AoPS:

1
AoPS

• Introduction to Counting Probability

2
Chapter 1

• Counting is Arithmetic

3
Counting Lists of Numbers

• Problem1.1
• How many s are in the list
• 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18?
  
• Obviously there are 18 numbers. That was pretty
  easy. The counting was done for us!

4
Problem 1.2

• How many s are in the list
• 7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,
• 23,24,25,26,27,28,29?
• In other words, how many s are there between
• 7 and 29 inclusive? (include 7 29 in the count)

5
Solution

• A clever way to approach this problem is to
  convert it to a problem like problem 1.1, by
  subtracting 6 from every in the list

6

• A clever way to approach this problem is to
  convert it to a problem like problem 1.1, by
  subtracting 6 from every in the list
• 7 8 9 29
• -6 -6 -6 -6
• 1 2 3 23

7

• A clever way to approach this problem is to
  convert it to a problem like problem 1.1, by
  subtracting 6 from every in the list
• 7 8 9 29
• -6 -6 -6 -6
• 1 2 3 23
• You may notice that we found that there are
  29-71 23 s from 7 to 29, inclusive.

8
Concept

• Given 2 positive s, a and b, with b gt a, find
  a formula for how many s there are between a and
  b inclusive.

9
Concept

• Given 2 positive s, a and b, with b gt a, find
  a formula for how many s there are between a and
  b inclusive.
• We can subtract a-1 from our list of s from a
  to b to get a list of s starting at 1

10
Concept

• Given 2 positive s, a and b, with b gt a, find
  a formula for how many s there are between a and
  b inclusive.
• We can subtract a-1 from our list of s from a
  to b to get a list of s starting at 1
• a a1 a2 . . . b
• -(a-1) -(a-1) -(a-1) . . . -(a-1)
• 1 2 3 . . . b a
  1

11
Concept

• Given 2 positive s, a and b, with b gt a, find
  a formula for how many s there are between a and
  b inclusive.
• We can subtract a-1 from our list of s from a
  to b to get a list of s starting at 1
• a a1 a2 . . . b
• -(a-1) -(a-1) -(a-1) . . . -(a-1)
• 1 2 3 . . . b a
  1
• Our new list of s has b a 1 numbers in it.

12
Problem 1.3

• How many multiples of 3 are between 62 and 215?

13
Problem 1.3

• How many multiples of 3 are between 62 and 215?
• We see that 62/3 20 2/3, so the simplest
  multiple of 3 is 3 X 21 63. Similarly, 215/3
  71 2/3, so the largest multiple of 3 is 3 X 71
  213.

14
Problem 1.3

• How many multiples of 3 are between 62 and 215?
• We see that 62/3 20 2/3, so the simplest
  multiple of 3 is 3 X 21 63. Similarly, 215/3
  71 2/3, so the largest multiple of 3 is 3 X 71
  213.
• So our list is 63,66,69, ,213.
• Divide it by 3 to convert it to a list we know
  how to count

15
Problem 1.3

• How many multiples of 3 are between 62 and 215?
• We see that 62/3 20 2/3, so the simplest
  multiple of 3 is 3 X 21 63. Similarly, 215/3
  71 2/3, so the largest multiple of 3 is 3 X 71
  213.
• So our list is 63,66,69, ,213.
• Divide it by 3 to convert it to a list we know
  how to count
• 21, 22, 23, . . . , 71.

16

• We know how to count this list! Subtracting 20
  from each number in the list gives
• 1, 2, 3, . . . , 51

17

• We know how to count this list! Subtracting 20
  from each number in the list gives
• 1, 2, 3, . . . , 51
• DO NOT BE TEMPTED TO DO THIS
• 215 62 153 51
• 3 3
• IT DOESNT ALWAYS WORK!
• See the next problem
  

18
Problem 1.4

• How many multiples of 10 are between 9 101?
• How many multiples of 10 are between 11 103?
• We know that 101-9 103 11 92, so shouldnt
  you get the same answers? Why arent they the
  same?

