1. Math 1150
Summer 2012

13. Problem Solving and
Number Patterns

14. How do you solve a
problem?

15. George Polya
(1887-1985)

16. Polya’s 4-Step problem-solving process (pg. 4)
1) Understand the problem
2) Devise a plan
• Look for a pattern
• Examine related problems
• Look at a simpler case
• Make a table or list
• Draw a picture
• Write an equation
• Guess and Check
• Work backward
• Use direct or indirect reasoning

17. 3) Carry out the plan
4) Look back
• Does your answer make sense?
• Did you answer the question that was
asked?
• Is there another way to find the solution?

18. How do
you
solve a
math
problem?

21. 4th grade - 1959

22. 4th grade - 2008

23. 1. A pen at the zoo holds giraffes and ostriches.
Altogether, here are 30 eyes and 44 feet on the animals.
How many of each type of animal are there?
How would you solve this problem?
Let x = the number of giraffes
Let y = the number of ostriches
2x + 2y = 30  4x + 4y = 60
4x + 2y = 44 –4x – 2y = –44
2y = 16
y=8
x=7

24. 1. A pen at the zoo holds giraffes and ostriches.
Altogether, here are 30 eyes and 44 feet on the animals.
How many of each type of animal are there?
How could a 2nd grader do this
problem?

25. 1. A pen at the zoo holds giraffes and ostriches.
Altogether, here are 30 eyes and 44 feet on the animals.
How many of each type of animal are there?
Each animal has two eyes 30 2 15
Each animal has at least two feet
We still need 14 more feet (44 – 30)
7 giraffes
8 ostriches

26. 1. A pen at the zoo holds giraffes and ostriches.
Altogether, here are 30 eyes and 44 feet on the animals.
How many of each type of animal are there?
How did a 7th grader do this problem?

27. 1. A pen at the zoo holds giraffes and ostriches.
Altogether, here are 30 eyes and 44 feet on the animals.
How many of each type of animal are there?

30. 2. Find the sum: 1 + 2 + 3 + … + 48 + 49 + 50

31. Gauss’ Method
2. Find the sum: 1 + 2 + 3 + … + 48 + 49 + 50
51
51
Sum of each pair = 51
Number of pairs = Number of numbers / 2 = 25
Sum of numbers = (Number of pairs)(Sum of each pair)
= (25 )(51)
= 1275

34. 3. If ten people are in a room, how many handshakes can they
exchange?
People Handshakes
1 0 +1
2 1
+2
3 3
+3
4 6
+4
5 10
6 15 +5
7 21 +6
8 28 +7
9 36 +8 45 handshakes
10 45 +9

39. 4. Mark and Bob began reading the same novel on the same
day. Mark reads 6 pages a day, and Bob reads 5 pages a day.
If Mark is on page 78, what page is Bob on?
Mark: 78 / 6 = 13 days of reading
Bob: 5 (13) = page 65

40. 4. Mark and Bob began reading the same novel on the
same day. Mark reads 6 pages a day, and Bob reads 5
pages a day. If Mark is on page 78, what page is Bob
on?

43. 5. If a bag of potato chips and a Snickers bar together
cost $3.00, and the chips cost $1.50 more than the
candy bar, how much does each item cost separately?

44. 5. If a bag of potato chips and a Snickers bar together cost $3.00,
and the chips cost $1.50 more than the candy bar, how much
does each item cost separately?
Guess-and-check must show at least one incorrect
guess and the correct guess
Chips Candy
$1.50 $1.50 Chips and candy cost the same
$2.50 $ .50 Chips cost $2.00 more
$2.25 $ .75 Chips cost $1.50 more
Chips: $2.25
Candy: $0.75

46. 6. Billy spent 2/3 of his money on baseball cards and 1/6 of his
money on a candy bar, and after that he still had 50 cents left in
his pocket. How much money did he start with?
Billy’s whole box of money
50₵ 50₵ 50₵ 50₵ 50₵ 50₵
Baseball cards Candy
bar
6 x 50₵ = $3.00

49. 6. Billy spent 2/3 of his money on baseball cards and
1/6 of his money on a candy bar, and after that he still
had 50 cents left in his pocket. How much money did he
start with?

51. Number
Patterns

55. From MODESE Model Curriculum Unit (6th grade)
The student council plans to build a flower box with
several sections in front of the school cafeteria. Each
section will consist of squares made with railroad ties of
equal length. Due to money constraints, the box will be
built in several phases. Below is the plan for the first
three phases. (Each side of the square represents one
tie.)
Phase 1 Phase 2 Phase 3
Create a table to show how many ties will be needed for each of
the first six phases. Explain how you know whether the pattern
represents a linear or a nonlinear function.

