIONS
COMPUTER INVESTIGATION 8.1
Coin Toss Simulation
1.
a.
Answers will vary but for 100 trials the empirical probabilities will be fairly close to the following
theoretical probabilities, which are given to the nearest hundredth:
Probability of obtaining exactly 2 heads in a toss of 4 coins, .38
Probability of obtaining exactly 3 heads in a toss of 6 coins, .31
Probability of obtaining exactly 4 heads in a toss of 8 coins, .27
2.
b.
As the number of coins increases, the probability of obtaining exactly one-half heads decreases and
approaches zero. The theoretical probability of obtaining exactly 25 heads in a toss of 50 coins is .11.
a.
Answers will vary, but for large numbers of trials they will be fairly close to these theoretical probabilities:
Probability of obtaining at least 1 head in a toss of 2 coins, .75
Probability of obtaining at least 2 heads in a toss of 4 coins, .69
Probability of obtaining at least 3 heads in a toss of 6 coins, .66
3.
b.
As the number of coins increases, the probability of obtaining at least 50% heads decreases and
approaches .5. The theoretical probability of obtaining at least 10 heads in a toss of 20 coins is .59.
a.
The theoretical probability of obtaining at least 3 consecutive heads in a toss of 5 coins is .2, to the
nearest tenth.
b.
The theoretical probability of obtaining at least 3 consecutive heads in a toss of 6 coins is .3, to the
nearest tenth.
c.
The probability of obtaining at least 3 consecutive heads appears to be close to 50% for a toss of 9 coins.
LABORATORY INVESTIGATION 8.2
Probability Machines
1.
Compartment A: 1 way and the probability is 1/16 . Compartment B: 4 ways and the probability is 1/4.
Compartment C: 6 ways and the probability is 3/8. Compartment D: 4 ways and the probability is 1/4.
Compartment E: 1 way and the probability is 1/16.
2.
For three pegs in the bottom row there are four compartments and the number of ways the balls can fall into
these compartments from left to right is 1, 3, 3, and 1. The probabilities of a ball falling into these
compartment are 1/8, 3/8, 3/8, and 1/8, respectively.
For a probability machine with n pegs in the bottom row, the number of ways a ball can fall into the n + 1
compartments are the n + 1 numbers in row n of Pascal's triangle.
3.
4.
The numbers of ways a ball can fall into the 11 compartments of a probability machine with 10 pegs in the
bottom row is given by the following 11 numbers from row 10 of Pascal's triangle.
1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1
The total of these numbers is 210 = 1024 and the probability of a ball falling into the center compartment
is 252/1024 ≈ 25% .
ANSWERS FOR INVESTIGATIONS
COMPUTER INVESTIGATION 9.1
Properties of Triangles
1.
a,b,c,d. If the original triangle is equilateral, the six triangles will be congruent and their perimeters will be
equal. Also, pairs of triangles (see polygons 2 and 3, 4 and 5, and 6 and 7 in the triangle shown below)
which are symmetric about the median of an isosceles triangle will be congruent and have the same
perimeter. For any given triangle, the six triangles formed by the medians will have the same area.
Distance (A to B) = 2.53 inches
Distance (C to A) = 2.53 inches
Area (Polygon 2) = 0.28 square inches
Perimeter (Polygon 2) = 3.42 inches
Area (Polygon 3) = 0.28 square inches
Perimeter (Polygon 3) = 3.42 inches
Area (Polygon 4) = 0.28 square inches
Perimeter (Polygon 4) = 2.85 inches
Area (Polygon 5) = 0.28 square inches
Perimeter (Polygon 5) = 2.85 inches
Area (Polygon 6) = 0.28 square inches
Perimeter (Polygon 6) = 2.55 inches
Area (Polygon 7) = 0.28 square inches
Perimeter (Polygon 7) = 2.55 inches
2.
a,b,c,d The figure below on the left shows the six triangles created by the medians of an equilateral triangle
and the six centroids (A, B, C, D, E, F) of these triangles. The same figure is shown at the right with the
lines that were used to obtain the centroids hidden. If the original triangle is equilateral, the hexagon whose
vertices are the six centroids will have two sets of three congruent sides each (see measurements); the
vertex angles of the hexagon will each have a measure of 120˚ (see measurements); and the hexagon will
have three lines of symmetry which are the three medians of the original triangle. This hexagon will never
be a regular hexagon.