19
Problem 1.4

• List 1 the multiples of 10 are 10, 20, 30, ,
  100,
• so there are 10 multiples.

20
Problem 1.4

• List 1 the multiples of 10 are 10, 20, 30, ,
  100,
• so there are 10 multiples.
• List 2 the multiples of 10 are 20, 30, , 100,
• so there are 9 multiples.

21
Problem 1.4

• List 1 the multiples of 10 are 10, 20, 30, ,
  100,
• so there are 10 multiples.
• List 2 the multiples of 10 are 20, 30, , 100,
• so there are 9 multiples.
• The shortcut doesnt work
• 101 9 103 11 92 9.2
• 10 10
  10
• So how would you know whether the answer is 9 or
  10?

22
Problem 1.5

• How many 4-digit numbers are perfect cubes?

23
Solution

• How many 4-digit numbers are perfect cubes?
• The smallest 4-digit cube is 1000 103

24
Solution

• How many 4-digit numbers are perfect cubes?
• The smallest 4-digit cube is 1000 103
• The largest 4-digit perfect cube is a little
  harder to find and requires a little
  experimentation. Start by noting 203 8000. By
  trial error,
• 213 9261
• 223 10,648
• So 9261 213 is the largest 4-digit cube the
  list is 1000, . . . , 9261.

25
Solution

• There is a much better way that we can write
  this list
• 103, 113, 123, . . . , 203, 213
• So the number of s in the list is the same
  as
• 10, 11, 12, . . . , 20, 21
• and that means there are 12 numbers in the
  list!

26
Now its your turn!

• 1. How many numbers are in the list
• 36, 37, 38, , 92, 93?
• 2. How many numbers are in the list
• 4, 6, 8, . . . , 128, 130?
• 3. How many numbers are in the list
• -33, -28, -23, , 52, 57?

27
Solutions

1. 58
2. Dividing each member of the list by 2, we get 2,
   3, 4, , 64,65, and then subtracting 1, we get
   1,2,3, ,63,64, so there are 64 numbers.
3. We could add 3 to each member in the list to get
   -30,-25,-20,,55,60, and divide by 5 to get
   -6,-5,-4,,11,12. Then using the integer formula,
   we get 12 (-6) 1 19.

28
Try some more!

• 4. How many numbers are in the list
• 147, 144, 141, . . . , 42, 39?
• 5. How many numbers are in the list
• 3 2/3, 4 1/3, 5, 5 2/3, , 26 1/3, 27?
• 6. How many positive multiples of 7 are less than
  150?

29
Solutions

• 4. First, reverse the list, then divide by 3 to
  get
• 13, 14, , 48, 49, so 49 13 1 37.
• 5. First multiply each number by 3 to get
• 11, 13, 15, , 79, 81. Then we can subtract 1
  and divide by 2 to get 5, 6, 7, , 39, 40. So we
  get 40 5 1 36.
• 6. 7 x 21 147 lt 150 lt 154 7 x 22, so 21
  positive multiples of 7 are less than 150.

30
OK, one last time!

• 7. How many perfect squares are between
• 50 and 250?
• 8. How many odd perfect squares are between
• 5 and 211?
• 9. How many sets of four consecutive positive
  integers are there such that the product of the
  four integers is less than 100,000?

31
Solutions

• 7. Since 72 lt 50 lt 82 and 152 lt 250 lt 162, the
  squares between 50 250 are 82,92,102,,152. So
  there are 15 8 1 8.
• 8. Since 12 lt 5 lt 32 and 132 lt 211 lt 152, we have
  the list 32,52,72,,132, which has the same of
  members as 3,5,7,,13 which 6.
• 9. Note that 174 83521 lt 100,000 lt 104,976
  184. Since 17.54 16 x 17 x 18 x 19, we check
  16 x 17 x 18 x 19 93,024. Also 17 x 18 x 19 x
  20 116,280, so 16 x 17 x 18 x 19 is the
  largest product of 4 consec. pos. integers which
  is less than 100,000. So there are 16 sets.