56. Phase 1 Phase 2 Phase 3
Phase Ties
1 4 Linear
+3
2 7 because the
3 10
+3 phase number
+3 increases at a
4 13
constant rate of
5 16 +3
1, and the ties
6 19 +3
increase at a
constant rate of 3

57. A sequence is a pattern of numbers or symbols.
A term is a number in a sequence.
Phase Ties
In an arithmetic
1 4 sequence, we add a
2 7 constant number (called
3 10 the common difference)
4 13 to find the next term in
5 16 the sequence.
6 19
For this sequence, d = 3

58. A sequence is a pattern of numbers or symbols.
A term is a number in a sequence.
Term Ties
1 4
2 7
3 10
4 13
5 16
6 19

59. A sequence is a pattern of numbers or symbols.
A term is a number in a sequence.
Term Ties
a1 4
a2 7
a3 10
a4 13
a5 16
a6 19

60. A sequence is a pattern of numbers or symbols.
A term is a number in a sequence.
Term Value
a1 4
a2 7
a3 10
a4 13
a5 16
a6 19

61. A sequence is a pattern of numbers or symbols.
A term is a number in a sequence.
Term Value Pattern
a1 4 4
+3
a2 7 4 + 1(3)
+3
a3 10 4 + 2(3)
a4 13 +3 4 + 3(3)
a5 16 +3 4 + 4(3)
+3
a6 19 4 + 5(3)
an 4 + (n-1)(3)

62. What is the 100th term in the sequence?
4 + (100 – 1)(3) = 4 + 297 = 301
Term Value Pattern
a1 4 4
+3
a2 7 4 + 1(3)
+3
a3 10 4 + 2(3)
a4 13 +3 4 + 3(3)
a5 16 +3 4 + 4(3)
+3
a6 19 4 + 5(3)
an 4 + (n-1)(3)

63. nth term formula for an arithmetic sequence:
an = a1 + (n – 1)d
Term Pattern
a1 4
a2 4 + 1(3)
a3 4 + 2(3) a1 = 4
a4 4 + 3(3)
a5 4 + 4(3) d = 3
a6 4 + 5(3)
an 4 + (n-1)(3)

64. 5, 9, 13, 17, … d=4
Find the next three terms 21, 25, 29
Find an nth term formula for the sequence
an = a1 + (n – 1)d a1 = 5 d=4
an = 5 + (n – 1)4
an = 5 + 4n – 4
an = 4n + 1
Find the 100th term of the sequence
a100 = 4(100) + 1 = 401

65. 5, 9, 13, 17, …
What position in the sequence is the number
609?
an = 4n + 1
609 = 4n + 1
-1 -1
608 = 4n
4 4
609 is the 152nd term
152 = n

66. DESE sample MAP item, 3rd grade

67. 4, 12, 36, 108, … In a geometric sequence, we
multiply by a constant (called
X3 X3 X3 the common ratio) to find the
next term in the sequence.
For this sequence, r = 3
Find the common ratio for the sequence
128, 64, 32, 16, …
16 1
=
32 2

68. 4, 12, 36, 108, …
Term Value Pattern
a1 4 4
x3
a2 12 4 x (3)
x3
a3 36 4 x (3)2
a4 108 x3 4 x (3)3
an 4 x (3)n - 1
What is the 100th term of the sequence?
a100 = 4 x 3100 – 1 = 4 x 399

69. Term Pattern
a1 4
a2 4 x (3) a = 4
2
1
a3 4 x (3) r=3
a4 4 x (3)3
an 4 x (3)n - 1
nth term formula for a geometric sequence
an = a1 · rn - 1

70. What are the next three terms in this sequence:
1, 3, 4, 7, 11, 18, 29, 47
In a Fibonacci sequence, we add two
consecutive terms to find the next term.

71. The first two terms of a sequence are 1 and 5.
Find the next two terms if the sequence is
a) Arithmetic 1, 5, 9, 13
+4 +4 +4
b) Geometric 1, 5, 25, 125
x5 x5 x5
c) Fibonacci 1, 5, 6, 11

72. A finite sequence has a limited number of terms.
One way to count the number of terms in a
sequence is to find a one-to-one correspondence
with the counting numbers.
How many terms are in the sequence?
A) 4, 8, 12, …, 40
divide each term by 4
1, 2, 3, …, 10
The sequence has 10 terms

73. How many terms are in the sequence?
B) 4, 9, 14, …, 59
add 1 to each term
5, 10, 15, …, 60
divide each term by 5
1, 2, 3, …, 12
The sequence has 12 terms

74. How many terms are in the sequence?
C) 62 , 63 , 64, …, 620
subtract 1 from each exponent
61 , 62 , 63, …, 619
look at exponents
1, 2, 3, …, 19
The sequence has 19 terms