Distance (A to B) = 0.76 inches
Distance (E to F) = 0.76 inches
Distance (D to C) = 0.76 inches
Angle (ABC) = 120 degrees
Angle (CDE) = 120 degrees
Angle (EFA) = 120 degrees
Distance (F to A) = 0.38 inches
Distance (D to E) = 0.38 inches
Distance (C to B) = 0.38 inches
Angle (BCD) = 120 degrees
Angle (DEF) = 120 degrees
Angle (FAB) = 120 degrees
ANSWERS FOR INVESTIGATIONS
3.
a.
For any 5-pointed star that is formed in this manner, the sum of the five angles at the five points of the
star is 180˚.
Angle (CAD) = 60 degrees
Angle (EBD) = 29 degrees
Angle (ACE) = 30 degrees
Angle (BDA) = 34 degrees
Angle (CEB) = 27 degrees
Angle (CAD) + Angle (EBD) + Angle (ACE) + Angle (BDA) + Angle (CEB) = 180.00
A
B
E
C
b.
D
For any 6-pointed star that is formed in this manner, the sum of the six angles at the six points of the
star is 360˚.
Angle (EAC) = 59 degrees
Angle (FBD) = 43 degrees
Angle (ACE) = 83 degrees
Angle (BDF) = 56 degrees
Angle (CEA) = 38 degrees
Angle (DFB) = 81 degrees
Angle (EAC) + Angle (FBD) + Angle (ACE) + Angle (BDF) + Angle (CEA) + … = 360.00
F
A
E
B
D
C
c.
The sums of the seven angles and the eight angles at the points of 7-pointed and 8-pointed stars is 540˚
and 720˚, respectively. Conjecture: The sum of the n angles at the n points of an n-pointed star is
(n - 4) × 180, for n ≥ 5.
ANSWERS FOR INVESTIGATIONS
COMPUTER INVESTIGATION 9.2
Inscribed Angles
1.
The angles all have the same measure.
Angle (AVB) = 37 degrees
2.
3.
Angle (AVB) = 37 degrees
Angles (AVB) = 37 degrees
a
The measure of ∠AVB is one-half the measure of ∠AOB. In general, the measure of any inscribed
angle is one-half the measure of its corresponding central angle.
b.
The measure of ∠AVB is always 90˚.
The sum of the measures of pairs of opposite angles for a quadrilateral that is inscribed in a circle is always
180˚. This can be proven by noting that such angles are inscribed angles and that the sum of the numbers
of their two corresponding central angles is 360˚. For an arbitrary quadrilateral, the sum of the measures of
pairs of opposite angles is not necessarily 180˚.
LABORATORY INVESTIGATION 9.3
Pyramid Patterns
1.
The vertices of the triangular flaps lie on lines which are perpendicular to the sides of the polygon
and pass through point P. Here are the steps for locating the vertex point for the flap on side AD.
a.
b.
Construct the perpendicular line from P to AD .
Label its intersection with AD as K.
To locate the vertex point on the triangular flap,
we will need to find the length from K to the vertex.
Label this length d.
d
A
K
D
P
B
C
ANSWERS FOR INVESTIGATIONS
c.
2.
The length d is the hypotenuse of a right triangle
with legs of length PK and h, where h is the
altitude of the pyramid. Measure length PK.
d.
Use the Pythagorean theorem to find d .
e.
Mark off the vertex point on PK so that its
distance to K is d.
h
d
↔
P
K
The vertices for the triangular flaps are located by the method described in #1.
LABORATORY INVESTIGATION 9.4
Mirror Cards
1.
Here are the placements of the mirror on the given card to obtain the figures.
2. The following figures can also be obtained by placing a mirror on the given card.
COMPUTER INVESTIGATION 10.1
Angles and Areas
1.
a.
The inner triangle is similar to the larger triangle. Each of its sides is 1/2 the length of the
corresponding side of the larger triangle.
ANSWERS FOR INVESTIGATIONS
b.
The smaller quadrilateral is a parallelogram whose area is 1/2 the area of the larger quadrilateral.
Area (Polygon 1) = 1.91 square inches
Area (Polygon 2) = 3.82 square inches
Area (Polygon 1)/Area (Polygon 2) = 0.50
c.
The ratio of the area of the inner polygon to the area of the outer polygon is shown below for a
pentagon, hexagon, and heptagon.