32
Counting with Addition and Subtraction

• Problem 1.6
• At Northshore High School there are 12 players
  on the basketball team. All of the players are
  taking at least one foreign language class. The
  school offers only Spanish French as its
  foreign language classes. 8 of the players are
  taking Spanish and 5 of the players are taking
  both languages. How many players are taking
  French?

33
Solution

• The players taking French fall into 2
  categories those who take Spanish and those who
  dont. The of players taking French and Spanish
  5 (given in the problem).

34
Solution

• The players taking French fall into 2 categories
  those who take Spanish and those who dont. The
  of players taking French and Spanish 5 (given
  in the problem).
• Next count the of players taking French but not
  Spanish. There are 12 players on the team in
  total, and 8 of them take Spanish, so there are
  12 8 4 not taking Spanish. Since every player
  must take at least one language, there are 4
  taking French.

35
Solution

• So the of players taking French is the sum of
  the of players in each of the two categories,
• 5 4 9.
• There is another way to solve this problem
• Draw a Venn Diagram. Use a Venn Diagram
  whenever you wish to count things or people which
  occur in two or three overlapping groups.
• See the next page.

36

• Place points in the circles to represent the
  players. A point that is in the French circle
  that is not in the Spanish circle represents one
  player taking French but not Spanish.

French
Spanish
37

• A point in the region that is in both circles
  represents a player taking both languages.

French
Spanish
38

• A player taking Spanish but not French is
  represented by a point inside the Spanish circle
  but not French one.

French
Spanish
39

• Finally a point placed outside both circles
  represents a player who is in neither class.

French
Spanish
40

• Now we can use the diagram to solve the
  problem. Put 5 points in the intersection of both
  circles because there are 5 players in both
  classes.

French
Spanish
41

• Now, since there are 8 players taking Spanish,
  and 5 points are already inside the Spanish
  circle on the right, there must be 3 more points
  inside the Spanish circle not in the French
  circle. Add 3.

French
Spanish
42

• Since we have 12 total points and we know
  there arent any outside both circles, there must
  be 4 left inside the French circle but not inside
  the Spanish circle so add 4 points.

French
Spanish
43

• So now we can just read off the answer there
  are 9 points inside the French circle on the left.

French
Spanish
4
5
3
44
Problem 1.7
There are 27 cats at the pound. 14 of them are
short-haired. 11 of them are kittens. 5 of them
are long-haired adult cats. How many of them are
short-haired kittens?
45
Solution
Draw a Venn Diagram, with one circle for cats
with short hair and one circle for cats which
are kittens.
46
Which s do we want to place in the regions?
Since 5 cats dont have short hair are not
kittens, we know there are 5 cats outside both
circles.
Short Hair
Kittens
5
47

• At this point we cant immediately fill any of
  the other numbers, because none of our s
  corresponds exactly to a region of the diagram.
  For example, we know there are 11 kittens, but
  theres no single region of the diagram that
  corresponds to kittens theres a region for
  short-haired kittens and a region for
  long-haired kittens. So were going to have to
  use a little bit of thought. (Be very careful
  here)

48
The part of the right circle that does not
intersect with short hair must represent
long-haired kittens.
Short Hair
Kittens
Short Hair
Kittens
This is the region that represents long-haired ki
ttens.
This region represents short-hair kittens.
5
5
49
Introduce a variable x. Call the of cats in one
of the regions inside the circles x and try to
find other regions in terms of x. Let the of
short-haired kittens be x.
Short Hair
Kittens
Short Hair
Kittens
x
5
5
50
Since there are a total of 14 short-haired cats,
and x of them are kittens, we know that 14 x of
them are not kittens. Then we have 11 x kittens
that are not short-haired.
Short Hair
Kittens
Short Hair
Kittens
x
14 - x
11 - x
5
5
51
Theres one more piece of information that we
havent used yet the total of cats 27. So
everything must add up to 27 (14
x) (11 x) x 5 27
so x 3
Short Hair
Kittens
Short Hair
Kittens
3
11
8
5
5
52
You Try!

• 1. There are 20 cars in my buildings parking
  lot. All of the cars are red or white. 12 of them
  are red, 15 of them are 4 door, and 4 of them are
  4 door and white. How many of the cars are 4 door
  and red?

53

• Let the of red 4-door cars be x. Since there
  are 12 red cars and 15 4-door cars, the of red
  2-door cars is 12 x, while the of white
  4-door cars is 15 x.

Red
4-Door
x
12 - x
15 - x
4
54

• The sum of the of red 4-door cars, red 2-door
  cars, white 4-door cars, and white 2-door cars is
  the total of cars, 20, because each white
  4-door car is contained in exactly one of these
  categories.

Red
4-Door
x
12 - x
15 - x
4
55

• Since the number of white 2-doors is 4, we have
• x (12 x) (15 x) 4 20,
• which makes x 11.

Red
4-Door
x
12 - x
15 - x
4
56
Another one!

• 2. Going back to the 12-person basketball team,
  all 12 players are taking at least one of biology
  or chemistry. If 7 players are taking biology and
  2 are 2 players are taking both sciences, how
  many players are taking chemistry?

57

• Going back to the 12-person basketball team,
  all 12 players are taking at least one of biology
  or chemistry. If 7 players are taking biology and
  2 are 2 players are taking both sciences, how
  many players are taking chemistry?
• Solution 7 players are taking biology, so 12 7
  5 players are not taking biology, which means 5
  players are taking chemistry alone. Since 2 are
  taking both, 5 2 7 players taking chemistry.

58
Problem 3

• There are 30 students in Mrs. Taylors
  kindergarten class. If there are twice as many
  students with blond hair as with blue eyes, 6
  students with blond hair and blue eyes, and 3
  students with neither blond hair nor blue eyes,
  how many students have blue eyes?

59
Solution

• Let the number of blue-eyed students be x, so
  the of blond students is 2x. Since the of
  blue-eyed blond students is 6, the of blue-eyed
  non-blond students is x 6, while the of blond
  non-blue-eyed students is 2x 6.

60

• Since the of non-blue-eyed non-blond
  students is 3, we can add up these four exclusive
  categories to sum to 30 students in the class. So
• (x 6) (2x - 6) 6 3 30 and x 11.

Blue-eyed
Blond
x - 6
2x - 6
x
3
61
Problem 4

• At the Good-dog Obedience School, dogs can
  learn to do 3 tricks sit, stay, and roll over.
  Of the dogs at the school
• 50 dogs can sit 17 dogs can sit
  stay
• 29 dogs can stay 12 dogs can stay
  roll over
• 34 dogs can roll over 18 can sit roll
  over
• 9 dogs can do all three 9 dogs can do none
• How many dogs are in the school? How many dogs
  can do exactly 2 tricks?

62
Sit
There are 9 dogs that can do all 3 tricks and
there are 9 dogs that can do none.
9
9
Stay
Roll Over
63
Since 18 dogs can sit and roll over (and possibly
stay) 9 dogs can sit, stay, roll over, there
are 18 9 9 dogs that sit and roll over, but
not stay.
Sit
9
9
9
Stay
Roll Over
64
Using the same reasoning, there are 12 9 3
dogs that can stay and roll over but not sit,
and 17 9 8 dogs that can sit stay, but not
roll over.
Sit
9
9
8
9
3
Stay
Roll Over
65
So now we know how many can do multiple tricks,
exactly what tricks they can do. Since 50 dogs
can sit, 9 dogs can sit roll over only, 8 dogs
can sit stay only, 9 dogs can do all three
tricks, the remaining dogs that cant do multiple
tricks can only sit, and there are 50 9 8 9
24.
Sit
9
24
9
8
9
3
Stay
Roll Over
66
Using the same reasoning, we find that 29 3
8 - 9 9 dogs can only stay and 34 9 3 9
13 dogs can only roll over.
Sit
9
24
9
8
9
9
13
3
Stay
Roll Over
67
Since 9 dogs can do no tricks, we can add each
category in the Venn Diagram to find that there
are a total of 9938241399 84 dogs and
8 9 3 20 dogs that can do exactly 2 tricks.
Sit
9
24
9
8
9
9
13
3
Stay
Roll Over
68
Problem 5