Area (Polygon 1) = 2.12 square inches)
Area (Polygon 2) = 3.24 square inches
Area (Polygon 1)/Area (Polygon 2) = 0.65
Area (Polygon 1) = 2.28 square inches)
Area (Polygon 2) = 3.04 square inches)
Area (Polygon 1)/Area(Polygon 2) = 0.75
Area (Polygon 1) = 2.63 square inches
Area (Polygon 2) = 3.24 square inches
Area (Polygon 1)/Area (Polygon 2) = 0.81
As the number of sides of a polygon increases, the ratio of the area of the inner polygon to the area of
the outer polygon increases. For example, for a nonagon and a decagon these ratios are .88 and .90 to
the nearest hundredth, respectively. This ratio can be computed for a regular n-gon by using the
following formula, where v is the measure of the vertex angle of the polygon:
ANSWERS FOR INVESTIGATIONS
v⎞
⎛
Ratio of area of inner polygon to area of outer polygon is 1 − ⎜ cos ⎟
2⎠
⎝
2
The diagram at the right shows part of
an n-gon with center C and v denoting
the measure of the vertex angle. The
ratio of the area of the inner polygon to
the area of the outer polygon is the
same as the ratio of the area of ACB
to the area of MCN, and this ratio is
the same as the square of x to the square
of k, which is 1 - (cos v/2)2.
2.
a.
The computer printout below shows that the areas of the three regions of the square are
approximately equal and that the measure of ∠XBY is 22˚. (Note: Polygon 1 in this figure is
quadrilateral XBYD.) The lengths of segments AD and CD are each 1.15 inches, so points X and Y
are approximately 1/3 of the distance from D to A and D to C, respectively.
Area (Polygon 1) = 0.43 square inches
Area (Polygon 2) = 0.44 square inches
Area (Polygon 3) = 0.44 square inches
Angle (XBY) = 22 degrees
b.
The regular pentagon, hexagon, and decagon shown below are each divided into three regions of
approximately equal areas and the measures of ∠XBY in these polygons are 28˚, 28˚, and 30˚,
respectively. (Note: The reference to Polygon 1 in the computer printout above each figure is the
polygon in the figure that has ∠XBY as a vertex angle.) A conjecture at this point might be that for
regular polygons with greater numbers of sides, ∠XBY will have a measure of at least 30˚.
Area (Polygon 1) = 0.22 square inches
Area (Polygon 2) = 0.22 square inches
Area (Polygon 3) = 0.21 square inches
Angle (XBY) = 28 degrees
Area (Polygon 1) = 0.59 square inches
Area (Polygon 2) = 0.58 square inches
Area (Polygon 3) = 0.58 square inches
Angle (XBY) = 28 degrees
ANSWERS FOR INVESTIGATIONS
Area (Polygon 1) = 1.06 square inches
Area (Polygon 2) = 1.07 square inches
Area (Polygon 3) = 1.06 square inches
Angle (XBY) = 30 degrees
The measure of ∠XBY in the following regular 20-sided polygon is also 30˚. The conjecture in part b
might now be revised to the following: For every regular polygon of 20 or more sides, the measure of
∠XBY will be 30˚. This investigation leads naturally to the question: What is the measure of ∠XBY
for points X, B, and Y on a circle?
Area (Polygon 1) = 1.79 square inches
Area (Polygon 2) = 1.79 square inches
Area (Polygon 3) = 1.80 square inches
Angle (XBY) = 30 degrees
ANSWERS FOR INVESTIGATIONS
COMPUTER INVESTIGATION 10.2
Area Relationships
1.
a.
The area of kite ABCD is 1/2 the
product of the lengths of its
diagonals: AC × BD. This
relationship holds for all kites.
(Note: In the computer printout
shown here, segment p is segment
AC and segment q is segment BD.)
Length (Segment p) = 1.49 inches
Length (Segment q) = 0.85 inches
Area (Polygon 1) = 0.63 square inches
(1/2)*Length (Segment p)*Length (Segment q) = 0.63
b.
The relationship in part a holds for squares, because a square is a special case of a kite, but it does not
hold for rectangles, parallelograms, and trapezoids, as shown by the following computer printouts.
Length (Segment q) = 1.57 inches
Length (Segment p) = 1.57 inches
Area (Polygon 1) = 0.96 square inches
(1/2)* Length (Segment p)*Length (Segment q) = 1.24
Length (Segment q) = 2.19 inches
Length (Segment p) = 1.26 inches
Area (Polygon 1) = 1.22 square inches
(1/2)*Length (Segment p)*Length (Segment q) = 1.38
Length (Segment q) = 2.20 inches
Length (Segment p) = 2.80 inches
Area (Polygon 1) = 1.93 square inches
(1/2)*Length (Segment p)*Length (Segment q) = 3.08
ANSWERS FOR INVESTIGATIONS
2.
a.
The following computer printout suggests that the areas of all four triangles are equal. Measurements
of other such figures support this conjecture.