• Every student in my school is in either French
  or Spanish class or both. Let x be the number of
  students in French class and y be the number of
  students in Spanish class, and z be the number os
  students in both classes. Find an expression in
• x, y, and z for how many students there are
  in my school.

69
Solution

• Since x people are in French and z people are
  in both, x z are only in French. Similarly,
• y z are only in Spanish. Everyone in the
  school is in either French only, Spanish only, or
  both, so the total of people in the school is
  (x z) (y z) z x y z.

70
Counting Multiple Events

• You have three shirts and four pairs of pants.
  How many outfits consisting of one shirt and one
  pair of pants can you make?

71
Counting Multiple Events

• You have three shirts and four pairs of pants.
  How many outfits consisting of one shirt and one
  pair of pants can you make?
• Easy 3 x 4 12 outfits.
• You could also make a tree diagram, but you
  already know how to do that!

72
Concept

• We use multiplication to count a series of
  independent events.
• By independent, we mean that each decision
  does not depend on the others.

73
2nd Example

• In how many ways can we form a license plate
  if there are 7 characters, none of which is the
  letter O, the first of which is a numeral digit
  (0-9), the second of which is a letter, and the
  remaining five of which can be either a digit or
  a letter (but not the letter O)?

74

• Each character is independent of any other.
  There are 10 choices for the 1st character
  (0-9), 25 choices for the 2nd character (A-Z
  except O), and there are 35 choices for each of
  the other five characters (any digit 0-9 or any
  letter A-Z, except O.
• Since the choices are independent, we have
• 10 x 25 x 35 x 35 x 35 x 35 x 35 10 x 25 x
  355
• 13, 130, 468, 750.

75
Arranging Things

• In how many ways can I arrange four different
  books on a shelf?

76

• There are 4 choices for the 1st book, with 3
  books remaining. So there are 3 choices for the
  2nd book, with 2 books remaining. Then there are
  2 choices for the 3rd book, with 1 book
  remaining. So we have only 1 choice for the last
  book.
• 4 x 3 x 2 x 1 24 choices for all four
  books.

77
What happens when choices are not independent?

• Your math club has 20 members. In how many
  ways can it select a president, a vice-president,
  and a treasurer if no member can hold more than
  one office?

78

• Once a student is chosen for president, he/she
  is not available to be chosen for the other
  offices.
• We have 20 choices for president, then 19
  choices left for vice-president, and last, 18
  choices left for treasurer.
• Therefore, there are 20 x 19 x 18 6840
• ways to fill the three offices.

79
The last 2 problems are examples of permutations.

• A permutation occurs whenever we have to
  choose several items one at a time from a larger
  groups of items.
• In the 1st problem, we are asked to order four
  different books. In the 2nd problem, we are asked
  the of permutations of 3 people out of 20
  people, or how to fill 3 different slots from a
  group of 20 people.

80
Factorial!

• We can order 4 different objects in 4!
  different ways.
• 4! 4 x 3 x 2 x 1 24 ways
• Important 0! means the of ways to arrange 0
  objects in a row. There is only one way to
  arrange zero objects, do nothing. So
• 0! 1

81
Exercises

1. For each of 8 colors, I have one shirt one tie
   of that color. How many shirt-and-tie outfits can
   I make if I refuse to wear a shirt tie of the
   same color?
2. How many license plates consist of 3 letters,
   followed by 2 even digits, followed by 2 odd
   digits?
3. In how many ways can I stack 5 books on a shelf?

82

• 1. There are 8 options for the shirt and only
  7 choices for the tie, or 8 x 7 56.
• 2. There are 26 choices of letters for each of
  the 1st two spots 10 choices of digits for each
  of the next 3, for a total of 262 x 103
  676,000.
• 3. There are 26 choices of letters for the 1st
  3 spots 5 choices for each of the last 4 spots
• (5 even or odd digits). 262 x 54 10, 985, 000.

83
Exercises

• 4. Suppose I have 6 different books, 2 of which
  are math books. In how many ways can I stack my 6
  books on a shelf if I want a math book on both
  ends of the stack?
• 5. There are 8 sprinters in the Olympic 100-meter
  finals. The gold medal goes to 1st place, silver
  to 2nd, and bronze to 3rd. In how many ways can
  the medals be awarded?

84

• 4. Place the math books 1st. We have 2 choices
  for the bottom book and 1 choice for the top math
  book. Then we place the other four books in the
  middle. There are 4 choices for the 1st, 3 for
  the 2nd, 2 for the 3rd, and only 1 for the 4th.
  So the total is 2 x 1 x 4 x 3 x 2 x 1 48.
• 5. There are 8 possible sprinters for gold, then
  7 left for silver, and last, 6 left for bronze,
  for
• 8 x 7 x 6 336 ways to award the medals.

85
Compute each of the following

• 9! 8!
• 42! 40!
• 8! 7!

86

• 9 x 8! 9
• 8!
• 42 x 41 x 40! 42 x 41 1722
• 40! 1
• 8 x 7! 7! 7!(8 1) 7! X 7 5040 x 7
• 
  35, 280

87
Permutations

• A club has n members, where n is a positive
  integer. In how many ways can we choose r
  different officers of the club (where r is a
  positive integer, and r lt n) such that no member
  holds more than one office?

88

• A club has n members, where n is a positive
  integer. In how many ways can we choose r
  different officers of the club (where r is a
  positive integer, and r lt n) such that no member
  holds more than one office?
• There are n choices for the 1st office, n 1
  for the 2nd, n 2 for the 3rd, and so on. When
  we get to the rth office, weve already chosen r
  1 members for the previous r 1 offices, so we
  have n (r 1) n r 1.

89

• There is an easier way to write this
• The number of permutations of n objects taken
  r at a time is
• P(n,r) n!
• (n r)!
• P(30,3) 30! 30! 30 x 29 x 28 x 27!
• (30-3)! 27!
  27!
• 30 x 29 x
  28 25, 360

90

• Slidell is running a lottery. In the lottery,
  25 balls numbered 1 25 are placed in a bin.
  Four balls are drawn one at a time their s are
  recorded. The winning combination consists of the
  four selected s in the order they are selected.
  How many winning combinations are there if
• (a) each ball is discarded after it is
  removed?
• (b) each ball is replaced in the bin after it
  is removed before the next ball is drawn?

91

• 25 choices for the 1st, 24 for the 2nd, 23 for
• the 3rd, 22 for the 4th.
• 25 x 24 x 23 x 22 303,600
• (b) 25 choices for each of the four balls, or
• 25 x 25 x 25 x 25 254 390, 625

92

• The difference between these 2 examples is the
  difference making selections without replacement
  and with replacement.

93
Review

• How many s are in the list
• 2.5, 5.5, 8.5, 11.5, , 80.5,83.5?
• How many 3-digit s are divisible by 7?
• There are 20 people in the 7th grade school band.
  8 of them are left-handed. 15 of them like jazz
  music. 2 of them are right-handed and dislike
  jazz music. How many club members are left-handed
  and like jazz music?

94

• Add 0.5, then divide by 3 to get
• 1,2,3,4,27,28 so there are 28 numbers.
• 7 x 14 98 lt 100 lt 105 7 x 15 and
• 7 x 142 994 lt 1000 lt 1001 7 x 143. That
  means the list of 3-digit s divisible by 7 is
  105,112,,994, and when we divide by 7, we get
  the list 15,16,17,,141,142 which has 142 15
  1 128 numbers.