Area (Polygon 1) = 0.42 square inches
Area (Polygon 2) = 0.42 square inches
Area (Polygon 3) = 0.43 square inches
Area (Polygon 4) = 0.43 square inches
b.
In the following figures, triangle T is a right triangle and an obtuse triangle. The area measurements
for these figures support the conjecture that the three triangles in each figure have the same area.
Area (Polygon 1) = 0.42 square inches
Area (Polygon 2) = 0.43 square inches
Area (Polygon 3) = 0.44 square inches
Area (Polygon 4) = 0.43 square inches
Area (Polygon 1) = 0.41 square inches
Area (Polygon 2) = 0.42 square inches
Area (Polygon 3) = 0.42 square inches
Area (Polygon 4) = 0.42 square inches
ANSWERS FOR INVESTIGATIONS
COMPUTER INVESTIGATION 10.3
Areas and Volumes
1.
If 88.9% of the diameter of the circle is used as the length of the side of a square, the area of the square will
be approximately equal to the area of the circle. (Note: the square in the second figure below was obtained
by applying a dilation with a scale factor of .89 to the square in the first figure.)
Area (Circle 1) = 2.41 square inches
Area (Polygon 1) = 3.09 square inches
Area (Circle 1) = 2.41 square inches
Area (Polygon 2) = 2.44 square inches
Historical note: The ancient Egyptians approximated the area of a circle by finding the area of a square
whose side has a length which is 8/9 the diameter of the circle.
Area of Circle
with Diameter D
⎛D⎞
π⎜ ⎟
⎝2⎠
2
=
πD 2
4
Area of Square Having
a Side of Length 8/9 D
8
8
64 2
D × D =
D
9
9
81
Setting these expressions equal to each other, produces an approximate value for π of 3.16.
πD 2
64 2
=
D
4
81
π=
64 × 4
81
π = 3.16
This approximation for π indicates that using 8/9 of the diameter of a circle leads to a reasonably good
approximation for the area of a circle.
ANSWERS FOR INVESTIGATIONS
2.
If 80.6% of the diameter of a sphere is used as the length of the side of a cube, the volume of the cube will
be approximately equal to the volume of the sphere. One method of determining this percent is by
experimenting with specific spheres and cubes. Another approach is to let x represent the length of the side
of the cube. Then x3 is the volume of the cube and solving the following equation for x shows that x is
approximately 80.6% of the diameter of the sphere.
4 D3
Volume of sphere: 3 π 8 = x3 which implies x =
3
3
π
4 π∆3
= D 6
24
≈ D × .8059966053
≈ D × 80.6%
LABORATORY INVESTIGATION 11.1
Constructions on Triangles
1.
The three perpendicular lines intersect in a point.
2.
These three lines intersect in a point.
3.
These three lines intersect in a point.
4.
These three points of intersection lie on a straight line.
LABORATORY INVESTIGATION 11.2
Tessellations
1.
Rotation and translation.
ANSWERS FOR INVESTIGATIONS
2.
Yes. The following figure was obtained by rotating the curve from A to B about point B; the curve from C
to D about point D; and the curve from E to F about point F. A portion of its tessellation is shown below.
Notice that each vertex of the original triangle occurs at the vertex points of the tessellation twice. Thus,
the three pairs of corresponding sides of the original figure fit together to form the tessellation.
3.
No. Figure a was created by point symmetry on each side. Beginning with any side, it is possible to fit
every other side of the original figure to it, as shown below in figure b, but not every side. Unlike the
tessellation in #2 which was created from an equilateral triangle, a tessellation with regular hexagons has
only three vertices of the hexagon meeting at each vertex of the tessellation. So, it is not possible to match
pairs of corresponding sides.
Figure a
Figure b
ANSWERS FOR INVESTIGATIONS
LABORATORY INVESTIGATION 11.3
Pantographs
1.
Point B should be 1/4 of the distance from O to A, and point C should be 1/4 of the distance from A to P'.
For this setting point P is 1/4 of the distance from O to P' and the distance from point O to point P' is 4
times the distance from point O to point P. The area of the enlargement for this setting will be 42 = 16
times greater than the area of the original figure.
2.
The scale factor increases.
3.
Point B should be 1/8 of the distance from O to A and
point C should be 1/8 of the distance from A to P'.
4.
The scale factor decreases and gets closer to 1. For
example, if point B is 3/4 of the distance from O to A
and C is 3/4 of the distance from A to P', then P will
be 3/4 of the distance from O to P' and the scale
factor is 4/3. That is, the distance from O to P' is 4/3
times the distance from O to P.