95

• 3. Let x left-handed jazz lovers, then 8 x
  left-handed people who dislike jazz and 15 x
  jazz lovers are right-handed. Since the of
  righty jazz dislikers 2 and the total of
  members of the club 20, we can add these four
  exclusive regions to get x (8 x)
• (15 - x) 2 20, so x 5 (lefty jazz
  lovers).

96

• 4. How many 3-letter combinations can be formed
  if the 2nd letter must be a vowel (a,e,i,o,u) and
  the 3rd letter must be different from the 1st
  letter?
• 5. The local theater has one ticket window. In
  how many ways can six people line up to buy a
  ticket? (Source MATHCOUNTS)
• 6. Our basketball team has 12 members, each of
  whom can play any position. In how many ways can
  we chose a starting lineup consisting of a
  center, a power forward, a shooting forward, a
  point guard, and a shooting guard?

97

• 4. There are 26 options for the 1st letter, and
  only 5 options for the 2nd, and only 25 options
  for the 3rd. This gives 26 x 5 x 25 3,250.

98

• 5. This is a permutation of 6 people 6! 720.

99

• 6. This is a permutation of 5 players being
  chosen in order out of 12, so the answer is
• P(12,5) 12! 12 x 11 x 10 x 9 x 8 x 7!
  
• (12-5)! 7!
• 12 x 11 x 10 x 9
  x 8
• 95, 040

100
Challenge!

• 1. How many positive integers less than 500 can
  be written as the sum of perfect cubes?

101
Challenge!

• 73 lt 500 lt 83, so a3 b3 must be
• 1 a 7 and 1 b 7.
• Make a chart of the sum of 2 cubes.
• There are 26 such numbers.

13 23 33 43 53
63 73 13 2 9 28
65 126 217 344 23 16
35 72 133 224 351 33
54 91 152 243
370 43
128 189 280 407 53
250 341
468 63
432 559 73
686
102
Challenge 2

• What is the greatest common factor of 5!,
  10!, and 15!?

103
Challenge 2

• Since 5! divides 10! and 15! and 5! has no
  factor larger than 5!, and 5! is a factor of all
  three, the answer is 5!.

104
Challenge 3

• What is the units digit of sum
• 1! 2! 3! 4! 5! 1000!?

105
Challenge 3

• The units digit of 1! is 1, the units digit
  of 2! is 2, the units digit of 3! is 6, the units
  digit of 4! 24 is 4, the units digit of 5!
  120 is 0.

106
Challenge 3

• The units digit of 1! is 1, the units digit
  of 2! is 2, the units digit of 3! is 6, the units
  digit of 4! 24 is 4, the units digit of 5!
  120 is 0.
• For all n 5, n! is a multiple of 5!, which
  is a multiple of 10, so for all n 5, the units
  digit of n! is 0.

107
Challenge 3

• The units digit of 1! is 1, the units digit
  of 2! is 2, the units digit of 3! is 6, the units
  digit of 4! 24 is 4, the units digit of 5!
  120 is 0.
• For all n 5, n! is a multiple of 5!, which
  is a multiple of 10, so for all n 5, the units
  digit of n! is 0.
• This means the units digit of the sum is
• 1 2 6 4 0 0 13, so the answer
  is 3

108
Challenge 4

• How many of the factorials from 1! to 100!
  are divisible 9?

109
Challenge 4

• To have a factor of 9, n! must have two
  factors of 3. The 1st such n for which this is
  true is 6, since 6! 6 x 5 x 4 x 3 x 2 x 1.

110
Challenge 4

• To have a factor of 9, n! must have two
  factors of 3. The 1st such n for which this is
  true is 6, since 6! 6 x 5 x 4 x 3 x 2 x 1.
• Since 9 is a factor of 6! and 6! is a factor
  n! for all n 6, the numbers 6!, 7!, 8!, , 99!,
  100! are all divisible by 9.

111
Challenge 4

• To have a factor of 9, n! must have two
  factors of 3. The 1st such n for which this is
  true is 6, since 6! 6 x 5 x 4 x 3 x 2 x 1.
• Since 9 is a factor of 6! and 6! is a factor
  n! for all n 6, the numbers 6!, 7!, 8!, , 99!,
  100! are all divisible by 9.
• There are 100 6 1 95 numbers in the list.

112
Challenge 5

• Which integers n satisfy
• 1 gt 1 gt 3
• 2 n 100
• and how many such integers are there?

113
Challenge 5

• Multiplying the inequality by 100n, we get
• 50n gt 100 gt 3n.
  

114
Challenge 5

• Multiplying the inequality by 100n, we get
• 50n gt 100 gt 3n.
• Since 50n gt 100, n gt 2 and 100 gt 3n, 100/3 gt n.
  

115
Challenge 5

• Multiplying the inequality by 100n, we get
• 50n gt 100 gt 3n.
• Since 50n gt 100, n gt 2 and 100 gt 3n, 100/3 gt n.
• The integers satisfying both inequalities are
  3, 4, 5, , 32, 33, and there 33 3 1
  31.

116
Challenge 6

• My classroom has 11 rows of chairs, with 11
  chairs in each row. The chairs in each row are
  numbered from 1 11.
• (a) How many chairs have odd numbers?
• (b) Suppose we replaced 11 with n. Can you
  find a formula in terms of n for the number of
  chairs with odd numbers?

117
Challenge 6

• (a) Each row has odd-numbered chairs 1,3,5,7,9,11
  for a total of 6 odd-numbered chairs per row.
  Since there are 11 rows, there are 6 x 11 66
  chairs with odd numbers.

118
Challenge 6

• (b) There are 2 cases n is odd, or n is even.
• If n is odd, each row has odd-numbered chairs
• 1,3,5,, n 2, n.

119
Challenge 6

• (b) There are 2 cases n is odd, or n is even.
• If n is odd, each row has odd-numbered chairs
• 1,3,5,, n 2, n. Adding 1 and dividing by
  2, we get
• 1, 2, 3, , n 1, n 1 .
• 2 2

120
Challenge 6

• (b) There are 2 cases n is odd, or n is even.
• If n is odd, each row has odd-numbered chairs
• 1,3,5,, n 2, n. Adding 1 and dividing by
  2, we get
• 1, 2, 3, , n 1, n 1 .
• 2 2
• So there are n 1 odd numbered chairs in each
  row
• 2
• times n rows, for a total of n(n 1) .
• 
  2

121
Challenge 6

• (b) There are 2 cases n is odd, or n is even.
• If n is even, each row has odd-numbered
  chairs
• 1,3,5,, n 3, n - 1.

122
Challenge 6

• (b) There are 2 cases n is odd, or n is even.
• If n is even, each row has odd-numbered
  chairs
• 1,3,5,, n 3, n - 1. Adding 1 and
  dividing by 2, we get
• 1, 2, 3, , n 2, n .
• 2 2

123
Challenge 6

• (b) There are 2 cases n is odd, or n is even.
• If n is even, each row has odd-numbered
  chairs
• 1,3,5,, n 3, n - 1. Adding 1 and
  dividing by 2, we get
• 1, 2, 3, , n 2, n .
• 2 2
• So there are n odd numbered chairs in each row
• 2
• times n rows, for a total of n(n) n2 .
• 2
  2

124
Challenge 7

• We connect dots with toothpicks in a grid as
  shown. If there are 10 horizontal toothpicks in
  each row and 20 vertical toothpicks in each
  column, how many total toothpicks are there?

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125
Challenge 7

• Notice that there are 21 rows of dots and 11
  columns of dots. Since there are 20 vertical
  toothpicks in each column, there are 20 x 11
• 220 vertical toothpicks.

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126
Challenge 7

• Similarly there are 10 horizontal toothpicks
  in each row and 21 rows, for 21 x 10 210
  vertical toothpicks.

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127
Challenge 7

• This gives a total of 220 210 430
  toothpicks.

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Fini!